problem solving on linear function

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• A linear equation represents a straight line on a coordinate plane. It can be written in the form: y = mx + b where m is the slope of the line and b is the y-intercept.
• How do you find the linear equation?
• To find the linear equation you need to know the slope and the y-intercept of the line. To find the slope use the formula m = (y2 - y1) / (x2 - x1) where (x1, y1) and (x2, y2) are two points on the line. The y-intercept is the point at which x=0.
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• The most basic linear equation is a first-degree equation with one variable, usually written in the form of y = mx + b, where m is the slope of the line and b is the y-intercept.

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How to Solve a Linear Function – A Step-by-Step Guide

To solve a linear function, I always begin by identifying its standard form, which is typically expressed as $y = mx + b$ . In this equation, (m) represents the slope of the line , and (b) denotes the y-intercept , where the line crosses the y-axis.

By knowing these components, I can graph the function or work with it algebraically. It’s crucial to remember that in a linear equation , the variables (x) and (y) are to the first power, indicating a constant rate of change and resulting in a straight line when graphed.

Understanding how to manipulate this equation to isolate the variable of interest is the key to finding solutions.

Whether I’m substituting values to find points on the line or reformatting the equation to point-slope or standard form depends on the context of the problem.

Encountering various linear equation word problems sharpens my skills, as they often require a tailored approach for each unique scenario.

Stay tuned, as I’m about to unveil the step-by-step process to confidently tackle any linear function that comes my way.

Steps for Solving Linear Functions

When I approach linear functions , I often think of them as a puzzle where my goal is to find the value of the variable that makes the function true. A linear function takes the form of $f(x) = mx + b$, where $m$ is the slope or rate of change , and $b$ is the y-intercept . To make things easier, I’ve broken down the process into clear steps:

Understanding Function Notation : A linear function is typically written as $f(x)$, which is equivalent to $y$. It represents the output for a given input $x$.

Identifying Key Components :

• Slope ($m$) : Determines how steep the line is on a graph.
• Y-intercept ($b$) : Where the line crosses the y-axis.

Writing the Equation : Start by placing the known values of the slope and y-intercept into their proper places in the equation $y = mx + b$.

Plotting the Graph : On a coordinate plane , plot the y-intercept and use the slope to find a second point. Connect these points to visualize the function as a straight line .

Solving for a Specific Value :

• If you need $y$ for a particular $x$, substitute the $x$ value into the equation and solve for $y$.
• To find an $x$ value for a given $y$, you can rearrange the equation to isolate $x$ and then substitute the $y$ value.

Remember, linear equations involving two lines can be parallel or perpendicular based on their slopes:

• Parallel lines share the same slope ($m$).
• Perpendicular lines have slopes that are negative reciprocals of each other.

The process of solving for a linear function is straightforward if you take it step by step. Just remember that you’re working with constants and a variable to find points that lie on a line, and that’s the heart of linear equations .

Applications in Real-World Contexts

In my day-to-day life, I often encounter situations where linear functions are incredibly useful.

 These functions, which can be written in the form $f(x) = mx + b$, where $m$ is the slope and $b$ is the y-intercept, model relationships with a constant rate of change. Here are a few examples of how I’ve seen linear functions shine in real-world contexts:

Budgeting and Finance: For instance, if I wanted to understand how my savings account grows over time, I could use a linear function . If I start with $200 and save $50 each week, the amount of money, $M$, after $t$ weeks is represented by $M(t) = 50t + 200$.

Cooking and Recipes: When I’m cooking and need to adjust the recipe according to the number of guests, I use linear functions to scale the ingredients. If a recipe calls for 2 cups of flour for 4 cookies, the equation would look like $c = \frac{1}{2}n$, where $c$ is cups of flour and $n$ is the number of cookies.

Travel and Distance: If I go for a jog and keep a steady pace, the distance I cover can be predicted with a linear function . The distance $d$ in miles, at a constant speed $s$, after jogging for $t$ hours, would be $d(t) = st$.

These are simple applications, but they show how understanding linear functions can apply to anything from managing finances to making dinner or exercising, making this concept a valuable tool in my everyday life.

Solving linear functions can be a satisfying experience, as it sharpens my problem-solving skills and enhances my understanding of algebra.

I’ve learned that by following a methodical approach, such as using the slope-intercept form, which is denoted as $f(x) = mx + b$ , handling linear functions becomes much more straightforward.

It’s important to remember the role of $m$ , which represents the slope, and $b$, the y-intercept, in graphing these linear functions .

When I embark on solving linear equations , it’s crucial to apply systematic methods—graphing, substitution, elimination, or matrices—based on the context of the problem. Each method has its own merits, and knowing when to use each can save me time and effort.

Moreover, understanding how to interpret the solutions in real-life scenarios is an integral part of the learning process.

Whether I am predicting outcomes or determining the relationship between variables, linear functions serve as a fundamental tool in various fields, from economics to engineering.

I encourage fellow learners to practice regularly, as this will certainly fortify their abilities to solve and apply linear functions effectively. The more I engage with these concepts, the more intuitive and rewarding they become.

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4.2 Modeling with Linear Functions

Learning objectives.

In this section, you will:

• Build linear models from verbal descriptions.
• Model a set of data with a linear function.

Elan is a college student who plans to spend a summer in Seattle. Elan has saved $3,500 for their trip and anticipates spending $400 each week on rent, food, and activities. How can we write a linear model to represent this situation? What would be the x -intercept, and what can Elan learn from it? To answer these and related questions, we can create a model using a linear function. Models such as this one can be extremely useful for analyzing relationships and making predictions based on those relationships. In this section, we will explore examples of linear function models.

Building Linear Models from Verbal Descriptions

When building linear models to solve problems involving quantities with a constant rate of change, we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them:

• Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system.
• Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value.
• Carefully read the problem to determine what we are trying to find, identify, solve, or interpret.
• Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem.
• When needed, write a formula for the function.
• Solve or evaluate the function using the formula.
• Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically.
• Clearly convey your result using appropriate units, and answer in full sentences when necessary.

Now let’s take a look at the student in Seattle. In Elan’s situation, there are two changing quantities: time and money. The amount of money they have remaining while on vacation depends on how long they stay. We can use this information to define our variables, including units.

So, the amount of money remaining depends on the number of weeks: M ( t ) M ( t ) .

Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week.

The rate of change is constant, so we can start with the linear model M ( t ) = m t + b . M ( t ) = m t + b . Then we can substitute the intercept and slope provided.

To find the t- intercept (horizontal axis intercept), we set the output to zero, and solve for the input.

The t -intercept (horizontal axis intercept) is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Elan will have no money left after 8.75 weeks.

When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be valid—almost no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make sense to talk about input values less than zero. A negative input value could refer to a number of weeks before Elan saved $3,500, but the scenario discussed poses the question once they saved $3,500 because this is when the trip and subsequent spending starts. It is also likely that this model is not valid after the t -intercept (horizontal axis intercept), unless Elan uses a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is 0 ≤ t ≤ 8.75. 0 ≤ t ≤ 8.75.

In this example, we were given a written description of the situation. We followed the steps of modeling a problem to analyze the information. However, the information provided may not always be the same. Sometimes we might be provided with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information we are given, and use it appropriately to build a linear model.

Using a Given Intercept to Build a Model

Some real-world problems provide the vertical axis intercept, which is the constant or initial value. Once the vertical axis intercept is known, the t -intercept (horizontal axis intercept) can be calculated. Suppose, for example, that Hannah plans to pay off a no-interest loan from her parents. Her loan balance is $1,000. She plans to pay $250 per month until her balance is $0. The y -intercept is the initial amount of her debt, or $1,000. The rate of change, or slope, is -$250 per month. We can then use the slope-intercept form and the given information to develop a linear model.

Now we can set the function equal to 0, and solve for x x to find the x -intercept.

The x -intercept is the number of months it takes her to reach a balance of $0. The x -intercept is 4 months, so it will take Hannah four months to pay off her loan.

Using a Given Input and Output to Build a Model

Many real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes instead be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output.

Given a word problem that includes two pairs of input and output values, use the linear function to solve a problem.

• Identify the input and output values.
• Convert the data to two coordinate pairs.
• Find the slope.
• Write the linear model.
• Use the model to make a prediction by evaluating the function at a given x -value.
• Use the model to identify an x -value that results in a given y -value.
• Answer the question posed.

Using a Linear Model to Investigate a Town’s Population

A town’s population has been growing linearly. In 2004, the population was 6,200. By 2009, the population had grown to 8,100. Assume this trend continues.

• ⓐ Predict the population in 2013.
• ⓑ Identify the year in which the population will reach 15,000.

The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y -intercept would correspond to the year 0, more than 2000 years ago!

