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 0 to 10 , then \( E(n)=\frac{2}{3} n-\frac{1}{90} n^{2} \) 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

To find the number of times a viewer should watch the commercial for maximum effectiveness, we need to maximize the function \( E(n) = \frac{2}{3} n - \frac{1}{90} n^{2} \). 1. First, take the derivative of \( E(n) \): \[ E'(n) = \frac{2}{3} - \frac{2}{90} n \] 2. Set the derivative equal to zero to find critical points: \[ \frac{2}{3} - \frac{2}{90} n = 0 \] 3. Solve for \( n \): \[ \frac{2}{90} n = \frac{2}{3} \implies n = \frac{2}{3} \cdot \frac{90}{2} = 45 \] 4. To confirm that this is a maximum, check the second derivative: \[ E''(n) = -\frac{2}{90} \] Since \( E''(n) < 0 \), the function has a maximum at \( n = 45 \). Thus, for maximum effectiveness, a viewer should watch the commercial **45 times**.