Q13.

Question

Due to increased production costs, the Daily News must increase its subscription rate. According to a recent survey, the number of subscriptions will decrease by about 1250 for each 25¢ increase in the subscription rate. What weekly subscription rate will maximize the newspaper's income from subscriptions?

Step-by-Step Solution

Verified

Answer

The weekly subscription rate that will maximize the newspaper's income is $8.75

1Step 1. Use the concept.

Consider the function $f (x) = a x^{2} + b x + c, a \neq 0$ , the x-coordinate of vertex is $\frac{- b}{2 a}$

The graph of $f (x) = a x^{2} + b x + c, a \neq 0$

opens up and has a minimum value when $a > 0$ , and
opens down and has a maximum value when $a < 0$

2Step 2. Given Information.

Given

Subscription Rate is $$ 7.50 / w k$

Current Circulation is 50,000

Number of subscriptions decrease by 1250 for each 25¢ increase in the subscription rate.

3Step 3. Solution.

Let x = the number of $$ 0.25$ price increases per week

Then $7.5 + 0.25 x$ = Subscription Rate per week per subscription and

$50, 000 - 1250 x$ = The number of subscriptions

Let I(x) = income as a function of x.

$\begin{array}{l} \overset{The Income}{\overset{⏞}{I (x)}} = \overset{T h e n u m b e r o f s u b s c r i p t i o n s}{\overset{⏞}{(50, 000 - 1250 x)}} \cdot \overset{S u b s c r i p t i o n R a t e p e r w e e k p e r s u b s c r i p t i o n}{\overset{⏞}{(7.5 + 0.25 x)}} \\ \Rightarrow I (x) = 3,75,000 + 12500 x - 9,375 x - 312.5 x^{2} \\ \Rightarrow I (x) = - 312.5 x^{2} + 3,125 x + 3,75,000 \end{array}$