To make computation a little nicer, we will define our input as the number of years since 2004.

To predict the population in 2013 ( t = 9 t = 9 ), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope.

To determine the rate of change, we will use the change in output per change in input.

The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to t = 0 , t = 0 , giving the point ( 0 , 6200 ) . ( 0 , 6200 ) . Notice that through our clever choice of variable definition, we have “given” ourselves the y -intercept of the function. The year 2009 would correspond to t = 5, t = 5, giving the point ( 5 , 8100 ) . ( 5 , 8100 ) .

The two coordinate pairs are ( 0 , 6200 ) ( 0 , 6200 ) and ( 5 , 8100 ) . ( 5 , 8100 ) . Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope.

We already know the y -intercept of the line, so we can immediately write the equation:

To predict the population in 2013, we evaluate our function at t = 9. t = 9.

If the trend continues, our model predicts a population of 9,620 in 2013.

To find when the population will reach 15,000, we can set P ( t ) = 15000 P ( t ) = 15000 and solve for t . t .

Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027.

A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It costs $0.25 to produce each doughnut.

ⓐ Write a linear model to represent the cost C C of the company as a function of x , x , the number of doughnuts produced. ⓑ Find and interpret the y -intercept.

A city’s population has been growing linearly. In 2008, the population was 28,200. By 2012, the population was 36,800. Assume this trend continues.

• ⓐ Predict the population in 2014.
• ⓑ Identify the year in which the population will reach 54,000.

Using a Diagram to Build a Model

It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate the variables. Often, geometrical shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful.

Using a Diagram to Model Distance Walked

Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after they start walking will they fall out of radio contact?

In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question:

"How long will it take them to be 2 miles apart"?

In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables.

Because it is not obvious how to define our output variable, we’ll start by drawing a picture such as Figure 2 .

Initial Value: They both start at the same intersection so when t = 0 , t = 0 , the distance traveled by each person should also be 0. Thus the initial value for each is 0.

Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for A A is 4 and the slope for E E is 3.

Using those values, we can write formulas for the distance each person has walked.

For this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the “starting point” at the intersection where they both started. Then we can use the variable, A , A , which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, E , E , to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure.

We can then define a third variable, D , D , to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful, as we can see from Figure 3 .

Recall that we need to know how long it takes for D , D , the distance between them, to equal 2 miles. Notice that for any given input t , t , the outputs A ( t ) , E ( t ) , A ( t ) , E ( t ) , and D ( t ) D ( t ) represent distances.

Figure 2 shows us that we can use the Pythagorean Theorem because we have drawn a right angle.

Using the Pythagorean Theorem, we get:

In this scenario we are considering only positive values of t , t , so our distance D ( t ) D ( t ) will always be positive. We can simplify this answer to D ( t ) = 5 t . D ( t ) = 5 t . This means that the distance between Anna and Emanuel is also a linear function. Because D D is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output D ( t ) = 2 D ( t ) = 2 and solve for t . t .

They will fall out of radio contact in 0.4 hour, or 24 minutes.

Should I draw diagrams when given information based on a geometric shape?

Yes. Sketch the figure and label the quantities and unknowns on the sketch.

Using a Diagram to Model Distance Between Cities

There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough?

It might help here to draw a picture of the situation. See Figure 4 . It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates ( 3 0 , 1 0 ) , ( 3 0 , 1 0 ) , and Eastborough at ( 2 0 , 0 ) . ( 2 0 , 0 ) .

Using this point along with the origin, we can find the slope of the line from Westborough to Agritown.

Now we can write an equation to describe the road from Westborough to Agritown.

From this, we can determine the perpendicular road to Eastborough will have slope m = – 3. m = – 3. Because the town of Eastborough is at the point (20, 0), we can find the equation.

We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal,

The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from Westborough to the junction.

One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points.

There is a straight road leading from the town of Timpson to Ashburn 60 miles east and 12 miles north. Partway down the road, it junctions with a second road, perpendicular to the first, leading to the town of Garrison. If the town of Garrison is located 22 miles directly east of the town of Timpson, how far is the road junction from Timpson?

Modeling a Set of Data with Linear Functions

Real-world situations including two or more linear functions may be modeled with a system of linear equations . Remember, when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are three types of answers possible, as shown in Figure 5 .

Given a situation that represents a system of linear equations, write the system of equations and identify the solution.

• Identify the input and output of each linear model.
• Identify the slope and y -intercept of each linear model.
• Find the solution by setting the two linear functions equal to another and solving for x , x , or find the point of intersection on a graph.

Building a System of Linear Models to Choose a Truck Rental Company

Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile 4 . When will Keep on Trucking, Inc. be the better choice for Jamal?

The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions in Table 1 .

A linear function is of the form f ( x ) = m x + b . f ( x ) = m x + b . Using the rates of change and initial charges, we can write the equations

Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when K ( d ) < M ( d ) . K ( d ) < M ( d ) . The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the K ( d ) K ( d ) function is smaller.

These graphs are sketched in Figure 6 , with K ( d ) K ( d ) in blue.

To find the intersection, we set the equations equal and solve:

This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that K ( d ) K ( d ) is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is d > 100 d > 100 .

Access this online resource for additional instruction and practice with linear function models.

• Interpreting a Linear Function

4.2 Section Exercises

Explain how to find the input variable in a word problem that uses a linear function.

Explain how to find the output variable in a word problem that uses a linear function.

Explain how to interpret the initial value in a word problem that uses a linear function.

Explain how to determine the slope in a word problem that uses a linear function.

Find the area of a parallelogram bounded by the y -axis, the line x = 3 , x = 3 , the line f ( x ) = 1 + 2 x , f ( x ) = 1 + 2 x , and the line parallel to f ( x ) f ( x ) passing through ( 2 , 7 ) . ( 2 , 7 ) .

Find the area of a triangle bounded by the x -axis, the line f ( x ) = 12 – 1 3 x , f ( x ) = 12 – 1 3 x , and the line perpendicular to f ( x ) f ( x ) that passes through the origin.

Find the area of a triangle bounded by the y -axis, the line f ( x ) = 9 – 6 7 x , f ( x ) = 9 – 6 7 x , and the line perpendicular to f ( x ) f ( x ) that passes through the origin.

Find the area of a parallelogram bounded by the x -axis, the line g ( x ) = 2 , g ( x ) = 2 , the line f ( x ) = 3 x , f ( x ) = 3 x , and the line parallel to f ( x ) f ( x ) passing through ( 6 , 1 ) . ( 6 , 1 ) .

For the following exercises, consider this scenario: A town’s population has been decreasing at a constant rate. In 2010 the population was 5,900. By 2012 the population had dropped to 4,700. Assume this trend continues.

Predict the population in 2016.

Identify the year in which the population will reach 0.

For the following exercises, consider this scenario: A town’s population has been increased at a constant rate. In 2010 the population was 46,020. By 2012 the population had increased to 52,070. Assume this trend continues.

Identify the year in which the population will reach 75,000.

For the following exercises, consider this scenario: A town has an initial population of 75,000. It grows at a constant rate of 2,500 per year for 5 years.

Find the linear function that models the town’s population P P as a function of the year, t , t , where t t is the number of years since the model began.

Find a reasonable domain and range for the function P . P .

If the function P P is graphed, find and interpret the x - and y -intercepts.

If the function P P is graphed, find and interpret the slope of the function.

When will the population reach 100,000?

What is the population in the year 12 years from the onset of the model?

For the following exercises, consider this scenario: The weight of a newborn is 7.5 pounds. The baby gained one-half pound a month for its first year.

Find the linear function that models the baby’s weight W W as a function of the age of the baby, in months, t . t .

Find a reasonable domain and range for the function W . W .

If the function W W is graphed, find and interpret the x - and y -intercepts.

If the function W is graphed, find and interpret the slope of the function.

When did the baby weight 10.4 pounds?

What is the output when the input is 6.2?

For the following exercises, consider this scenario: The number of people afflicted with the common cold in the winter months steadily decreased by 205 each year from 2005 until 2010. In 2005, 12,025 people were inflicted.

Find the linear function that models the number of people inflicted with the common cold C C as a function of the year, t . t .

Find a reasonable domain and range for the function C . C .

If the function C C is graphed, find and interpret the x - and y -intercepts.

If the function C C is graphed, find and interpret the slope of the function.

When will the output reach 0?

In what year will the number of people be 9,700?

For the following exercises, use the graph in Figure 7 , which shows the profit, y , y , in thousands of dollars, of a company in a given year, t , t , where t t represents the number of years since 1980.

Find the linear function y , y , where y y depends on t , t , the number of years since 1980.

