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 0 to 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?
Answer
Explanation:
Step1: Recall derivative - based optimization.
To find the maximum of a function $E(n)=\frac{3}{7}n-\frac{1}{72}n^{2}$, we take its first - derivative. The power rule for differentiation states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. $E^\prime(n)=\frac{d}{dn}(\frac{3}{7}n-\frac{1}{72}n^{2})=\frac{3}{7}-\frac{2}{72}n=\frac{3}{7}-\frac{1}{36}n$
Step2: Set the first - derivative equal to zero.
At the maximum or minimum of a function, the first - derivative is zero. So we set $E^\prime(n) = 0$. $\frac{3}{7}-\frac{1}{36}n=0$
Step3: Solve for $n$.
First, add $\frac{1}{36}n$ to both sides of the equation: $\frac{1}{36}n=\frac{3}{7}$. Then multiply both sides by 36 to isolate $n$: $n=\frac{3\times36}{7}=\frac{108}{7}\approx15.43$. Since $n$ represents the number of times a viewer watches a commercial and it must be a non - negative integer, we can also check the second - derivative of $E(n)$ to confirm it's a maximum. The second - derivative $E^{\prime\prime}(n)=\frac{d}{dn}(\frac{3}{7}-\frac{1}{36}n)=-\frac{1}{36}<0$, which means the function is concave down and the critical point is a maximum. Rounding $n$ to the nearest whole number, we get $n = 15$.
Answer:
15