Specify what the variables in a cost/ revenue system of linear equations represent
Determine and apply an appropriate method for solving the system
(5.2.2) – Solve value problems with a system of linear equations
(5.2.3) – Solve mixture problems with a system of linear equations
(5.2.4) – Solve uniform motion problems with a system of linear equations
A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible?
(credit: Thomas Sørenes)
(5.2.1) – Solve cost and revenue problems
Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation [latex]R=xp[/latex], where [latex]x=[/latex] quantity and [latex]p=[/latex] price. The revenue function is shown in orange in the graph below.
The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The [latex]x[/latex] -axis represents quantity in hundreds of units. The y-axis represents either cost or revenue in hundreds of dollars.
The point at which the two lines intersect is called the break-even point. We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.
The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as [latex]P\left(x\right)=R\left(x\right)-C\left(x\right)[/latex]. Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.
Example
A business wants to manufacture bike frames. Before they start production, they need to make sure they can make a profit with the materials and labor force they have. Their accountant has given them a cost equation of [latex]y=0.85x+35,000[/latex] and a revenue equation of [latex]y=1.55x[/latex]:
Interpret x and y for the cost equation
Interpret x and y for the revenue equation
Show Solution
Cost: [latex]y=0.85x+35,000[/latex]
Revenue:[latex]y=1.55x[/latex]
The cost equation represents money leaving the company, namely how much it costs to produce a given number of bike frames. If we use the skateboard example as a model, x would represent the number of frames produced (instead of skateboards) and y would represent the amount of money it would cost to produce them (the same as the skateboard problem).
The revenue equation represents money coming into the company, so in this context x still represents the number of bike frames manufactured, and y now represents the amount of money made from selling them. Let’s organize this information in a table:
Equation Type
x represents
y represents
Revenue Eqn.
number of frames
amount of money made selling frames
Cost Eqn.
number of frames
cost for making frames
Example: Finding the Break-Even Point and the Profit Function Using Substitution
Given the cost function [latex]C\left(x\right)=0.85x+35,000[/latex] and the revenue function [latex]R\left(x\right)=1.55x[/latex], find the break-even point and the profit function.
Solution
Write the system of equations using [latex]y[/latex] to replace function notation.
The profit function is [latex]P\left(x\right)=0.7x - 35,000[/latex].
Analysis of the Solution
The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units.
We see from the graph below that the profit function has a negative value until [latex]x=50,000[/latex], when the graph crosses the x-axis. Then, the graph emerges into positive y-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.
(5.2.2) – Solve value problems with a system of linear equations
It is rare to be given equations that neatly model behaviors that you encounter in business, rather, you will probably be faced with a situation for which you know key information as in the example above. Below, we summarize three key factors that will help guide you in translating a situation into a system.
How To: Given a situation that represents a system of linear equations, write the system of equations and identify the solution.
1) Identify unknown quantities in a problem represent them with variables.
2) Write a system of equations which models the problem’s conditions.
3) Solve the system.
4) Check proposed solution.
Now let’s practice putting these key factors to work. In the next example, we determine how many different types of tickets are sold given information about the total revenue and amount of tickets sold to an event.
Example: Writing and Solving a System of Equations in Two Variables
The cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets?
Solution
Let c = the number of children and a = the number of adults in attendance.
The total number of people is [latex]2,000[/latex]. We can use this to write an equation for the number of people at the circus that day.
[latex]c+a=2,000[/latex]
The revenue from all children can be found by multiplying $25.00 by the number of children, [latex]25c[/latex]. The revenue from all adults can be found by multiplying $50.00 by the number of adults, [latex]50a[/latex]. The total revenue is $70,000. We can use this to write an equation for the revenue.
[latex]25c+50a=70,000[/latex]
We now have a system of linear equations in two variables.
In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either [latex]c[/latex] or [latex]a[/latex]. We will solve for [latex]a[/latex].
We find that [latex]1,200[/latex] children and [latex]800[/latex] adults bought tickets to the circus that day.
In this video example we show how to set up a system of linear equations that represents the total cost for admission to a museum.
Try It
Meal tickets at the circus cost $4.00 for children and $12.00 for adults. If 1,650 meal tickets were bought for a total of $14,200, how many children and how many adults bought meal tickets?
