Chapter 1.4, Problem 13Q

Solution

To find: the price that the store should charge to maximize monthly revenue and find the maximum monthly revenue.

To maximize revenue, each DVD player should be sold at $120 and the maximum monthly revenue is $\$7,200$

Given information:

The store sells 50 new model DVD player per month at price $140 each.

For each $10 decrease in price, about 5 more DVD players per month are sold.

Property Used:

Factoring and Zeros:

To find the maximum or minimum value of a quadratic function, first use factoring to write the function in intercept form $y=a\left(x-p\right)\left(x-q\right)$ . Because the function’s vertex lies on the axis of symmetry $x=\frac{p+q}{2}$ , the maximum or minimum occurs at the average of the zeros p and q .

Calculation:

Let x represent the price decrease and $R\left(x\right)$ represents the monthly revenue.

Now, the required verbal model is:

Monthly revenue(dollars) = Number of DVD players sold $\times$ price of each DVD player

So, the quadratic equation is:

  $\begin{array}{l}R\left(x\right)=\left(50+5x\right)\times \left(140-10x\right)\\ R\left(x\right)=\left(5x+50\right)\left(-10x+140\right)\\ R\left(x\right)=5\left(x+10\right)\left[-10\left(x-14\right)\right]\\ R\left(x\right)=-50\left(x+10\right)\left(x-14\right)\end{array}$

Now, it is clear that here the zeros of the revenue function are 14 and -10.

The average of zeros is $\frac{14+\left(-10\right)}{2}=2$

So, to maximize revenue, each subscription should cost

  $\begin{array}{c}140-10x=140-10\times 2\\ =\$120\end{array}$

The maximum monthly revenue is:

  $\begin{array}{c}R\left(2\right)=-50\left(2+10\right)\left(2-14\right)\\ =-50\left(12\right)\left(-12\right)\\ =\$7,200\end{array}$