the effectiveness of a television commercial depends on how many times a viewer watches it. after some…

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

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

Answer

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