Maximizing profit given cost and demand functions

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SUMMARY

The discussion focuses on maximizing profit using the cost function C(x) = 13000 + 600x − 0.6x² + 0.004x³ and the demand function p(x) = 1800 − 6x. The key approach involves calculating the revenue function R(x) = xp(x) = 1800x - 6x², deriving it to find marginal revenue R'(x) = 1800 - 12x, and equating it to the marginal cost C'(x) = 600 - 1.2x + 0.012x². The resulting equation 1200 = 10.8x + 0.012x² leads to a quadratic equation that can be solved for the production level x that maximizes profit.

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Homework Statement



If
C(x) = 13000 + 600x − 0.6x62 + 0.004x3 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.)

Homework Equations



R(x)=xp(x)

The Attempt at a Solution


Because of the hint, How I thought this question should be answered was to find the revenue function and take the derivative of that which is the marginal revenue and then set it equal to the derivative of the cost function which is marginal cost
So,
R(x)=x(1800-6x)
R(x)=1800x-6x^2
R'(x)=1800-12x

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

is this right? Where do I go from here, can I solve for x at this point? If so how?
 
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