Find and interpret the y -intercept.

Find and interpret the x -intercept.

Find and interpret the slope.

For the following exercises, use the graph in Figure 8 , which shows the profit, y , y , in thousands of dollars, of a company in a given year, t , t , where t t represents the number of years since 1980.

For the following exercises, use the median home values in Mississippi and Hawaii (adjusted for inflation) shown in Table 2 . Assume that the house values are changing linearly.

In which state have home values increased at a higher rate?

If these trends were to continue, what would be the median home value in Mississippi in 2010?

If we assume the linear trend existed before 1950 and continues after 2000, the two states’ median house values will be (or were) equal in what year? (The answer might be absurd.)

For the following exercises, use the median home values in Indiana and Alabama (adjusted for inflation) shown in Table 3 . Assume that the house values are changing linearly.

If these trends were to continue, what would be the median home value in Indiana in 2010?

Real-World Applications

In 2004, a school population was 1001. By 2008 the population had grown to 1697. Assume the population is changing linearly.

• ⓐ How much did the population grow between the year 2004 and 2008?
• ⓑ How long did it take the population to grow from 1001 students to 1697 students?
• ⓒ What is the average population growth per year?
• ⓓ What was the population in the year 2000?
• ⓔ Find an equation for the population, P , P , of the school t years after 2000.
• ⓕ Using your equation, predict the population of the school in 2011.

In 2003, a town’s population was 1431. By 2007 the population had grown to 2134. Assume the population is changing linearly.

• ⓐ How much did the population grow between the year 2003 and 2007?
• ⓑ How long did it take the population to grow from 1431 people to 2134 people?
• ⓔ Find an equation for the population, P , P , of the town t t years after 2000.
• ⓕ Using your equation, predict the population of the town in 2014.

A phone company has a monthly cellular plan where a customer pays a flat monthly fee and then a certain amount of money per minute used for voice or video calling. If a customer uses 410 minutes, the monthly cost will be $71.50. If the customer uses 720 minutes, the monthly cost will be $118.

• ⓐ Find a linear equation for the monthly cost of the cell plan as a function of x , the number of monthly minutes used.
• ⓑ Interpret the slope and y -intercept of the equation.
• ⓒ Use your equation to find the total monthly cost if 687 minutes are used.

A phone company has a monthly cellular data plan where a customer pays a flat monthly fee of $10 and then a certain amount of money per megabyte (MB) of data used on the phone. If a customer uses 20 MB, the monthly cost will be $11.20. If the customer uses 130 MB, the monthly cost will be $17.80.

• ⓐ Find a linear equation for the monthly cost of the data plan as a function of x , x , the number of MB used.
• ⓒ Use your equation to find the total monthly cost if 250 MB are used.

In 1991, the moose population in a park was measured to be 4,360. By 1999, the population was measured again to be 5,880. Assume the population continues to change linearly.

• ⓐ Find a formula for the moose population, P since 1990.
• ⓑ What does your model predict the moose population to be in 2003?

In 2003, the owl population in a park was measured to be 340. By 2007, the population was measured again to be 285. The population changes linearly. Let the input be years since 2003.

• ⓐ Find a formula for the owl population, P . P . Let the input be years since 2003.
• ⓑ What does your model predict the owl population to be in 2012?

The Federal Helium Reserve held about 16 billion cubic feet of helium in 2010 and is being depleted by about 2.1 billion cubic feet each year.

• ⓐ Give a linear equation for the remaining federal helium reserves, R , R , in terms of t , t , the number of years since 2010.
• ⓑ In 2015, what will the helium reserves be?
• ⓒ If the rate of depletion doesn’t change, in what year will the Federal Helium Reserve be depleted?

Suppose the world’s oil reserves in 2014 are 1,820 billion barrels. If, on average, the total reserves are decreasing by 25 billion barrels of oil each year:

• ⓐ Give a linear equation for the remaining oil reserves, R , R , in terms of t , t , the number of years since now.
• ⓑ Seven years from now, what will the oil reserves be?
• ⓒ If the rate at which the reserves are decreasing is constant, when will the world’s oil reserves be depleted?

You are choosing between two different prepaid cell phone plans. The first plan charges a rate of 26 cents per minute. The second plan charges a monthly fee of $19.95 plus 11 cents per minute. How many minutes would you have to use in a month in order for the second plan to be preferable?

You are choosing between two different window washing companies. The first charges $5 per window. The second charges a base fee of $40 plus $3 per window. How many windows would you need to have for the second company to be preferable?

When hired at a new job selling jewelry, you are given two pay options:

Option A: Base salary of $17,000 a year with a commission of 12% of your sales

Option B: Base salary of $20,000 a year with a commission of 5% of your sales

How much jewelry would you need to sell for option A to produce a larger income?

When hired at a new job selling electronics, you are given two pay options:

Option A: Base salary of $14,000 a year with a commission of 10% of your sales

Option B: Base salary of $19,000 a year with a commission of 4% of your sales

How much electronics would you need to sell for option A to produce a larger income?

Option A: Base salary of $20,000 a year with a commission of 12% of your sales

Option B: Base salary of $26,000 a year with a commission of 3% of your sales

Option A: Base salary of $10,000 a year with a commission of 9% of your sales

Option B: Base salary of $20,000 a year with a commission of 4% of your sales

• 4 Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/

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Linear Equations

A linear equation is an equation for a straight line

These are all linear equations:

Let us look more closely at one example:

Example: y = 2x + 1 is a linear equation:

The graph of y = 2x+1 is a straight line

• When x increases, y increases twice as fast , so we need 2x
• When x is 0, y is already 1. So +1 is also needed
• And so: y = 2x + 1

Here are some example values:

Check for yourself that those points are part of the line above!

Different Forms

There are many ways of writing linear equations, but they usually have constants (like "2" or "c") and must have simple variables (like "x" or "y").

Examples: These are linear equations:

But the variables (like "x" or "y") in Linear Equations do NOT have:

• Exponents (like the 2 in x 2 )
• Square roots , cube roots , etc

Examples: These are NOT linear equations:

Slope-intercept form.

The most common form is the slope-intercept equation of a straight line :

Example: y = 2x + 1

• Slope: m = 2
• Intercept: b = 1

Point-Slope Form

Another common one is the Point-Slope Form of the equation of a straight line:

Example: y − 3 = (¼)(x − 2)

It is in the form y − y 1 = m(x − x 1 ) where:

General Form

And there is also the General Form of the equation of a straight line:

Example: 3x + 2y − 4 = 0

It is in the form Ax + By + C = 0 where:

There are other, less common forms as well.

As a Function

Sometimes a linear equation is written as a function , with f(x) instead of y :

And functions are not always written using f(x):

The Identity Function

There is a special linear function called the "Identity Function":

And here is its graph:

It is called "Identity" because what comes out is identical to what goes in:

Constant Functions

Another special type of linear function is the Constant Function ... it is a horizontal line:

No matter what value of "x", f(x) is always equal to some constant value.

Using Linear Equations

You may like to read some of the things you can do with lines:

• Finding the Midpoint of a Line Segment
• Finding Parallel and Perpendicular Lines
• Finding the Equation of a Line from 2 Points

Online Equation Solver

Solve linear, quadratic and polynomial systems of equations with wolfram|alpha.

• Natural Language

More than just an online equation solver

Wolfram|Alpha is a great tool for finding polynomial roots and solving systems of equations. It also factors polynomials, plots polynomial solution sets and inequalities and more.

Learn more about:

• Equation solving

Tips for entering queries

Enter your queries using plain English. To avoid ambiguous queries, make sure to use parentheses where necessary. Here are some examples illustrating how to formulate queries.

• find roots to quadratic x^2-7x+12
• plot inequality x^2-7x+12<=0
• solve {3x-5y==2,x+2y==-1}
• plot inequality 3x-5y>=2 and x+2y<=-1
• solve 3x^2-y^2==2 and x+2y^2==5
• plot 3x^2-y^2>=2 and x+2y^2<=5
• View more examples

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Get immediate feedback and guidance with step-by-step solutions and Wolfram Problem Generator

• Step-by-step solutions
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About solving equations

A value is said to be a root of a polynomial if ..

The largest exponent of appearing in is called the degree of . If has degree , then it is well known that there are roots, once one takes into account multiplicity. To understand what is meant by multiplicity, take, for example, . This polynomial is considered to have two roots, both equal to 3.

One learns about the "factor theorem," typically in a second course on algebra, as a way to find all roots that are rational numbers. One also learns how to find roots of all quadratic polynomials, using square roots (arising from the discriminant) when necessary. There are more advanced formulas for expressing roots of cubic and quartic polynomials, and also a number of numeric methods for approximating roots of arbitrary polynomials. These use methods from complex analysis as well as sophisticated numerical algorithms, and indeed, this is an area of ongoing research and development.

