Cost: [latex]y=0.85x+35,000[/latex]

Revenue:[latex]y=1.55x[/latex]

1. 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, [latex]x[/latex] would represent the number of frames produced (instead of skateboards) and [latex]y[/latex] would represent the amount of money it would cost to produce them (the same as the skateboard problem).
2. The revenue equation represents money coming into the company, so in this context [latex]x[/latex] still represents the number of bike frames manufactured, and [latex]y[/latex] now represents the amount of money made from selling them.
3. We must solve the system of equations

   [latex]\left\{\begin{array}{l}y=0.85x+35,000\\ y=1.55x \end{array}\right.[/latex]

   Substitution is the best method since both equations are already solved for [latex]y[/latex]. Substitute the expression [latex]0.85x+35,000[/latex] from the first equation into the second equation and solve for [latex]x[/latex].

   [latex]\begin{array}{rl}0.85x+35,000&=\;1.55x\\ 35,000&=\;0.7x\\ 50,000&=\;x\end{array}[/latex]

   Then, we substitute [latex]x=50,000[/latex] into either the cost function or the revenue function.

   [latex]y=1.55\left(50,000\right)=77,500[/latex]

   The break-even point is [latex]\left(50,000,77,500\right),[/latex] which represents [latex]50,000[/latex] bike frames sold and both cost and revenue of [latex]\$77,500.[/latex] If more than [latex]50,000[/latex] bike frames are sold, a profit will result.

Analysis of the Solution

The profit function is found using the formula [latex]P\left(x\right)=R\left(x\right)-C\left(x\right)[/latex].

[latex]\begin{array}{l}P\left(x\right)=1.55x-\left(0.85x+35,000\right)\hfill \\ \text{ }=0.7x - 35,000\hfill \end{array}[/latex]

The profit function is [latex]P\left(x\right)=0.7x - 35,000[/latex]. The above problem could have also been solved by setting [latex]P\left(x\right)=0[/latex] and solving. You may want to verify this yourself.

The cost to produce [latex]50,000[/latex] units is [latex]\$77,500[/latex], and the revenue from the sales of [latex]50,000[/latex] units is also [latex]\$77,500.[/latex] To make a profit, the business must produce and sell more than [latex]50,000[/latex] 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 [latex]x[/latex]-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 [latex]0[/latex]. The area to the left of the break-even point represents operating at a loss.

Let [latex]c=[/latex] the number of children and [latex]a=[/latex] 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 [latex]\$25.00[/latex] by the number of children, [latex]25c[/latex]. The revenue from all adults can be found by multiplying [latex]\$50.00[/latex] by the number of adults, [latex]50a[/latex]. The total revenue is [latex]\$70,000[/latex]. 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.

[latex]\left\{ \begin{array}{llll}c&+&a&=\;2,000\\ 25c&+&50a&=\;70,000\end{array} \right.[/latex]

In the first equation the coefficient of both variables is [latex]1[/latex] so we solve by substitution (elimination would also be possible.) We will solve the first equation for [latex]a[/latex].

[latex]\begin{array}{rl}c+a&=\;2,000\\ a&=\;2,000-c\end{array}[/latex]

Substitute the expression [latex]2,000-c[/latex] in the second equation for [latex]a[/latex] and solve for [latex]c[/latex].

[latex]\begin{array}{rl} 25c+50\left(2,000-c\right)&=\;70,000\hfill \\ 25c+100,000 - 50c&=\;70,000\hfill \\ \text{ }-25c&=\;-30,000\hfill \\ \text{ }c&=\;1,200\hfill \end{array}[/latex]

Substitute [latex]c=1,200[/latex] into the first equation to solve for [latex]a[/latex].

[latex]\begin{array}{rl}1,200+a&=\;2,000\hfill \\ \text{ }\text{}a&=\;800\hfill \end{array}[/latex]

We find that [latex]1,200[/latex] children and [latex]800[/latex] adults bought tickets to the circus that day.

Let’s make sure we understand the problem and the variables we will be using first before starting to write equations.

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 [latex]70[/latex] mL of [latex]50\%[/latex] methane and solution 2 is the unknown amount with [latex]80\%[/latex] methane.  We can call our unknown amount [latex]x[/latex]. We know the final mixture has a concentration of [latex]60\%[/latex].

Write and Solve:  Set up the mixture table. Remember that concentrations are written as decimals before we can perform mathematical operations on them.

First complete the first row by adding [latex]70+x[/latex]. Then complete the last row by multiplying Amount by Concentration in each column.

The last row now represents an equation we can solve since the parts must add to the total.

[latex]\begin{align}70\cdot 0.5+x\cdot 0.8&=(70+x)\cdot 0.6\\ 35+0.8x&=42+0.6x&&\color{blue}{\textsf{simplify}}\\ 0.8x&=7+0.6x&&\color{blue}{\textsf{subtract }35\textsf{ from both sides}}\\ 0.2x&=7&&\color{blue}{\textsf{subtract }0.6x\textsf{ from both sides}}\\ x&=35&&\color{blue}{\textsf{divide by }0.2\textsf{ on both sides}}\end{align}[/latex]

We must add [latex]35[/latex] mL of the [latex]80\%[/latex] methane solution to the original [latex]70[/latex] mL of [latex]50\%[/latex] methane in order to get the desired mixture of [latex]60\%[/latex] methane.

