you are running a business selling homemade bread. your weekly revenue from the sale of q loaves of bread is…

you are running a business selling homemade bread. your weekly revenue from the sale of q loaves of bread is r(q)=66q - 0.1q² dollars, and the weekly cost of making q loaves of bread is c(q)=28 + 20q dollars. (note that p(q)=r(q)-c(q)). a. find the weekly profit function p(q). p(q)=66q - 0.1q² - 28 + 20q b. find the production level q that maximizes the weekly profit. q = loaves of bread

you are running a business selling homemade bread. your weekly revenue from the sale of q loaves of bread is r(q)=66q - 0.1q² dollars, and the weekly cost of making q loaves of bread is c(q)=28 + 20q dollars. (note that p(q)=r(q)-c(q)). a. find the weekly profit function p(q). p(q)=66q - 0.1q² - 28 + 20q b. find the production level q that maximizes the weekly profit. q = loaves of bread

Answer

Explanation:

Step1: Recall profit - revenue - cost formula

Given $P(q)=R(q)-C(q)$, with $R(q)=66q - 0.1q^{2}$ and $C(q)=28 + 20q$. Substitute the revenue and cost functions into the profit - function formula. $P(q)=(66q - 0.1q^{2})-(28 + 20q)$

Step2: Simplify the profit function

$P(q)=66q - 0.1q^{2}-28 - 20q=-0.1q^{2}+(66q - 20q)-28=-0.1q^{2}+46q - 28$

Step3: Find the derivative of the profit function

For a quadratic function $y = ax^{2}+bx + c$, its derivative $y^\prime=2ax + b$. Here, $a=-0.1$, $b = 46$, $c=-28$. So $P^\prime(q)=-0.2q+46$

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

$-0.2q + 46=0$. Solve for $q$: $-0.2q=-46$, then $q=\frac{46}{0.2}=230$

Step5: Check the second - derivative to confirm it's a maximum

The second - derivative $P^{\prime\prime}(q)=\frac{d}{dq}(-0.2q + 46)=-0.2<0$. Since the second - derivative is negative, the function $P(q)$ is concave down at $q = 230$, and $q = 230$ is a point of maximum.

Answer:

a. $P(q)=-0.1q^{2}+46q - 28$ b. $230$ loaves of bread