Chapter 1.1, Problem 8GP

Solution

To find: how the weekly revenue can be maximized.

The weekly revenue is maximized when each racer is charged $\$22.25$ .

Given information:

The track has about 380 racers per week and charges $35 each racer to race.

Each $1 reduction in price brings in 40 more racers per week.

Concept used:

The vertex of the quadratic function represents the maximum/minimum point of the function.

The x-coordinate of the vertex of a parabola of the form $y=a{x}^2+bx+c$ is given $x=-\frac{b}{2a}$ .

Calculation:

Let the price be decreased by $ $x$ .

Then, the equation modeling the situation can be given as $R(x)=(35-x)\cdot (380+40x)$ .

Rewrite the equation as follows:

  $\begin{array}{c}R(x)=(35-x)\cdot (380+40x)\\ R(x)=35\cdot 380+35\cdot 40x-x\cdot 380-x\cdot 40x\\ =13300+1020x-40x^2\\ =-40x^2+1020x+13300\end{array}$

This represents a downward parabola. So, it has a maximum value at its vertex.

The x-coordinate of the vertex of a parabola of the form $y=a{x}^2+bx+c$ is given $x=-\frac{b}{2a}$ .

For $R(x)=-40x^2+1020x+13300$ , $a=-40$ and $b=1020$ . Use this information to find the x-coordinate of the vertex.

  $\begin{array}{c}x=-\frac{1020}{2\left(-40\right)}\\ =\frac{51}{4}\\ =12.75\end{array}$

This shows that the maximum weekly revenue of the track occurs when $x=12.75$ .

That means, the weekly revenue is maximized by decreasing the price by $12.75.

That is, the weekly revenue is maximized when it charges $\$\left(35-12.75\right)=\$22.25$ .

Conclusion:

The weekly revenue is maximized when it charges $\$22.25$ .