Solution
700 children, 950 adults
Sometimes, a system can inform a decision. In our next example, we help answer the question, “Which truck rental company will give the best value?”
Example: 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.[1] When will Keep on Trucking, Inc. be the better choice for Jamal?
Solution
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.
Input
d, distance driven in miles
Outputs
K(d): cost, in dollars, for renting from Keep on TruckingM(d) cost, in dollars, for renting from Move It Your Way
Initial Value
Up-front fee: K(0) = 20 and M(0) = 16
Rate of Change
K(d) = $0.59/mile and P(d) = $0.63/mile
A linear function is of the form [latex]f\left(x\right)=mx+b[/latex]. 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 [latex]K\left(d\right)<M\left(d\right)[/latex]. The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the [latex]K\left(d\right)[/latex] function is smaller.
These graphs are sketched above, with 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 [latex]K\left(d\right)[/latex] 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 [latex]d>100[/latex].
(5.2.3) – Solve mixture problems with a system of linear equations
One application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. A solution is a mixture of two or more different substances like water and salt or vinegar and oil. Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand. There are many other disciplines that use solutions as well.
The concentration or strength of a liquid solution is often described as a percentage. This number comes from the ratio of how much mass is in a specific volume of liquid. For example if you have 50 grams of salt in a 100mL of water you have a 50% salt solution based on the following ratio:
[latex]\frac{50\text{ grams }}{100\text{ mL }}=0.50\frac{\text{ grams }}{\text{ mL }}=50\text{ % }[/latex]
Solutions used for most purposes typically come in pre-made concentrations from manufacturers, so if you need a custom concentration, you would need to mix two different strengths. In this section, we will practice writing equations that represent the outcome from mixing two different concentrations of solutions.
We will use the following table to help us solve mixture problems:
Amount
Concentration (%)
Total
Solution 1
Solution 2
Final Solution
To demonstrate why the table is helpful in solving for unknown amounts or concentrations of a solution, consider two solutions that are mixed together, one is 120mL of a 9% solution, and the other is 75mL of a 23% solution. If we mix both of these solutions together we will have a new volume and a new mass of solute and with those we can find a new concentration.
First, find the total mass of solids for each solution by multiplying the volume by the concentration.
Amount
Concentration (%)
Total Mass
Solution 1
120 mL
0.09 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex]
[latex]\left(120\cancel{\text{ mL}}\right)\left(0.09\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=10.8\text{ grams }[/latex]
Solution 2
75 mL
0.23 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex]
[latex]\left(75\cancel{\text{ mL}}\right)\left(0.23\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=17.25\text{ grams }[/latex]
Final Solution
Next we add the new volumes and new masses.
Amount
Concentration (%)
Total Mass
Solution 1
120 mL
0.09 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex]
[latex]\left(120\cancel{\text{ mL}}\right)\left(0.09\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=10.8\text{ grams }[/latex]
Solution 2
75 mL
0.23 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex]
[latex]\left(75\cancel{\text{ mL}}\right)\left(0.23\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=17.25\text{ grams }[/latex]
Final Solution
195 mL
[latex]\frac{28.05\text{ grams }}{ 195 \text{ mL }}=0.14=14\text{ % }[/latex]
Now we have used mathematical operations to describe the result of mixing two different solutions. We know the new volume, concentration and mass of solute in the new solution. In the following examples, you will see that we can use the table to find an unknown final volume or concentration. These problems can have either one or two variables. We will start with one variable problems, then move to two variable problems.
Example
A chemist has 70 mL of a 50% methane solution. How much of an 80% solution must she add so the final solution is 60% methane?
Show Solution
Let’s use the problem solving process outlined in Module 1 to help us work through a solution to the problem.
Read and Understand: We are looking for a new amount – in this case a volume – based on the words “how much”. We know two starting concentrations and the final concentration, as well as one volume.
Define and Translate: Solution 1 is the 70 mL of 50% methane and solution 2 is the unknown amount with 80% methane. We can call our unknown amount x.
Write and Solve: Set up the mixture table. Remember that concentrations are written as decimals before we can perform mathematical operations on them.
