Question

a marine manufacturer will sell n(x) power boats after spending $x thousand on advertising, as given by n(x)=950 - \frac{3,810}{x}, 5\leq x\leq30. (a) find n(x). (b) find n(10) and n(20). write a brief verbal interpretation of these results. (a) n(x)=\square

Explanation:

Step1: Rewrite the function

Rewrite $N(x)=950 - \frac{3810}{x}$ as $N(x)=950-3810x^{- 1}$.

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$. For $N(x)=950-3810x^{-1}$, the derivative of a constant 950 is 0, and the derivative of $-3810x^{-1}$ is $(-3810)\times(-1)x^{-1 - 1}$. So $N^\prime(x)=3810x^{-2}=\frac{3810}{x^{2}}$.

Step3: Calculate $N^\prime(10)$

Substitute $x = 10$ into $N^\prime(x)$: $N^\prime(10)=\frac{3810}{10^{2}}=\frac{3810}{100}=38.1$. This means that when the advertising spending is $10,000, the number of power - boats sold is increasing at a rate of 38.1 boats per additional $1,000 spent on advertising.

Step4: Calculate $N^\prime(20)$

Substitute $x = 20$ into $N^\prime(x)$: $N^\prime(20)=\frac{3810}{20^{2}}=\frac{3810}{400}=9.525$. This means that when the advertising spending is $20,000, the number of power - boats sold is increasing at a rate of 9.525 boats per additional $1,000 spent on advertising.

Answer:

(A) $N^\prime(x)=\frac{3810}{x^{2}}$
(B) $N^\prime(10) = 38.1$, which means when advertising spending is $10,000, the number of boats sold is increasing at a rate of 38.1 boats per additional $1,000 spent on advertising. $N^\prime(20)=9.525$, which means when advertising spending is $20,000, the number of boats sold is increasing at a rate of 9.525 boats per additional $1,000 spent on advertising.