Models and Applications

2.3 Models and Applications

Topics covered in this section are:

  1. Set up a linear equation to solve a real-world application.
  2. Use a formula to solve a real-world application.

Josh is hoping to get an A in his college algebra class. He has scores of $75, 82, 95, 91$, and $94$ on his first five tests. Only the final exam remains, and the maximum of points that can be earned is $100$. Is it possible for Josh to end the course with an A? A simple linear equation will give Josh his answer.

Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talk-time; it costs a widget manufacturer a certain amount to produce $x$ widgets per month plus monthly operating charges; a car rental company charges a daily fee plus an amount per mile driven. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems.

2.3.1 Setting up a Linear Equation to Solve a Real-World Application

To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $\$0.10$/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write $0.10x$. This expression represents a variable cost because it changes according to the number of miles driven.

If a quantity is independent of a variable, we usually just add or subtract it, according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $\$0.10$/mi plus a daily fee of $\$50$. We can use these quantities to model an equation that can be used to find the daily car rental cost $C$.

$C=0.10x+50$

When dealing with real-world applications, there are certain expressions that we can translate directly into math. Table 1 lists some common verbal expressions and their equivalent mathematical expressions.

VerbalTranslation to Math Operations
One number exceeds another by $a$$x, x+a$
Twice a number$2x$
Once number is $a$ more than another number$x, x+a$
Once number is $a$ less than another number$x, 2x-a$
The product of a number and $a$, decreased by $b$$ax-b$
The quotient of a number and the number plus $a$ is three times the number$\frac{x}{x+a}=3x$
The product of three times a number and the number decreased by $b$ is $c$$3x(x-b)=c$
Table 1

HOW TO: Given a real-world problem, model a linear equation to fit it.

  1. Identify known quantities.
  2. Assign a variable to represent the unknown quantity.
  3. If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.
  4. Write an equation interpreting the words as mathematical operations.
  5. Solve the equation. Be sure the solution can be explained in words, including the units of measure.

Example 1

Find a linear equation to solve for the following unknown quantities: One number exceeds another number by $17$ and their sum is $31$. Find the two numbers.

Solution

Let $x$ equal the first number. Then, as the second number exceeds the first by $17$, we can write the second number as $x+17$. The sum of the two numbers is $31$. We usually interpret the word is as an equal sign.

$\begin{align*} x+(x+17)&=31 \\ 2x+17&=31 \\ 2x&=14 \\ x&=7 \\ \\ x+17&=7+17 \\ &=24 \end{align*}$Simplify and solve.

The two numbers are $7$ and $24$.

Example 2

There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $\$34$ plus $\$.05$/min talk-time. Company B charges a monthly service fee of $\$40$ plus $\$.04$/min talk-time.

  • Write a linear equation that models the packages offered by both companies.
  • If the average number of minutes used each month is $1,160$, which company offers the better plan?
  • If the average number of minutes used each month is $420$, which company offers the better plan?
  • How many minutes of talk-time would yield equal monthly statements from both companies?
Solution

Part 1

The model for Company A can be written as $A=0.05x+34$. This includes the variable cost of $0.05x$ plus the monthly service charge of $\$34$. Company B’s package charges a higher monthly fee of $\$40$, but a lower variable cost of $0.04x$. Company B’s model can be written as $B=0.04x+40$.

Part 2

If the average number of minutes used each month is $1,160$, we have the following:

Company A$=0.05(1,160)+34$
$=58+34$
$=92$
Company B$=0.04(1,160)+40$
$=46.4+40$
$=86.4$

So Company B offers the lower monthly cost of $\$86.40$ as compared with the $\$92$ monthly cost offered by Company A when the average number of minutes used each month is $1,160$.

Part 3

If the average number of minutes used each month is $420$, we have the following:

Company A$=0.05(420)+34$
$=21+34$
$=55$
Company B$=0.04(420)+40$
$=16.8+40$
$=56.8$

If the average number of minutes used each month is $420$, then Company A offers a lower monthly cost of $\$55$ compared to Company B’s monthly cost of $\$56.80$.

Part 4

To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of $(x, y)$ $x$-value and the $y$-value equal? We can find this point by setting the equations equal to each other and solving for $x$.

$\begin{align*} 0.05x+34&=0.04x+40 \\ 0.01x&=6 \\ x&=600 \end{align*}$

Check the $x$-value in each equation.

$0.05(600)+34=64$
$0.04(600)+40=64$

Therefore, a monthly average of $600$ talk-time minutes renders the plans equal. See Figure 2.

