Answer
a. $ \frac{c-b}{2e} $
b. $ \frac{b+c}{2} $
c. $\frac{(c-b)^2}{4e} -a$
d. $\frac{c-b-t}{2e} $, $\frac{t}{2}$
Work Step by Step
a. Step 1. Given the cost of $y=a+bx$ for $x$ units per week and the selling price of $P=c-ex$ per unit per week, we can write the profit as
$T=xP-y=cx-ex^2-a-bx=-ex^2+(c-b) x-a$
Step 2. To maximize the profit, let $T’=0$; we have $-2ex+(c-b)=0$, which gives $x= \frac{c-b}{2e} $. Check $T’’=-2e\lt0$, which indicates a local maximum.
b. The corresponding price is given as
$P_0=c-ex=c-\frac{c-b}{2}= \frac{b+c}{2} $
c. The weekly profit is then
$T=-e(\frac{c-b}{2e})^2+(c-b)(\frac{c-b}{2e})-a=\frac{(c-b)^2}{4e} -a$
d. If there is a tax of $t$ dollars per unit sold, the profit will be
$T=xP-y-xt=cx-ex^2-a-bx-xt=-ex^2+(c-b-t)x-a$
Let $T’=0$, we have $-2ex+(c-b-t)=0$ which gives $x= \frac{c-b-t}{2e} $ and the price becomes $P_1=c-e(\frac{c-b-t}{2e})=\frac{b+c+t}{2} $. We can find the difference between the two prices as $\Delta P=P_1-P_0=\frac{t}{2}$, indicating the price need to be raised to half of the tax value in order to maximize the profit.
