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anonymous

  • one year ago

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!

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  1. Jacob902
    • one year ago
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    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.

  2. anonymous
    • one year ago
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    so 100 would be my answer? @Jacob902 i typed that in and it was wrong...

  3. anonymous
    • one year ago
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    oh okay

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