An electronics store sells an average of 60 entertainment systems per month at an average of $800 more than the cost price. For every $20 increase in the selling price, the store sells one fewer system. What amount over the cost price will maximize revenue?

Question

anonymous · Answer

c = cost price
s = selling price
n = number of systems
sn = total revenue

sn - c = income/profit/loss
s = (c+800)
n = 60
c?

if s goes up 20
n = n -1

sorry this is all i got so far, hope it helps in any way

anonymous · Answer

Q = 60 + (800-P)/20
R= QP = (100 -P/20)P
so find the max of R = -P^2 /20 +100P

and find the Price at that max...  (set the derivative equal to zero and solve for P)

anonymous · Answer

First thing you want is number sold in terms of price sold at
lets call number sold n, and price sold p
$$n = 60-\frac{p-800}{20}$$
$$n = 60+\frac{800-p}{20}$$

revenue (r) = number sold x price sold
$$r = p(60+\frac{800-p}{20})$$
$$r = 60p+40p-\frac{p^2}{20}$$
$$r = 100p - \frac{p^2}{20}$$

To find the max, we find the stationary point
(we know since this is a quadratic, and that the coefficient of p^2 is negative the max is the stationary point)

$$\frac{dr}{dp} = 100 -\frac{2p}{20}$$
$$\frac{dr}{dp} = 100 -\frac{p}{10}$$
$$0 = 100 - \frac{p}{10}$$
\[0 = 1000 - p]
\[p = 1000]

anonymous · Answer

Thank you very much! I finally understand it.

anonymous · Answer

that's great :)