If C(x) = 12000 + 600x − 0.6x^2 + 0.004x^3

Question

is the cost function and p(x) = 1800 − 6x 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.)

Answer 1

R(x)=x(1800-6x)
R(x)=1800x-6x^2
R'(x)=1800-12x

C'(x)=600-1.2x+.012x^2

Marginal revenue = marginal cost

1800-12x=600-1.2x+.012x^2

1200=10.8x+.012x^2

1200=.012x(900+x)

100,000=x(900+x)

100,000=900x+x^2

x^2+900x-100000 =0

(x-100)(x-1000) =0

x = 100

x+1000) = 0
x = 100

At 100 units, the demand price is $1800-6(100) = $1200
The cost to make them will be $[.004(1000000) - .6(10000) + 600(100) + 12000] = $70000
The revenue is $1800(100) - 6(100^2) = $120000
The profit is $(120000 - 70000)= $50000, which is the maximum profit.

Answer 2

the answer above is correct. you have to put 100 units as the solution.