Chapter 1.4, Problem 22GP

Solution

To find: the price that the magazine should charge to maximize annual revenue and find the maximum annual revenue.

To maximize revenue, each subscription should cost $12.5 and the maximum annual revenue is $\$312,000$ .

Given information:

The magazine has 28,000 subscribers when it charges $11 per annual subscription.

For each $1 increase in price, the magazine loses about 2000 subscribers.

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 increase and $R\left(x\right)$ represents the annual revenue.

Now, the required verbal model is:

Annual revenue(dollars) = Number of subscribers $\times$ Subscription price(dollars/person)

So, the quadratic equation is:

  $\begin{array}{l}R\left(x\right)=\left(28000-2000x\right)\times \left(11+x\right)\\ R\left(x\right)=\left(-2000x+28000\right)\left(x+11\right)\\ R\left(x\right)=-2000\left(x-14\right)\left(x+11\right)\end{array}$

Now, it is clear that here the zeros of the revenue function are 14 and -11.

The average of zeros is $\frac{14+\left(-11\right)}{2}=1.5$ .

So, to maximize revenue, each subscription should cost

  $\$11+\$1.5=\$12.5$

The maximum annual revenue is:

  $\begin{array}{c}R\left(1.5\right)=-2000\left(1.5-14\right)\left(1.5+11\right)\\ =-2000\left(-12.5\right)\left(12.5\right)\\ =\$312,000\end{array}$