Systems of linear equations are often solved using Gaussian elimination or related methods. This too is typically encountered in secondary or college math curricula. More advanced methods are needed to find roots of simultaneous systems of nonlinear equations. Similar remarks hold for working with systems of inequalities: the linear case can be handled using methods covered in linear algebra courses, whereas higher-degree polynomial systems typically require more sophisticated computational tools.

How Wolfram|Alpha solves equations

For equation solving, Wolfram|Alpha calls the Wolfram Language's Solve and Reduce functions, which contain a broad range of methods for all kinds of algebra, from basic linear and quadratic equations to multivariate nonlinear systems. In some cases, linear algebra methods such as Gaussian elimination are used, with optimizations to increase speed and reliability. Other operations rely on theorems and algorithms from number theory, abstract algebra and other advanced fields to compute results. These methods are carefully designed and chosen to enable Wolfram|Alpha to solve the greatest variety of problems while also minimizing computation time.

Although such methods are useful for direct solutions, it is also important for the system to understand how a human would solve the same problem. As a result, Wolfram|Alpha also has separate algorithms to show algebraic operations step by step using classic techniques that are easy for humans to recognize and follow. This includes elimination, substitution, the quadratic formula, Cramer's rule and many more.

Solving Linear Equations

Solving linear equations means finding the value of the variable(s) given in the linear equations. A linear equation is a combination of an algebraic expression and an equal to (=) symbol. It has a degree of 1 or it can be called a first-degree equation. For example, x + y = 4 is a linear equation. Sometimes, we may have to find the values of variables involved in a linear equation. When we are given two or more such linear equations, we can find the values of each variable by solving linear equations. There are a few methods to solve linear equations. Let us discuss each of these methods in detail.

Solving Linear Equations in One Variable

A linear equation in one variable is an equation of degree one and has only one variable term. It is of the form 'ax+b = 0', where 'a' is a non zero number and 'x' is a variable. By solving linear equations in one variable, we get only one solution for the given variable. An example for this is 3x - 6 = 0. The variable 'x' has only one solution, which is calculated as 3x - 6 = 0 3x = 6 x = 6/3 x = 2

For solving linear equations with one variable, simplify the equation such that all the variable terms are brought to one side and the constant value is brought to the other side. If there are any fractional terms then find the LCM ( Least Common Multiple ) and simplify them such that the variable terms are on one side and the constant terms are on the other side. Let us work out a small example to understand this.

4x + 8 = 8x - 10. To find the value of 'x', let us simplify and bring the 'x' terms to one side and the constant terms to another side.

4x - 8x = -10 - 8 -4x = -18 4x = 18 x = 18/4 On simplifying, we get x = 9/2.

Solving Linear Equations by Substitution Method

The substitution method is one of the methods of solving linear equations. In the substitution method , we rearrange the equation such that one of the values is substituted in the second equation. Now that we are left with an equation that has only one variable, we can solve it and find the value of that variable. In the two given equations, any equation can be taken and the value of a variable can be found and substituted in another equation. For solving linear equations using the substitution method, follow the steps mentioned below. Let us understand this with an example of solving the following system of linear equations. x + y = 6 --------------(1) 2x + 4y = 20 -----------(2)

Step 1: Find the value of one of the variables using any one of the equations. In this case, let us find the value of 'x' from equation (1). x + y = 6 ---------(1) x = 6 - y Step 2: Substitute the value of the variable found in step 1 in the second linear equation. Now, let us substitute the value of 'x' in the second equation 2x + 4y = 20.

x = 6 - y Substituting the value of 'x' in 2x + 4y = 20, we get,

2(6 - y) + 4y = 20 12 - 2y + 4y = 20 12 + 2y = 20 2y = 20 - 12 2y = 8 y = 8/2 y = 4 Step 3: Now substitute the value of 'y' in either equation (1) or (2). Let us substitute the value of 'y' in equation (1).

x + y = 6 x + 4 = 6 x = 6 - 4 x = 2 Therefore, by substitution method, the linear equations are solved, and the value of x is 2 and y is 4.

Solving Linear Equations by Elimination Method

The elimination method is another way to solve a system of linear equations. Here we make an attempt to multiply either the 'x' variable term or the 'y' variable term with a constant value such that either the 'x' variable terms or the 'y' variable terms cancel out and gives us the value of the other variable. Let us understand the steps of solving linear equations by elimination method . Consider the given linear equations: 2x + y = 11 ----------- (1) x + 3y = 18 ---------- (2) Step 1: Check whether the terms are arranged in a way such that the 'x' term is followed by a 'y' term and an equal to sign and after the equal to sign the constant term should be present. The given set of linear equations are already arranged in the correct way which is ax+by=c or ax+by-c=0.

Step 2: The next step is to multiply either one or both the equations by a constant value such that it will make either the 'x' terms or the 'y' terms cancel out which would help us find the value of the other variable. Now in equation (2), let us multiply every term by the number 2 to make the coefficients of x the same in both the equations. x + 3y = 18 ---------- (2) Multiplying all the terms in equation (2) by 2, we get,

2(x) + 2(3y) = 2(18). Now equation (2) becomes, 2x + 6y = 36 -----------(2)

Therefore, y = 5. Step 4: Using the value obtained in step 3, find out the value of another variable by substituting the value in any of the equations. Let us substitute the value of 'y' in equation (1). We get, 2x + y = 11 2x + 5 = 11 2x = 11 - 5 2x = 6 x = 6/2 x = 3

Therefore, by solving linear equations, we get the value of x = 3 and y = 5.

Graphical Method of Solving Linear Equations

Another method for solving linear equations is by using the graph. When we are given a system of linear equations, we graph both the equations by finding values for 'y' for different values of 'x' in the coordinate system. Once it is done, we find the point of intersection of these two lines. The (x,y) values at the point of intersection give the solution for these linear equations. Let us take two linear equations and solve them using the graphical method.

x + y = 8 -------(1)

y = x + 2 --------(2)

Let us take some values for 'x' and find the values for 'y' for the equation x + y = 8. This can also be rewritten as y = 8 - x.

Let us take some values for 'x' and find the values for 'y' in the equation y = x + 2.

Plotting these points on the coordinate plane, we get a graph like this.

Now, we find the point of intersection of these lines to find the values of 'x' and 'y'. The two lines intersect at the point (3,5). Therefore, x = 3 and y = 5 by using the graphical method of solving linear equations .

This method is also used to find the optimal solution of linear programming problems. Let us look at one more method of solving linear equations, which is the cross multiplication method.

Cross Multiplication Method of Solving Linear Equations

The cross multiplication method enables us to solve linear equations by picking the coefficients of all the terms ('x' , 'y' and the constant terms) in the format shown below and apply the formula for finding the values of 'x' and 'y'.

Topics Related to Solving Linear Equations

Check the given articles related to solving linear equations.

• Linear Equations
• Application of Linear Equations
• Two-Variable Linear Equations
• Linear Equations and Half Planes
• One Variable Linear Equations and Inequations

Solving Linear Equations Examples

Example 1: Solve the following linear equations by the substitution method.

3x + y = 13 --------- (1) 2x + 3y = 18 -------- (2)

By using the substitution method of solving linear equations, let us take the first equation and find the value of 'y' and substitute it in the second equation.

From equation (1), y = 13-3x. Now, substituting the value of 'y' in equation (2), we get, 2x + 3 (13 - 3x) = 18 2x + 39 - 9x = 18 -7x + 39 = 18 -7x = 18 - 39 -7x = -21 x = -21/-7 x = 3 Now, let us substitute the value of 'x = 3' in equation (1) and find the value of 'y'. 3x + y = 13 ------- (1) 3(3) + y = 13 9 + y = 13 y = 13 - 9 y = 4

Therefore, by the substitution method, the value of x is 3 and y is 4.

Example 2: Using the elimination method of solving linear equations find the values of 'x' and 'y'.

3x + y = 21 ------ (1) 2x + 3y = 28 -------- (2)

By using the elimination method, let us make the 'y' variable to be the same in both the equations (1) and (2). To do this let us multiply all the terms of the first equation by 3. Therefore equation (1) becomes,

3(3x) + 3(y) = 63 9x + 3y = 63 ---------- (3) The second equation is, 2x + 3y = 28 Now let us cancel the 'y' terms and find the value of 'x' by subtracting equation (2) from equation (3). This is done by changing the signs of all the terms in equation (2).

