the revenue from selling q items is r(q)=500q - q², and the total cost is c(q)=50 + 7q. write a function…

the revenue from selling q items is r(q)=500q - q², and the total cost is c(q)=50 + 7q. write a function that gives the total profit earned, and find the quantity which maximizes the profit. profit p(q)= quantity maximizing profit q=

the revenue from selling q items is r(q)=500q - q², and the total cost is c(q)=50 + 7q. write a function that gives the total profit earned, and find the quantity which maximizes the profit. profit p(q)= quantity maximizing profit q=

Answer

Explanation:

Step1: Recall profit - revenue - cost formula

Profit $P(q)=R(q)-C(q)$. Given $R(q) = 500q - q^{2}$ and $C(q)=50 + 7q$, then $P(q)=(500q - q^{2})-(50 + 7q)$.

Step2: Simplify the profit function

$P(q)=500q - q^{2}-50 - 7q=-q^{2}+493q - 50$.

Step3: Find the derivative of the profit function

The derivative $P'(q)=\frac{d}{dq}(-q^{2}+493q - 50)=-2q + 493$.

Step4: Set the derivative equal to zero to find critical points

$-2q + 493 = 0$. Solving for $q$ gives $2q=493$, so $q=\frac{493}{2}=246.5$.

Step5: Find the second - derivative of the profit function

$P''(q)=\frac{d}{dq}(-2q + 493)=-2<0$. Since the second - derivative is negative, the profit function is concave down at $q = 246.5$, and this critical point is a maximum.

Answer:

The profit function is $P(q)=-q^{2}+493q - 50$, and the quantity maximizing profit $q = 246.5$