Chapter 1.1, Problem 56PS

Solution

To find: the value of $x$ at which the store’s revenue is maximized.

The function that models the situation of the problem is $R(x)=(320-20x)\cdot (70+5x)$ .

The store can maximize the monthly revenue by charging $\$300$ for each camera.

Given information:

The store sells 70 of new modal camera per month when the price is $320, and for each $20 decrease in the price, the sale increases by 5 camera per month.

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 $x$ times.

Then, the equation modeling the situation can be given as $R(x)=(320-20x)\cdot (70+5x)$ .

Rewrite the equation as follows:

  $\begin{array}{c}R(x)=(320-20x)\cdot (70+5x)\\ R(x)=-100x^2+200x+22400\end{array}$

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 $R(x)=-100x^2+200x+22400$ , $a=-100$ and $b=200$ . Use this information to find the x-coordinate of the vertex.

  $\begin{array}{c}x=-\frac{b}{2a}\\ =-\frac{200}{2\left(-100\right)}\\ =1\end{array}$

This shows that the maximum revenue of the store occurs when $x=1$ .

That means, the store can maximize its revenue by decreasing the price $20.

That is, the store’s revenue is maximum when it charges $\$\left(320-20\right)=\$300$ .

Conclusion:

The store can maximize the monthly revenue by charging $\$300$ for each camera.