Example 3: Using the cross multiplication method of solving linear equations, solve the following equations.

x + 2y - 16 = 0 --------- (1) 4x - y - 10 = 0 ---------- (2)

Compare the given equation with \(a_{1}\)x + \(b_{1}\)y + \(c_{1}\) = 0, and \(a_{2}\)x+\(b_{2}\)y+\(c_{2}\) = 0. From the given equations,

\(a_{1}\) = 1, \(a_{2}\) = 4, \(b_{1}\) = 2, \(b_{2}\) = -1, \(c_{1}\) = -16, and \(c_{2}\) = -10.

By cross multiplication method,

x = \(b_{1}\)\(c_{2}\) - \(b_{2}\)\(c_{1}\)/\(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\) y = \(c_{1}\)\(a_{2}\) - \(c_{2}\)\(a_{1}\) / \(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\)

Substituting the values in the formula we get,

x = ((2)(-10)) - ((-1)(-16)) / ((1)(-1)) - ((4)(2)) x = (-20-16)/(-1-8) x = -36/-9 x = 36/9 x = 4 y = ((-16)(4)) - ((-10)(1)) / ((1)(-1)) - ((4)(2)) y = (-64 + 10) / (-1 - 8) y = -54 / -9 y = 54/9 y = 6 Therefore, by the cross multiplication method, the value of x is 4 and y is 6.

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Practice Questions on Solving Linear Equations

Faqs on solving linear equations, what does it mean by solving linear equations.

An equation that has a degree of 1 is called a linear equation. We can have one variable linear equations , two-variable linear equations , linear equations with three variables, and more depending on the number of variables in it. Solving linear equations means finding the values of all the variables present in the equation. This can be done by substitution method, elimination method, graphical method, and the cross multiplication method . All these methods are different ways of finding the values of the variables.

How to Use the Substitution Method for Solving Linear Equations?

The substitution method of solving equations states that for a given system of linear equations, find the value of either 'x' or 'y' from any of the given equations and then substitute the value found of 'x' or 'y' in another equation so that the other unknown value can be found.

How to Use the Elimination Method for Solving Linear Equations?

In the elimination method of solving linear equations, we multiply a constant or a number with one equation or both the equations such that either the 'x' terms or the 'y' terms are the same. Then we cancel out the same term in both the equations by either adding or subtracting them and find the value of one variable (either 'x' or 'y'). After finding one of the values, we substitute the value in one of the equations and find the other unknown value.

What is the Graphical Method of Solving Linear Equations?

In the graphical method of solving linear equations, we find the value of 'y' from the given equations by putting the values of x as 0, 1, 2, 3, and so on, and plot a graph in the coordinate system for the line for various values of 'x' for both the system of linear equations. We will see that these two lines intersect at a point. This point is the solution for the given system of linear equations. If there is no intersection point between two lines, then we consider them as parallel lines , and if we found that both the lines lie on each other, those are known as coincident lines and have infinitely many solutions.

What are the Steps of Solving Linear Equations that has One Variable?

A linear equation is an equation with degree 1. To solve a linear equation that has one variable we bring the variable to one side and the constant value to the other side. Then, a non-zero number may be added, subtracted, multiplied, or divided on both sides of the equation. For example, a linear equation with one variable will be of the form 'x - 4 = 2'. To find the value of 'x', we add the constant value '4' to both sides of the equation. Therefore, the value of 'x = 6'.

What are the Steps of Solving Linear Equations having Three Variables?

To solve a system of linear equations that has three variables, we take any two equations and variables. We then take another pair of linear equations and also solve for the same variable. Now that, we have two linear equations with two variables, we can use the substitution method or elimination method, or any other method to solve the values of two unknown variables. After finding these two variables, we substitute them in any of the three equations to find the third unknown variable.

What are the 4 Methods of Solving Linear Equations?

The methods for solving linear equations are given below:

• Substitution method
• Elimination method
• Cross multiplication method
• Graphical method

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2.4: Modeling with Linear Functions

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Learning Objectives

In this section, you will:

• Identify steps for modeling and solving.
• Build linear models from verbal descriptions.
• Build systems of linear models.

Emily is a college student who plans to spend a summer in Seattle. She has saved $3,500 for her trip and anticipates spending $400 each week on rent, food, and activities. How can we write a linear model to represent her situation? What would be the x -intercept, and what can she learn from it? To answer these and related questions, we can create a model using a linear function. Models such as this one can be extremely useful for analyzing relationships and making predictions based on those relationships. In this section, we will explore examples of linear function models.

Identifying Steps to Model and Solve Problems

When modeling scenarios with linear functions and solving problems involving quantities with a constant rate of change , we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them:

• Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system.
• Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value.
• Carefully read the problem to determine what we are trying to find, identify, solve, or interpret.
• Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem.
• When needed, write a formula for the function.
• Solve or evaluate the function using the formula.
• Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically.
• Clearly convey your result using appropriate units, and answer in full sentences when necessary.

Building Linear Models

Now let’s take a look at the student in Seattle. In her situation, there are two changing quantities: time and money. The amount of money she has remaining while on vacation depends on how long she stays. We can use this information to define our variables, including units.

• Output: M , M , money remaining, in dollars
• Input: t , t , time, in weeks

So, the amount of money remaining depends on the number of weeks: M ( t ) M ( t )

We can also identify the initial value and the rate of change.

• Initial Value: She saved $3,500, so $3,500 is the initial value for M . M .
• Rate of Change: She anticipates spending $400 each week, so –$400 per week is the rate of change, or slope.

Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week.

The rate of change is constant, so we can start with the linear model M ( t ) = m t + b . M ( t ) = m t + b . Then we can substitute the intercept and slope provided.

To find the x - x - intercept, we set the output to zero, and solve for the input.

0 = − 400 t + 3500 t = 3500 400 = 8.75 0 = − 400 t + 3500 t = 3500 400 = 8.75

The x - x - intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Emily will have no money left after 8.75 weeks.

When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be valid—almost no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make sense to talk about input values less than zero. A negative input value could refer to a number of weeks before she saved $3,500, but the scenario discussed poses the question once she saved $3,500 because this is when her trip and subsequent spending starts. It is also likely that this model is not valid after the x - x - intercept, unless Emily will use a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is 0 ≤ t ≤ 8.75. 0 ≤ t ≤ 8.75.

In the above example, we were given a written description of the situation. We followed the steps of modeling a problem to analyze the information. However, the information provided may not always be the same. Sometimes we might be provided with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information we are given, and use it appropriately to build a linear model.

Using a Given Intercept to Build a Model

Some real-world problems provide the y - y - intercept, which is the constant or initial value. Once the y - y - intercept is known, the x - x - intercept can be calculated. Suppose, for example, that Hannah plans to pay off a no-interest loan from her parents. Her loan balance is $1,000. She plans to pay $250 per month until her balance is $0. The y - y - intercept is the initial amount of her debt, or $1,000. The rate of change, or slope, is -$250 per month. We can then use the slope-intercept form and the given information to develop a linear model.

f ( x ) = m x + b = − 250 x + 1000 f ( x ) = m x + b = − 250 x + 1000

Now we can set the function equal to 0, and solve for x x to find the x - x - intercept.

0 = − 250 x + 1000 1000 = 250 x 4 = x x = 4 0 = − 250 x + 1000 1000 = 250 x 4 = x x = 4

The x - x - intercept is the number of months it takes her to reach a balance of $0. The x x -intercept is 4 months, so it will take Hannah four months to pay off her loan.

Using a Given Input and Output to Build a Model

Many real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes instead be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output.

Given a word problem that includes two pairs of input and output values, use the linear function to solve a problem.

• Identify the input and output values.
• Convert the data to two coordinate pairs.
• Find the slope.
• Write the linear model.
• Use the model to make a prediction by evaluating the function at a given x - x - value.
• Use the model to identify an x - x - value that results in a given y - y - value.
• Answer the question posed.

Using a Linear Model to Investigate a Town’s Population

A town’s population has been growing linearly. In 2004 the population was 6,200. By 2009 the population had grown to 8,100. Assume this trend continues.

• ⓐ Predict the population in 2013.
• ⓑ Identify the year in which the population will reach 15,000.

The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y - y - intercept would correspond to the year 0, more than 2000 years ago!

To make computation a little nicer, we will define our input as the number of years since 2004:

• Input: t , t , years since 2004
• Output: P ( t ) , P ( t ) , the town’s population

To predict the population in 2013 ( t = 9 ), ( t = 9 ), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope.

To determine the rate of change, we will use the change in output per change in input.

m = change in output change in input m = change in output change in input

The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to t = 0 , t = 0 , giving the point ( 0 , 6200 ) . ( 0 , 6200 ) . Notice that through our clever choice of variable definition, we have “given” ourselves the y -intercept of the function. The year 2009 would correspond to t = 5, t = 5, giving the point ( 5 , 8100 ) . ( 5 , 8100 ) .

