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:

Write the system of equations using [latex]y[/latex] to replace function notation.

[latex]\begin{array}{l}\begin{array}{l}\\ y=0.85x+35,000\end{array}\hfill \\ y=1.55x\hfill \end{array}[/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}{c}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]1.55\left(50,000\right)=77,500[/latex]

The break-even point is [latex]\left(50,000,77,500\right)[/latex].

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].

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.

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.

[latex]\begin{array}{c}c+a=2,000\\ 25c+50a=70,000\end{array}[/latex]

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].

[latex]\begin{array}{c}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}{l} 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}{l}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.

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 [latex]f\left(x\right)=mx+b[/latex]. Using the rates of change and initial charges, we can write the equations

[latex]\begin{array}{l}K\left(d\right)=0.59d+20\\ M\left(d\right)=0.63d+16\end{array}[/latex]

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)

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

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

[latex]\begin{array}{l}K\left(d\right)=M\left(d\right)\hfill \\ 0.59d+20=0.63d+16\hfill \\ 4=0.04d\hfill \\ 100=d\hfill \\ d=100\hfill \end{array}[/latex]

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].

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.

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.

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.

[latex]35+0.8x=42+0.6x[/latex]

[latex]\begin{array}{c}35+0.8x=42+0.6x\\\underline{-0.6x}\,\,\,\,\,\,\,\underline{-0.6x}\\35+0.2x=42\\\end{array}[/latex]

Subtract 35 from both sides

[latex]\begin{array}{c}35+0.2x=42\\\underline{-35}\,\,\,\,\,\,\,\underline{-35}\\0.2x=7\end{array}[/latex]

Divide both sides by 0.2

[latex]\begin{array}{c}0.2x=7\\\frac{0.2x}{0.2}=\frac{7}{0.2}\end{array}[/latex]
[latex]x=35[/latex]

Answer

35mL must be added to the original 70 mL to gain a solution with a concentration of 60%

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.

Find the total mass by multiplying the amount of each solution by the concentration. The total mass of the final solution comes from

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]

[latex]\begin{array}{cc}-0.18(x+y) &= (42)(-0.18) \\ -0.18x-0.18y &= -7.56 \end{array}[/latex]

Now our system of equations looks like this:

[latex]\begin{array}{cc} -0.18x-0.18y &= -7.56\\0.24x + 0.18y &= 8.4 \end{array}[/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.

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.

 [latex]\begin{array}{c}4(s+c)=60 \\ 5(s-c) = 60\end{array}[/latex]

Solve the system of equations. Distribute to put both equations in standard form, then solve by elimination.

 [latex]\begin{array}{c}4s+4c=60 \\ 5s-5c = 60\end{array}[/latex]

Multiply the top equation by 5 and the bottom equation by 4. Add the equations, then solve for [latex]s[/latex].

[latex]\begin{array}{cc}20s+20c &= 300 \\ 20s-20c &= 240 \\ \hline \\ 40s &= 540 \\ s &= 13.5 \end{array}[/latex]

Substitute [latex]s=13.5[/latex] into one of the original equations.

[latex]\begin{array}{cc}4(s+c) &= 60 \\ 4(13.5+c) &= 60 \\ 54 + 4c &= 60 \\ 4c &= 6 \\ c &= 1.5 \end{array}[/latex]

Check the answer in the problem. The downstream rate would be:

[latex]13.5+1.5 = 15[/latex] mph

In 4 hours the ship would travel:

[latex]15 \cdot 4 = 60[/latex] miles.

The upstream rate would be

[latex]13.5 - 1.5 = 12[/latex] mph.

In 5 hours the ship would travel

[latex]12\cdot 5[/latex] miles.