Amount
Concentration (%)
Total Mass
Solution 1
70
0.5
Solution 2
x
0.8
Final Solution
0.6
Multiply amount by concentration to get total, be sure to distribute on the last row: [latex]\left(70 + x\right)0.6[/latex]Add the entries in the amount column to get final amount. The concentration for this amount is 0.6 because we want the final solution to be 60% methane.
Amount
Concentration (%)
Total Mass
Solution 1
70
0.5
35
Solution 2
x
0.8
0.8x
Final Solution
70+x
0.6
[latex]42+0.6x[/latex]
Add the total mass for solution 1 and solution 2 to get the total mass for the 60% solution. This is our equation for finding the unknown volume.
35mL must be added to the original 70 mL to gain a solution with a concentration of 60%
The above problem illustrates how we can use the mixture table to define an equation to solve for an unknown volume. In the next example we will start with two known concentrations and use a system of equations to find two starting volumes necessary to achieve a specified final concentration.
Example
A farmer has two types of milk, one that is 24% butterfat and another which is 18% butterfat. How much of each should he use to end up with 42 gallons of 20% butterfat?
Show Solution
Read and Understand: We are asked to find two starting volumes of milk whose concentrations of butterfat are both known. We also know the final volume is 42 gallons. There are two unknowns in this problem.
Define and Translate: We will call the unknown volume of the 24% solution x, and the unknown volume of the 18% solution y.
Write and Solve: Fill in the table with the information we know.
Amount
Concentration (%)
Total Mass
Solution 1
x
0.24
Solution 2
y
0.18
Final Solution
42
0.2
Find the total mass by multiplying the amount of each solution by the concentration. The total mass of the final solution comes from
Amount
Concentration (%)
Total Mass
Solution 1
x
0.24
0.24x
Solution 2
y
0.18
0.18y
Final Solution
x+y=42
0.2
8.4
When you sum the amount column you get one equation: [latex]x+ y = 42[/latex]
When you sum the total column you get a second equation: [latex]0.24x + 0.18y = 8.4[/latex]
Use elimination to find a value for [latex]x[/latex], and [latex]y[/latex].
Multiply the first equation by [latex]-0.18[/latex]
Adding the two equations together to eliminate the y terms gives this equation:
[latex]0.06x = 8.4[/latex]
Divide by 0.06 on each side:
[latex]x = 14[/latex]
Now substitute the value for x into one of the equations in order to solve for y.
[latex]\begin{array}{cc} (14) + y &= 42\\ y &= 28 \end{array}[/latex]
Answer
This can be interpreted as 14 gallons of 24% butterfat milk added to 28 gallons of 18% butterfat milk will give 42 gallons of 20% butterfat milk.
In the following video you will be given an example of how to solve a mixture problem without using a table, and interpret the results.
(5.2.4) – Solve uniform motion problems with a system of linear equations
Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.
Let’s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.
The images below show how a river current affects the speed at which a boat is actually travelling. We’ll call the speed of the boat in still water [latex]b[/latex] and the speed of the river current [latex]c[/latex].
The boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is [latex]b+c[/latex].
Now, the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat’s actual speed is slower than its speed in still water. The actual speed of the boat is [latex]b-c[/latex].
We’ll put some numbers to this situation in the next example.
EXAMPLE
Translate to a system of equations and then solve.
A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.
Show Answer
Read the problem: This is a uniform motion problem and a picture will help us visualize the situation.
Identify what we are looking for: We are looking for the speed of the ship in still water and the speed of the current.
Name what we are looking for:
Let [latex]s=[/latex] the rate of the ship in still water.
Let [latex]c=[/latex] the rate of the current.
A chart will help us organize the information. The ship goes downstream and then upstream. Going downstream, the current helps the ship and so the ship’s actual rate is [latex]s+c[/latex]. Going upstream, the current slows the ship and so the actual rate is [latex]s-c[/latex]. Downstream it takes 4 hours. Upstream it takes 5 hours. Each way the distance is 60 miles.
Translate into a system of equations. Since rate times time is distance, we can write the system of equations.
Question ID 29699. Authored by: McClure, Caren. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL
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Ex: System of Equations Application - Mixture Problem.. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning.. Located at: https://youtu.be/4s5MCqphpKo.. License: CC BY: Attribution
Beginning and Intermediate Algebra Textbook. . Authored by: Tyler Wallace. Located at: . License: CC BY: Attribution