The two coordinate pairs are ( 0 , 6200 ) ( 0 , 6200 ) and ( 5 , 8100 ) . ( 5 , 8100 ) . Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope.

m = 8100 − 6200 5 − 0 = 1900 5 = 380 people per year m = 8100 − 6200 5 − 0 = 1900 5 = 380 people per year

We already know the y -intercept of the line, so we can immediately write the equation:

P ( t ) = 380 t + 6200 P ( t ) = 380 t + 6200

To predict the population in 2013, we evaluate our function at t = 9. t = 9.

P ( 9 ) = 380 ( 9 ) + 6 , 200 = 9 , 620 P ( 9 ) = 380 ( 9 ) + 6 , 200 = 9 , 620

If the trend continues, our model predicts a population of 9,620 in 2013.

To find when the population will reach 15,000, we can set P ( t ) = 15000 P ( t ) = 15000 and solve for t . t .

15000 = 380 t + 6200 8800 = 380 t t ≈ 23.158 15000 = 380 t + 6200 8800 = 380 t t ≈ 23.158

Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027.

A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It costs $0.25 to produce each doughnut.

• ⓐ Write a linear model to represent the cost C C of the company as a function of x , x , the number of doughnuts produced.
• ⓑ Find and interpret the y -intercept.

A city’s population has been growing linearly. In 2008, the population was 28,200. By 2012, the population was 36,800. Assume this trend continues.

• ⓐ Predict the population in 2014.
• ⓑ Identify the year in which the population will reach 54,000.

Using a Diagram to Model a Problem

It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate the variables. Often, geometrical shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful.

Using a Diagram to Model Distance Walked

Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after they start walking will they fall out of radio contact?

In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question:

“How long will it take them to be 2 miles apart?”

In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables.

• Input: t , t , time in hours.
• Output: A ( t ) , A ( t ) , distance in miles, and E ( t ) , E ( t ) , distance in miles

Because it is not obvious how to define our output variable, we’ll start by drawing a picture such as Figure 2 .

Initial Value: They both start at the same intersection so when t = 0 , t = 0 , the distance traveled by each person should also be 0. Thus the initial value for each is 0.

Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for A A is 4 and the slope for E E is 3.

Using those values, we can write formulas for the distance each person has walked.

A ( t ) = 4 t E ( t ) = 3 t A ( t ) = 4 t E ( t ) = 3 t

For this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the “starting point” at the intersection where they both started. Then we can use the variable, A , A , which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, E , E , to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure.

We can then define a third variable, D , D , to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful, as we can see from Figure 3 .

Recall that we need to know how long it takes for D , D , the distance between them, to equal 2 miles. Notice that for any given input t , t , the outputs A ( t ) , E ( t ) , A ( t ) , E ( t ) , and D ( t ) D ( t ) represent distances.

Figure 2 shows us that we can use the Pythagorean Theorem because we have drawn a right angle.

Using the Pythagorean Theorem, we get:

D ( t ) 2 = A ( t ) 2 + E ( t ) 2 = ( 4 t ) 2 + ( 3 t ) 2 = 16 t 2 + 9 t 2 = 25 t 2 D ( t ) = ± 25 t 2 Solve for D ( t ) using the square root = ± 5 | t | D ( t ) 2 = A ( t ) 2 + E ( t ) 2 = ( 4 t ) 2 + ( 3 t ) 2 = 16 t 2 + 9 t 2 = 25 t 2 D ( t ) = ± 25 t 2 Solve for D ( t ) using the square root = ± 5 | t |

In this scenario we are considering only positive values of t , t , so our distance D ( t ) D ( t ) will always be positive. We can simplify this answer to D ( t ) = 5 t . D ( t ) = 5 t . This means that the distance between Anna and Emanuel is also a linear function. Because D D is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output D ( t ) = 2 D ( t ) = 2 and solve for t . t .

D ( t ) = 2 5 t = 2 t = 2 5 = 0.4 D ( t ) = 2 5 t = 2 t = 2 5 = 0.4

They will fall out of radio contact in 0.4 hours, or 24 minutes.

Should I draw diagrams when given information based on a geometric shape?

Yes. Sketch the figure and label the quantities and unknowns on the sketch.

Using a Diagram to Model Distance between Cities

There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough?

It might help here to draw a picture of the situation. See Figure 4 . It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates ( 3 0 , 1 0 ) , ( 3 0 , 1 0 ) , and Eastborough at ( 2 0 , 0 ) . ( 2 0 , 0 ) .

Using this point along with the origin, we can find the slope of the line from Westborough to Agritown:

m = 10 − 0 30 − 0 = 1 3 m = 10 − 0 30 − 0 = 1 3

The equation of the road from Westborough to Agritown would be

W ( x ) = 1 3 x W ( x ) = 1 3 x

From this, we can determine the perpendicular road to Eastborough will have slope m = – 3. m = – 3. Because the town of Eastborough is at the point (20, 0), we can find the equation:

E ( x ) = − 3 x + b 0 = − 3 ( 20 ) + b Substitute in (20, 0) b = 60 E ( x ) = − 3 x + 60 E ( x ) = − 3 x + b 0 = − 3 ( 20 ) + b Substitute in (20, 0) b = 60 E ( x ) = − 3 x + 60

We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal,

1 3 x = − 3 x + 60 10 3 x = 60 10 x = 180 x = 18 Substituting this back into W ( x ) y = W ( 18 ) = 1 3 ( 18 ) = 6 1 3 x = − 3 x + 60 10 3 x = 60 10 x = 180 x = 18 Substituting this back into W ( x ) y = W ( 18 ) = 1 3 ( 18 ) = 6

The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from Westborough to the junction.

distance = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( 18 − 0 ) 2 + ( 6 − 0 ) 2 ≈ 18.974 miles distance = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 = ( 18 − 0 ) 2 + ( 6 − 0 ) 2 ≈ 18.974 miles

One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points.

There is a straight road leading from the town of Timpson to Ashburn 60 miles east and 12 miles north. Partway down the road, it junctions with a second road, perpendicular to the first, leading to the town of Garrison. If the town of Garrison is located 22 miles directly east of the town of Timpson, how far is the road junction from Timpson?

Building Systems of Linear Models

Real-world situations including two or more linear functions may be modeled with a system of linear equations . Remember, when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are three types of answers possible, as shown in Figure 5 .

Given a situation that represents a system of linear equations, write the system of equations and identify the solution.

• Identify the input and output of each linear model.
• Identify the slope and y -intercept of each linear model.
• Find the solution by setting the two linear functions equal to one another and solving for x , x , or find the point of intersection on a graph.

Building a System of Linear Models to Choose a Truck Rental Company

Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile 4 . When will Keep on Trucking, Inc. be the better choice for Jamal?

The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.

A linear function is of the form f ( x ) = m x + b . f ( x ) = m x + b . Using the rates of change and initial charges, we can write the equations

K ( d ) = 0.59 d + 20 M ( d ) = 0.63 d + 16 K ( d ) = 0.59 d + 20 M ( d ) = 0.63 d + 16

Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when K ( d ) < M ( d ) . K ( d ) < M ( d ) . The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the K ( d ) K ( d ) function is smaller.

These graphs are sketched in Figure 6 , with K ( d ) K ( d ) in blue.

To find the intersection, we set the equations equal and solve:

K ( d ) = M ( d ) 0.59 d + 20 = 0.63 d + 16 4 = 0.04 d 100 = d d = 100 K ( d ) = M ( d ) 0.59 d + 20 = 0.63 d + 16 4 = 0.04 d 100 = d d = 100

This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that K ( d ) K ( d ) is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is d > 100. d > 100.

Access this online resource for additional instruction and practice with linear function models.

• Interpreting a Linear Function

2.3 Section Exercises

Explain how to find the input variable in a word problem that uses a linear function.

Explain how to find the output variable in a word problem that uses a linear function.

Explain how to interpret the initial value in a word problem that uses a linear function.

Explain how to determine the slope in a word problem that uses a linear function.

Find the area of a parallelogram bounded by the y -axis, the line x = 3 , x = 3 , the line f ( x ) = 1 + 2 x , f ( x ) = 1 + 2 x , and the line parallel to f ( x ) f ( x ) passing through ( 2 , 7 ) . ( 2 , 7 ) .

Find the area of a triangle bounded by the x -axis, the line f ( x ) = 12 – 1 3 x , f ( x ) = 12 – 1 3 x , and the line perpendicular to f ( x ) f ( x ) that passes through the origin.