Coordinate plane with the x-axis ranging from 0 to 1200 in intervals of 100 and the y-axis ranging from 0 to 90 in intervals of 10.  The functions A = 0.05x + 34 and B = 0.04x + 40 are graphed on the same plot.
Figure 2

2.3.2 Using a Formula to Solve a Real-World Application

Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem’s question is answered. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the area of a rectangular region, $A=LW$; the perimeter of a rectangle, $P=2L+2W$; and the volume of a rectangular solid, $V=LWH$. When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time.

Example 3

It takes Andrew $30$ min to drive to work in the morning. He drives home using the same route, but it takes $10$ min longer, and he averages $10$ mi/h less than in the morning. How far does Andrew drive to work?

Solution

This is a distance problem, so we can use the formula $d=rt$, where distance equals rate multiplied by time. Note that when rate is given in mi/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution.

First, we identify the known and unknown quantities. Andrew’s morning drive to work takes $30$ min, or $\frac{1}{2}$ h at rate $r$. His drive home takes $40$ min, or $\frac{2}{3}$ h, and his speed averages $10$ mi/h less than the morning drive. Both trips cover distance $d$. A table, such as Table 2, is often helpful for keeping track of informations in these types of problems.

$d$$r$$t$
To Work$d$$r$$\frac{1}{2}$
To Home$d$$r-10$$\frac{2}{3}$
Table 2

Write two equations, one for each trip.

$d=r(\frac{1}{2})$To work
$d=(r-10)(\frac{2}{3})$To home

As both equations equal the same distance, we set them equal to each other and solve for $r$.

$\begin{align*} r(\frac{1}{2})&=(r-10)(\frac{2}{3}) \\ \frac{1}{2}r&=\frac{2}{3}r-\frac{20}{3} \\ \frac{1}{2}r-\frac{2}{3}r&=-\frac{20}{3} \\ -\frac{1}{6}&=-\frac{20}{3} \\ r&=-\frac{20}{3}(-6) \\ r&=40 \end{align*}$

We have solved for the rate of speed to work, $40$ mph. Substituting $40$ into the rate on the return trip yields $30$ mi/h. Now we can answer the question. Substitute the rate back into either equation and solve for $d$.

$\begin{align*} d&=40(\frac{1}{2}) \\ &=20 \end{align*}$

The distance between home and work is $20$ mi.

Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for $r$.

$\begin{align*} r(\frac{1}{2})&=(r-10)(\frac{2}{3}) \\ 6 \times r(\frac{1}{2})&=6 \times (r-10)(\frac{2}{3}) \\ 3r&=4(r-10) \\ 3r&=4r-40 \\ -r&=-40 \\ r&=40 \end{align*}$

Example 4

The perimeter of a rectangular outdoor patio is $54$ ft. The length is $3$ ft greater than the width. What are the dimensions of the patio?

Solution

The perimeter formula is standard: $P=2L+2W$. We have two unknown quantities, length and width. However, we can write the length in terms of the width as $L=W+3$. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as in Figure 3.

A rectangle with the length labeled as: L = W + 3 and the width labeled as: W.
Figure 3

Now we can solve for the width and then calculate the length.

$\begin{align*} P&=2L+2W \\ 54&=2(W+3)+2W \\ 54&=2W+6+2W \\ 54&=4W+6 \\ 48&=4W \\ 12&=W \\ (12+3)&=L \\ 15&=L \end{align*}$

The dimensions are $L=15$ ft. and $W=12$ ft.

Example 5

The perimeter of a tablet of graph paper is $48$ in. The length is $6$ in. more than the width. Find the area of the graph paper.

Solution

The standard formula for area is $A=LW$; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one.

We know that the length is $6$ in. more than the width, so we can write length as $L=W+6$. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.

$\begin{align*} P&=2L+2W \\ 48&=2(W+6)+2W \\ 48&=2W+12+2W \\ 48&=4W+12 \\ 36&=4W \\ 9&=W \\ (9+6)&=L \\ 15&=L \end{align*}$

Now we find the area given the dimensions of $L=15$ in. and $W=9$ in.

$\begin{align*} A&=LW \\ A&=15(9) \\ &=135 \ in.^{2} \end{align*}$

The area is $135$ in$^{2}$.

Example 6

Find the dimensions of a shipping box given that the length is twice the width, the height is $8$ inches, and the volume is $1,600$ in.$^{3}$.

Solution

The formula for the volume of a box is given as $V=LWH$, the product of length, width, and height. We are given that $L=2W$, and $H=8$. The volume is $1,600$ cubic inches.

$\begin{align*} V&=LWH \\ 1,600 &=(2W)W(8) \\ 1,600&=16W^{2} \\ 100&=W^{2} \\ 10&=W \end{align*}$

The dimensions are $L=20$ in., $W=10$ in., and $H=8$ in.

Note that the square root of $W^{2}$ would result in a positive and a negative value. However, because we are describing width, we can use only the positive result.

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