Question

the effectiveness of a television commercial depends on how many times a viewer watches it. after some experiments an advertising agency found that if the effectiveness ( e ) is measured on a scale of 10, then ( e(n)=\frac{2}{3}n - \frac{1}{72}n^{2}) where ( n ) is the number of times a viewer watches a given commercial. for a commercial to have maximum effectiveness, how many times should a viewer watch it? ( \times ) times

Explanation:

Step1: Find the derivative of \(E(n)\)

We have \(E(n)=\frac{2}{3}n-\frac{1}{72}n^{2}\). Using the power - rule \((x^n)^\prime = nx^{n - 1}\), the derivative \(E^\prime(n)=\frac{2}{3}-\frac{1}{36}n\).

Step2: Set the derivative equal to zero

To find the critical points, we set \(E^\prime(n) = 0\). So, \(\frac{2}{3}-\frac{1}{36}n=0\).

Step3: Solve for \(n\)

First, add \(\frac{1}{36}n\) to both sides: \(\frac{1}{36}n=\frac{2}{3}\). Then multiply both sides by 36 to get \(n = 24\).

Step4: Confirm it's a maximum

We find the second - derivative \(E^{\prime\prime}(n)=-\frac{1}{36}<0\). Since the second - derivative is negative, \(n = 24\) is a point of maximum.

Answer:

24