Find the area of a triangle bounded by the y -axis, the line f ( x ) = 9 – 6 7 x , f ( x ) = 9 – 6 7 x , and the line perpendicular to f ( x ) f ( x ) that passes through the origin.

Find the area of a parallelogram bounded by the x -axis, the line g ( x ) = 2 , g ( x ) = 2 , the line f ( x ) = 3 x , f ( x ) = 3 x , and the line parallel to f ( x ) f ( x ) passing through ( 6 , 1 ) . ( 6 , 1 ) .

For the following exercises, consider this scenario: A town’s population has been decreasing at a constant rate. In 2010 the population was 5,900. By 2012 the population had dropped 4,700. Assume this trend continues.

Predict the population in 2016.

Identify the year in which the population will reach 0.

For the following exercises, consider this scenario: A town’s population has been increased at a constant rate. In 2010 the population was 46,020. By 2012 the population had increased to 52,070. Assume this trend continues.

Identify the year in which the population will reach 75,000.

For the following exercises, consider this scenario: A town has an initial population of 75,000. It grows at a constant rate of 2,500 per year for 5 years.

Find the linear function that models the town’s population P P as a function of the year, t , t , where t t is the number of years since the model began.

Find a reasonable domain and range for the function P . P .

If the function P P is graphed, find and interpret the x - and y -intercepts.

If the function P P is graphed, find and interpret the slope of the function.

When will the output reached 100,000?

What is the output in the year 12 years from the onset of the model?

For the following exercises, consider this scenario: The weight of a newborn is 7.5 pounds. The baby gained one-half pound a month for its first year.

Find the linear function that models the baby’s weight W W as a function of the age of the baby, in months, t . t .

Find a reasonable domain and range for the function W W .

If the function W W is graphed, find and interpret the x - and y -intercepts.

If the function W is graphed, find and interpret the slope of the function.

When did the baby weight 10.4 pounds?

What is the output when the input is 6.2? Interpret your answer.

For the following exercises, consider this scenario: The number of people afflicted with the common cold in the winter months steadily decreased by 205 each year from 2005 until 2010. In 2005, 12,025 people were afflicted.

Find the linear function that models the number of people inflicted with the common cold C C as a function of the year, t . t .

Find a reasonable domain and range for the function C . C .

If the function C C is graphed, find and interpret the x - and y -intercepts.

If the function C C is graphed, find and interpret the slope of the function.

When will the output reach 0?

In what year will the number of people be 9,700?

For the following exercises, use the graph in Figure 7 , which shows the profit, y, y, in thousands of dollars, of a company in a given year, t, t, where t t represents the number of years since 1980.

Find the linear function y , y , where y y depends on t , t , the number of years since 1980.

Find and interpret the y -intercept.

Find and interpret the x -intercept.

Find and interpret the slope.

For the following exercises, use the graph in Figure 8 , which shows the profit, y , y , in thousands of dollars, of a company in a given year, t , t , where t t represents the number of years since 1980.

For the following exercises, use the median home values in Mississippi and Hawaii (adjusted for inflation) shown in Table 2 . Assume that the house values are changing linearly.

In which state have home values increased at a higher rate?

If these trends were to continue, what would be the median home value in Mississippi in 2010?

If we assume the linear trend existed before 1950 and continues after 2000, the two states’ median house values will be (or were) equal in what year? (The answer might be absurd.)

For the following exercises, use the median home values in Indiana and Alabama (adjusted for inflation) shown in Table 3 . Assume that the house values are changing linearly.

If these trends were to continue, what would be the median home value in Indiana in 2010?

Real-World Applications

In 2004, a school population was 1,001. By 2008 the population had grown to 1,697. Assume the population is changing linearly.

• ⓐ How much did the population grow between the year 2004 and 2008?
• ⓑ How long did it take the population to grow from 1,001 students to 1,697 students?
• ⓒ What is the average population growth per year?
• ⓓ What was the population in the year 2000?
• ⓔ Find an equation for the population, P , P , of the school t years after 2000.
• ⓕ Using your equation, predict the population of the school in 2011.

In 2003, a town’s population was 1,431. By 2007 the population had grown to 2,134. Assume the population is changing linearly.

• ⓐ How much did the population grow between the year 2003 and 2007?
• ⓑ How long did it take the population to grow from 1,431 people to 2,134 people?
• ⓔ Find an equation for the population, P P of the town t t years after 2000.
• ⓕ Using your equation, predict the population of the town in 2014.

A phone company has a monthly cellular plan where a customer pays a flat monthly fee and then a certain amount of money per minute used on the phone. If a customer uses 410 minutes, the monthly cost will be $71.50. If the customer uses 720 minutes, the monthly cost will be $118.

• ⓐ Find a linear equation for the monthly cost of the cell plan as a function of x , the number of monthly minutes used.
• ⓑ Interpret the slope and y -intercept of the equation.
• ⓒ Use your equation to find the total monthly cost if 687 minutes are used.

A phone company has a monthly cellular data plan where a customer pays a flat monthly fee of $10 and then a certain amount of money per megabyte (MB) of data used on the phone. If a customer uses 20 MB, the monthly cost will be $11.20. If the customer uses 130 MB, the monthly cost will be $17.80.

• ⓐ Find a linear equation for the monthly cost of the data plan as a function of x x , the number of MB used.
• ⓒ Use your equation to find the total monthly cost if 250 MB are used.

In 1991, the moose population in a park was measured to be 4,360. By 1999, the population was measured again to be 5,880. Assume the population continues to change linearly.

• ⓐ Find a formula for the moose population, P since 1990.
• ⓑ What does your model predict the moose population to be in 2003?

In 2003, the owl population in a park was measured to be 340. By 2007, the population was measured again to be 285. The population changes linearly. Let the input be years since 1990.

• ⓐ Find a formula for the owl population, P. P. Let the input be years since 2003.
• ⓑ What does your model predict the owl population to be in 2012?

The Federal Helium Reserve held about 16 billion cubic feet of helium in 2010 and is being depleted by about 2.1 billion cubic feet each year.

• ⓐ Give a linear equation for the remaining federal helium reserves, R, R, in terms of t, t, the number of years since 2010.
• ⓑ In 2015, what will the helium reserves be?
• ⓒ If the rate of depletion doesn’t change, in what year will the Federal Helium Reserve be depleted?

Suppose the world’s oil reserves in 2014 are 1,820 billion barrels. If, on average, the total reserves are decreasing by 25 billion barrels of oil each year:

• ⓐ Give a linear equation for the remaining oil reserves, R, R, in terms of t, t, the number of years since now.
• ⓑ Seven years from now, what will the oil reserves be?
• ⓒ If the rate at which the reserves are decreasing is constant, when will the world’s oil reserves be depleted?

You are choosing between two different prepaid cell phone plans. The first plan charges a rate of 26 cents per minute. The second plan charges a monthly fee of $19.95 plus 11 cents per minute. How many minutes would you have to use in a month in order for the second plan to be preferable?

You are choosing between two different window washing companies. The first charges $5 per window. The second charges a base fee of $40 plus $3 per window. How many windows would you need to have for the second company to be preferable?

When hired at a new job selling jewelry, you are given two pay options:

• Option A: Base salary of $17,000 a year with a commission of 12% of your sales
• Option B: Base salary of $20,000 a year with a commission of 5% of your sales

How much jewelry would you need to sell for option A to produce a larger income?

When hired at a new job selling electronics, you are given two pay options:

• Option A: Base salary of $14,000 a year with a commission of 10% of your sales
• Option B: Base salary of $19,000 a year with a commission of 4% of your sales

How much electronics would you need to sell for option A to produce a larger income?

• Option A: Base salary of $20,000 a year with a commission of 12% of your sales
• Option B: Base salary of $26,000 a year with a commission of 3% of your sales
• Option A: Base salary of $10,000 a year with a commission of 9% of your sales
• Option B: Base salary of $20,000 a year with a commission of 4% of your sales
• 4 Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/

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Types of Linear Programming Problems

Linear programming is a mathematical technique for optimizing operations under a given set of constraints. The basic goal of linear programming is to maximize or minimize the total numerical value. It is regarded as one of the most essential strategies for determining optimum resource utilization. Linear programming challenges include a variety of problems involving cost minimization and profit maximization, among others. They will be briefly discussed in this article.

The purpose of this article is to provide students with a comprehensive understanding of the different types of linear programming problems and their solutions.

What is Linear Programming?

Linear programming (LP) is a mathematical optimization technique used to maximize or minimize a linear objective function, subject to a set of linear equality and inequality constraints. It is widely used in various fields such as economics, engineering, operations research, and management science to find the best possible outcome given limited resources.