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 [latex]42[/latex] gallons. There are two unknowns in this problem.

Define and Translate: We will call the unknown volume of the [latex]24\%[/latex] solution [latex]x[/latex], and the unknown volume of the [latex]18\%[/latex] solution [latex]y[/latex].

Write and Solve: Fill in the table with the information we know.

Complete the last row by multiplying Amount by Concentration in each column.

Remember the “Amount” row and the “Value” row represent the idea that two parts must add to a total. The first and third rows thus give us a system of equations:

[latex]\left\{ \begin{array}{rl}\phantom{0.24}x+\phantom{0.18}y&=\text{ }42\\ 0.24x+0.18y&=\text{ }8.4\end{array}\right.[/latex]

We can use either substitution or elimination to solve. In this example we demonstrate the elimination method and begin by eliminating [latex]x[/latex].

[latex]\begin{array}{rllrl}x+y&=42&\xrightarrow{\cdot \; -0.24} & -0.24x-0.24y &=-10.08\\ 0.24x+0.18y&=8.4&\xrightarrow{\phantom{\cdot -0.24}} & 0.24x+0.18y&=8.4 \\ & & & \text{__________}&\text{______} \\ & & & -0.06y&=-1.68 \\ & & & y &=28 \\ \end{array}[/latex]

We can substitute [latex]y=28[/latex] into either of the original equations to solve for [latex]x[/latex]. We use the first equation because it has easier numbers to work with.

[latex]\begin{array}{rl}x+(28)&=42\\ x&=14\\ \end{array}[/latex]

This can be interpreted as [latex]14[/latex] gallons of [latex]24\%[/latex] butterfat milk added to [latex]28[/latex] gallons of [latex]18\%[/latex] butterfat milk will give the desired [latex]42[/latex] gallons of [latex]20\%[/latex] butterfat milk.

Read and Understand: We are asked to find the starting quantities of cashews and raisins with known prices per pound. The total number of final pounds and final price per pound are known.

Define and Translate: We will set [latex]c[/latex] = pounds of cashews and [latex]r[/latex] = pounds of raisins.

Write and Solve: Fill in the table with the information we know.

Complete the last row by multiplying Amount by Price Per Pound in each column.

Remember the “Amount” row and the “Value” row represent the idea that two parts must add to a total. The first and third rows thus give us a system of equations:

[latex]\left\{ \begin{array}{rl}\phantom{6.5}c+\phantom{4}r&=\text{ }40\\ 6.5c+4r&=\text{ }190\end{array}\right.[/latex]

Substitution is probably easier since the first equation can easily be solved for either variable. We solve it for [latex]c.[/latex]

[latex]\begin{align}c+r&=40\\ c&=40-r\end{align}[/latex]

Now substitute [latex]40-r[/latex] for [latex]c[/latex] in the second equation.

[latex]\begin{align}6.5c+4r&=190\\ 6.5(40-r)+4r&=190\\ 260-6.5r+4r&=190\\ 260-2.5r&=190\\ -2.5r&=-70\\ r&=28\end{align}[/latex]

Substitute [latex]r=28[/latex] into the original first equation.

[latex]\begin{align}c+(28)&=40\\ c&=12 \end{align}[/latex]

To achieve the desired mixture, the store owner needs to use [latex]12[/latex] pounds of cashews and [latex]28[/latex] pounds of raisins.

Read and Understand: The trains’ speeds are known. We do not know their times but we have a relationship between the two times. We do not know the distances for either train but the distance will be equal when the second train is passing the first train.

Define and Translate: Let [latex]d[/latex] be the unknown distance from the station when the second train passes the first one. Let [latex]t[/latex] be the amount of time the second train is traveling. Hence the first train is traveling 4 hours longer, or [latex]t+4.[/latex]

Write and Solve: Fill in the table with the information we know. The table is not necessary, but can help you organize.

Each row represents a uniform motion, which can be modeled by [latex]d=rt.[/latex] Thus we have the system of equations

[latex]\left\{\begin{array}{ll}d\;=&56(t+4)\\ d\;=&84t\end{array}\right.[/latex]

This is best solved with substitution since [latex]d[/latex] is already solved for in both equations. The substitution will effectively set the right sides equal to each other.

[latex]\begin{align}56(t+4)&=84t\\ 56t+224&=84t\\ 224&=28t\\ 8&=t\end{align}[/latex]

The second train will pass the first after [latex]8[/latex] hours. The original question asked us for the distance from the station, which is found by using [latex]t=8[/latex] in either equation, for example

[latex]d=84t=84(8)=672.[/latex]

The second train will pass the first [latex]672[/latex] km from the station.

Read and Understand: A picture can help us visualize what is happening. Distance and time are known to us for both trips.

Define and Translate:  The unknowns are the speed of the ship [latex]b[/latex] in still water and the speed of the current [latex]c,[/latex] which are the same for both trips.

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]b+c[/latex]. Going upstream, the current slows the ship and so the actual rate is [latex]b-c[/latex].

Write and Solve: Since [latex]d=rt[/latex], we can write an equation for each direction.

[latex]\left\{\begin{array}{rl}60\;&=(b+c)4\\ 60\;&=(b-c)5\end{array}\right.[/latex]

Distribute in each equation, then solve by elimination.