Question

If C(x) = 16000 + 500x − 3.2x^2 + 0.004x^3
is the cost function and
p(x) = 2900 − 8x is the demand function, find the production level that will maximize profit. (Hint: If the profit is maximized, then the marginal revenue equals the marginal cost.) I have no idea hot to solve this, please help!

Answer 1

C(1000) = 16000 + 200(1000) + 4(1000^3/2)
...............= 16000 + 200000 + 126491
........cost = 342491

AC = 342491 / 1000 = 342.50
MC(x) = C'(x)
..........= 200 + 8x^1/2
..........= 200 + 8(31.62)
..........= 453

ii. graph the average cost function (AC) and the marginal cost function (MC)

AC = [(16000 + 200x + 4x^(3/2)] / x
MC = C' = 200 + 8x^(1/2)

find the x value at the intersection.
x = 252 parts

iii. minimum average cost (c) would be the y value at the intersection.
c = $327

2. First you find the revenue function R(x)
R(x) = x*p(x) = x(1700 - 7x)
R(x) = 1700x - 7x^2

max profit occurs when R'(x) = C'(x)
R'(x) = 1700 - 14x.....C'(x) = 500 - 3.2x + .012x^2
1700 - 14x = 500 - 3.2x + .012x^2
0 = .012x^2 + 10.8x - 1200
graphing this equation will give
x = 100 when y = 0 so 100 is the production level.

Answer 2

so 100 would be my answer? @Jacob902 i typed that in and it was wrong...

Answer 3

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CCEQFjAA&url=https%3A%2F%2Fanswers.yahoo.com%2Fquestion%2Findex%3Fqid%3D20090216152002AAwga1m&ei=BYaYVYDoC4S9ggSt5oL4AQ&usg=AFQjCNFwae_kN8dJuIdk-84SVucrnKT6EA&bvm=bv.96952980,d.eXY

Answer 4

oh okay