maximum profit using optimization?

Question

5. Maximum Profit: A computer dealer can sell 12 personal computers per week at a price of $2,000 each. He estimates that each $400 price decrease will result in three more sales per week. If the computers cost him $1,200 each, what price should he charge to maximize his profit? How many will he sell at that price? Show all work. I got a p(x)=2000-400x and q(x) 12+3x I was able to get a revenue function out of those. What im having trouble with is getting the cost function, and then using both of those to get the P(x) profit function.

Update:

The answer key says x= -1, p= 2,400, and q=9. Maybe we missed something.

Mike G · Accepted Answer

The answer key is correct.
I did it this way.
Let D be the number of sets demanded at price $x
D = 12 + (2000-x)*3/400
D = 27 - 3x/400
R = revenue = D*x
R = 27x-3x^2/400
Cost C = 1200*D = 1200*27 - 9x
Profit P = R-C
P = 27x-3x^2/400 -32,400+9x = -3x^2/400+36x-32,400
dP/dx = -6x/400 +36 = 0 for maximum
-6x = -14,400
x = $2,400 dollars
At this price he will get a demand of 9
Profit = 9 * 1,200 = $10,800

Iggy Rocko · Answer

P(x) = (2000 - 400x)(12 + 3x) - 1200(12 + x) = 24000 + 6000x - 4800x - 1200x^2 - 14000 - 1200x =
-1200x^2 + 10000
P's maximum occurs where P '(x) = 0 and P ''(x) < 0.
P '(x) = -2400x
0 = -2400x
x = 0
P ''(x) = -2400 < 0 for all x so we know maximum occurs at x = 0.
He should charge $2000 and sell 12 computers.

? · Answer

p(x) = price each
q(x) = units sold

Revenue = p(x) * q(x)
Costs = 1200 * q(x)

Profit = Revenue - Costs

... you should find the answer key correct

price = 2400
demand = 9

? · Answer

Fashion is a forever topic. Everyone loves beautiful things and wants to be beautiful or handsome.

Trending News

maximum profit using optimization?

4 Answers