Components of Linear Programming

Components of linear programming include:

• Objective Function: This is a linear function that needs to be optimized (maximized or minimized). It represents the quantity to be maximized or minimized, such as profit, cost, time, etc.
• Decision Variables: These are the variables that represent the choices or decisions to be made. They are the unknown quantities that the optimization process seeks to determine. Decision variables must be continuous and can take any real value within a specified range.
• Constraints: These are restrictions or limitations on the decision variables that must be satisfied. Constraints can be expressed as linear equations or inequalities. They represent the limitations imposed by available resources, capacity constraints, demand requirements, etc.
• Feasible Region: The feasible region is the set of all possible combinations of decision variables that satisfy all constraints. It is defined by the intersection of the constraint boundaries.
• Optimal Solution: This is the best possible solution that maximizes or minimizes the objective function while satisfying all constraints. In graphical terms, it is the point within the feasible region that maximizes or minimizes the objective function.

Linear programming provides a systematic and efficient approach to decision-making in situations where resources are limited and objectives need to be optimized.

Different Types of Linear Programming Problems

The following are the types of linear programming problems:

• Manufacturing problems
• Diet problems
• Transportation problems
• Optimal alignment problem

Let’s discuss more about each of them.

Manufacturing Problems

In these problems, we evaluate the number of units of various items that should be produced and sold by a company when each product requires a given number of workforce, machine hours, labour hours per unit of product, warehouse space per unit of output, and so on, to maximize profit.

Manufacturing problems involve maximizing the production rate or net profits of manufactured products, which might measure the available workspace, the number of workers, machine hours, packing materials used, raw materials required, the product’s market value, and other factors. These are commonly used in the industrial sector to anticipate a company’s future capital increase over time.

Diet Problems

In these challenges, we assess how many components or nutrients a diet should contain in order to lower the cost of the desired diet while guaranteeing the minimal amount of each vitamin.

As the name suggests, diet-related problems can be resolved by eating more particular foods that are rich in essential nutrients and can support the adoption of a particular diet plan. Finding a set of meals that will satisfy a set of daily nutritional demands for the least amount of money is the aim of a diet problem.

Transportation Problems

In these problems , we create a transportation schedule to discover the most cost-effective method of carrying a product from various plants/factories to various markets.

The study of transportation routes or how items from diverse production sources are transported to various markets to minimize the total transportation cost is linked to transportation difficulties. Analyzing such challenges is crucial for large firms with several production units and a broad customer base.

Optimal Assignment Problems

This problem addresses a company’s completion of a given task/assignment by selecting a specific number of employees to complete the assignment within the required timeframe, assuming that each person works on only one job. Event planning and management in major organizations, for example, are examples of such problems.

Constraints and Objective Function of Each Linear Programming Problem

Steps for solving linear programming problems.

Step 1: Identify the decision variables : The first step is to determine which choice factors control the behaviour of the objective function. A function that needs to be optimised is an objective function. Determine the decision variables and designate them with X, Y, and Z symbols.

Step 2: Form an objective function : Using the decision variables, write out an algebraic expression that displays the quantity we aim to maximize.

Step 3: Identify the constraints : Choose and write the given linear inequalities from the data.

Step 4: Draw the graph for the given data : Construct the graph by using constraints for solving the linear programming problem.

Step 5: Draw the feasible region : Every constraint on the problem is satisfied by this portion of the graph. Anywhere in the feasible zone is a viable solution for the objective function.

Step 6: Choosing the optimal point : Choose the point for which the given function has maximum or minimum values.

Solved Problems of Linear Programming Problems

Question 1. A factory manufactures two types of gadgets, regular and premium. Each gadget requires the use of two operations, assembly and finishing, and there are at most 12 hours available for each operation. A regular gadget requires 1 hour of assembly and 2 hours of finishing, while a premium gadget needs 2 hours of assembly and 1 hour of finishing. Due to other restrictions, the company can make at most 7 gadgets a day. If a profit of $20 is realized for each regular gadget and $30 for a premium gadget, how many of each should be manufactured to maximize profit?

We define our unknowns:

Let the number of regular gadgets manufactured each day = x

and the number of premium gadgets manufactured each day = y

The objective function is

P = 20x + 30y

We now write the constraints. The fourth sentence states that the company can make at most 7 gadgets a day. This translates as

Since the regular gadget requires one hour of assembly and the premium gadget requires two hours of assembly, and there are at most 12 hours available for this operation, we get

x + 2y ≤ 12

Similarly, the regular gadget requires two hours of finishing and the premium gadget one hour. Again, there are at most 12 hours available for finishing. This gives us the following constraint.

2x + y ≤ 12

The fact that x and y can never be negative is represented by the following two constraints:

x ≥ 0, and y ≥ 0.

We have formulated the problem as follows :

Maximize P=20x + 30y Subject to : x + y ≤ 7, x + 2y ≤ 122, x + y ≤ 12, x ≥ 0, y ≥ 0

In order to solve the problem, we next graph the constraints and feasible region.

Again, we have shaded the feasible region, where all constraints are satisfied.

Since the extreme value of the objective function always takes place at the vertices of the feasible region, we identify all the critical points. They are listed as (0, 0), (0, 6), (2, 5), (5, 2), and (6, 0). To maximize profit, we will substitute these points in the objective function to see which point gives us the maximum profit each day. The results are listed below.

FAQ on Linear programming

How many methods are there in lpp.

There are different methods to solve a linear programming problem. Such as Graphical method, Simplex method, Ellipsoid method, Interior point methods.

What are the four 4 special cases in linear programming?

Four special cases and difficulties arise at times when using the graphical approach to solving LP problems: (1) infeasibility, (2) unboundedness, (3) redundancy, and (4) alternate optimal solutions.

What are the 3 components of linear programming?

The basic components of the LP are as follows: Decision Variables. Constraints. Objective Functions.

What are the applications of LPP?

LPP applications may include production scheduling, inventory policies, investment portfolio, allocation of advertising budget, construction of warehouses, etc.

What are the limitations of LPP?

Constraints (limitations) should be expressed in mathematical form. Relationships between two or more variables should be linear. The values of the variables should always be non-negative or zero. There should always be finite and infinite inputs and output numbers.

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Analysis of a hybrid fractional coupled system of differential equations in $ n $-dimensional space with linear perturbation and nonlinear boundary conditions

• Salma Noor 1 , 
• Aman Ullah 1 , 
• Anwar Ali 1 , 
• Saud Fahad Aldosary 2 ,  , 
• 1. Department of Mathematics, University of Malakand, Dir(L), Khyber Pakhtunkhwa, Pakistan
• 2. Department of Mathematics, College of Science and Humanities in Alkharj, Prince Sattam bin Abdulaziz University, Alkharj 11942, Saudi Arabia
• Received: 30 January 2024 Revised: 06 April 2024 Accepted: 17 April 2024 Published: 08 May 2024

MSC : 26A33, 34A08, 34K38

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In this paper, we investigated $ n $-dimensional fractional hybrid differential equations (FHDEs) with nonlinear boundary conditions in a nonlinear coupled system. For this purpose, we used Dhage's fixed point theory, and applied the Krasnoselskii-type coupled fixed point theorem to construct existence conditions of the solution of the FHDEs. To illustrated this idea, suitable examples are presented in $ 3 $-dimensional space at the end of the paper.

• $ n $-dimensional nonlinear coupled system ,
• fractional hybrid differential equations ,
• Dhage's fixed point theory

Citation: Salma Noor, Aman Ullah, Anwar Ali, Saud Fahad Aldosary. Analysis of a hybrid fractional coupled system of differential equations in $ n $-dimensional space with linear perturbation and nonlinear boundary conditions[J]. AIMS Mathematics, 2024, 9(6): 16234-16249. doi: 10.3934/math.2024785

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• This work is licensed under a Creative Commons Attribution-NonCommercial-Share Alike 4.0 Unported License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-sa/4.0/ -->

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• © 2024 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License ( http://creativecommons.org/licenses/by/4.0 )

通讯作者: 陈斌, [email protected]

沈阳化工大学材料科学与工程学院 沈阳 110142

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Mathematics > Optimization and Control

Title: on existence of solutions to non-convex minimization problems.

Abstract: We provide new sufficient conditions for the finiteness of the optimal value and existence of solutions to a general problem of minimizing a proper closed function over a nonempty closed set. The conditions require an asymptotically bounded decay of a function, a relaxation of p-supercoercivity, and a certain relation for the asymptotic cone of the constraint set and the asymptotic function of the objective function. Our analysis combines these conditions with a regularization technique. We refine the notion of retractive directions of a set, extend its definition to functions, and establish some basic relations for such directions for both sets and functions. Using these tools, we provide existence of solutions results that generalize many of the results in the literature for both non-convex and convex problems.

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