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Calculus IChapter 4: Application of DerivativesSection 4-14

Business Applications — Question 10

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Question 10

A company manufactures and sells a product. The weekly revenue and cost functions (in thousands of dollars) are: R(x)=500x−5x2,C(x)=200x+500R(x) = 500x - 5x^2, \quad C(x) = 200x + 500

  • (a) Determine the profit function P(x)P(x).

  • (b) Find the production level xx that maximizes profit.

  • (c) Compute the maximum profit.

Original worksheet page 1: question and worked solution for 4-14-010
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Question 10 - Solution

(a) Profit Function: P(x)=R(x)−C(x)=(500x−5x2)−(200x+500)=−5x2+300x−500P(x) = R(x) - C(x) = (500x - 5x^2) - (200x + 500) = -5x^2 + 300x - 500

(b) Maximize Profit: P′(x)=−10x+300P'(x) = -10x + 300 Set P′(x)=0P'(x) = 0: −10x+300=0⇒x=30-10x + 300 = 0 \quad \Rightarrow \quad x = 30 Check the second derivative: P″(x)=−10<0P''(x) = -10 < 0 Thus, profit is maximized when x=30x = 30.

(c) Maximum Profit: P(30)=−5(30)2+300(30)−500=−4500+9000−500=4000P(30) = -5(30)^2 + 300(30) - 500 = -4500 + 9000 - 500 = 4000

Maximum profit occurs at x=30 units, Pmax=4000 thousand dollars\boxed{\text{Maximum profit occurs at } x = 30 \text{ units, } P_{\max} = 4000 \text{ thousand dollars}}

Original worksheet page 2: question and worked solution for 4-14-010

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