Question

What is their profit equation? Type it as p= (show your work)

Answer 1

1)

To find out the profit we must subtract the expense from the income. We have both functions, we know that the expense is

$20x+500$

And the income

$-x^2+100x$

The profit will be (income - expense), therefore:

$\begin{gathered} p=-x^2+100x-(20x+500)^{} \\ \\ p=-x^2+100x-20x-500 \\ \\ p=-x^2+80x-500 \end{gathered}$

The profit is given by

$p=-x^2+80x-500$

2)

Now we must discover the ticket price they would break even, the break-even point happens when the income and the expense are the same, then (expense = income)

$\begin{gathered} -x^2+100x=20x+500 \\ \\ -x^2+80x-500=0 \end{gathered}$

See that, it's the same as the profit, we knew that the profit was

$p=-x^2+80x-500$

The break-even happens when p = 0.

Now we must solve that quadratic function

$-x^2+80x-500=0$

Using the quadratic formula

$x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

Where a = -1, b = 80, c = -500.

$\begin{gathered} x=\frac{-80\pm\sqrt[]{80^2-4\cdot(-1)\cdot(-500)}}{2\cdot(-1)} \\ \\ x=\frac{-80\pm\sqrt[]{4400}}{-2} \end{gathered}$

If we simplify the expression

$\begin{gathered} x=\frac{-80\pm20\, \, \sqrt[]{11}}{-2} \\ \\ x=40\pm10\, \, \sqrt[]{11} \end{gathered}$

Therefore they would break-even at

$x=6.834\text{ and }x=73.166$

3)

The max profit will be the vertex of the profit function (parabola), the vertex of a parabola is

$V=\mleft(-\frac{b}{2a},-\frac{\Delta}{4a}\mright)$

Here, the values of "x" is the ticket price and the "y" values are the profit.

Using the x coordinate to find out the vertex ticket price we have:

$x=-\frac{b}{2a}$

Remember that a = -1, b = 80, c = -500, therefore

$\begin{gathered} x=-\frac{80}{2\cdot(-1)} \\ \\ x=\frac{80}{2}=40 \end{gathered}$

They reach the maximum profit when the ticket price is $40

4)

To find out the maximum profit we can use x = 40 in the profit formula

$p=-x^2+80x-500$

Put x = 40

$\begin{gathered} p=-(40)^2+80\cdot(40)-500 \\ \\ p=-1600+3200-500 \\ \\ p=3200-2100 \\ \\ p=1100 \end{gathered}$

The maximum profit is $1100

5)

If we want a determined profit and find the ticket price, we must take the profit function:

$p=-x^2+80x-500$

And solve that function for "x", which is the same to discover the ticket price for a generic profit.

So, we will solve the quadratic for a generic value of p:

$-x^2+80x-500-p=0$

Now, we have a = -1, b = 80, c = -500 - p.

Using again the quadratic formula, we can solve it for x

$\begin{gathered} x=\frac{-80\pm\sqrt[]{80^2-4\cdot(-1)\cdot(-500-p)}}{2\cdot(-1)} \\ \\ x=\frac{-80\pm\sqrt[]{4400-4p}}{-2} \\ \\ x=\frac{-80\pm2\, \sqrt[]{1100-p}}{-2} \\ \\ x=40\pm\, \sqrt[]{1100-p} \end{gathered}$

Therefore, given a profit value, the ticket price will be

$x=40+\sqrt[]{1100-p}\text{ or }x=40-\sqrt[]{1100-p}$

Examples:

Let's suppose that the profit is $600, we just have to put p = 600 and simplify the expression, some of them may be hard to find by hand so use a calculator. The "strange" part here is: there are two values that the ticket can be

$x=40+\sqrt[]{1100-p}\text{ or }x=40-\sqrt[]{1100-p}$

Let's put p = 600 and solve

$\begin{gathered} x=40+\sqrt[]{1100-p}\text{ or }x=40-\sqrt[]{1100-p} \\ x=40+\sqrt[]{1100-600}\text{ or }x=40-\sqrt[]{1100-600} \\ x=62.360\text{ or }x=17.639 \end{gathered}$

Therefore for p = 600 the ticket price and be 62.360 or 17.639.

Let's solve it using the quadratic function to check if our result is correct:

$600=-x^2+80x-500$

We must have zero on one side, then

$\begin{gathered} 600=-x^2+80x-500 \\ \\ 0=-x^2+80x-500-600 \\ \\ -x^2+80x-1100=0 \end{gathered}$

And now we solve using the quadratic formula

$\begin{gathered} x=\frac{-80\pm\sqrt[]{80^2-4\cdot(-1)\cdot(-1100)}}{2\cdot(-1)} \\ \\ x=\frac{-80\pm\sqrt[]{2000}}{-2} \\ \\ x=40\pm10\, \, \sqrt[]{5} \end{gathered}$

If you put that result in decimal we get

$\begin{gathered} x=62.361 \\ x=17.639 \end{gathered}$