Chapter 2.2, Problem 70E

To determine

The amount charged per camera to maximize daily revenue using verbal model and quadratic function.

Answer to Problem 70E

The store should sell 300 cameras to maximize daily revenue of $22500.

Explanation of Solution

Given Information:

The store sells 70 digital cameras per month at a price of $320 each. For each $20 decrease in price, 5 more cameras per month are sold.

  $\begin{array}{c}\text{Revenue}=\text{Price}\times \text{Sales}\\ \text{R}\left(x\right)=\left(320-20x\right)\times \left(70+5x\right)\end{array}$

Calculation:

The maximum sale of the model can be obtained by the y -coordinates. Write the model of revenue in standard from and find the vertex,

  $\begin{array}{c}\text{R}\left(x\right)=\left(320-20x\right)\times \left(70+5x\right)\\ =22,400+1600x-1400x-100x^2\\ =-100x^2+200x+22400\\ =-100\left(x^2-2x+1-1\right)+4000\\ =-100\left(x^2-2x+1\right)-\left(-100\right)+22400\\ =-100{\left(x-1\right)}^2+22500\end{array}$

Which is of the form $y=a{\left(x-h\right)}^2+k$

This gives the value of the vertex $\left(h,k\right)$ as $h=1,k=22500$ .

Therefore, the store should charge,

  $\begin{array}{c}70+5x=70+5\times 1\\ =75\end{array}$

The price of each camera is,

  $\begin{array}{c}320-20x=320-20\left(1\right)\\ =300\end{array}$

Maximum daily revenue earned is $22500.