Question

section 2.10: applied optimization (homework)
score: 120/150 answered: 12/15
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question 13
0/10 pts 4 99 details
demand for an electric fan is related to its selling price p (in dollars) by the equation
$n = 2880 - 90p$
where n is the number of fans that can be sold per month at a price p. find the selling price that will maximize the revenue.
the selling price that will yield the maximum revenue is $
question help: video

Explanation:

Step1: Define the revenue function

Revenue $R = p\times n$. Substitute $n = 2880 - 90p$ into the revenue - formula, so $R(p)=p(2880 - 90p)=2880p-90p^{2}$.

Step2: Find the derivative of the revenue function

The derivative of $R(p)$ with respect to $p$ is $R^\prime(p)=\frac{d}{dp}(2880p - 90p^{2})=2880-180p$.

Step3: Set the derivative equal to zero and solve for $p$

Set $R^\prime(p) = 0$, then $2880-180p = 0$. Add $180p$ to both sides: $2880=180p$. Divide both sides by 180: $p=\frac{2880}{180}=16$.

Step4: Confirm it's a maximum

The second - derivative of $R(p)$ is $R^{\prime\prime}(p)=\frac{d}{dp}(2880 - 180p)=-180<0$. Since $R^{\prime\prime}(p)<0$, when $p = 16$, the revenue function $R(p)$ has a maximum.

Answer:

$16$