STEM-Master
Calculus IChapter 4: Application of DerivativesSection 4-14

Business Applications — Question 2

PDF ↗

Question 2

A company manufactures and sells xx units of a product per week. The revenue (in thousands of dollars) and cost (in thousands of dollars) functions are given by: R(x)=50x−0.5x2,C(x)=20x+100R(x) = 50x - 0.5x^2, \quad C(x) = 20x + 100

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

  • (b) Determine the number of units xx that maximizes the profit.

  • (c) Compute the maximum profit.

Original worksheet page 1: question and worked solution for 4-14-002
Show solutionHide solution

Question 2 - Solution

(a) Profit Function:

P(x)=R(x)−C(x)=(50x−0.5x2)−(20x+100)=−0.5x2+30x−100P(x) = R(x) - C(x) = (50x - 0.5x^2) - (20x + 100) = -0.5x^2 + 30x - 100

(b) Maximize Profit:

P′(x)=−x+30P'(x) = -x + 30

Set P′(x)=0P'(x) = 0: −x+30=0⇒x=30-x + 30 = 0 \quad \Rightarrow \quad x = 30

Second derivative: P″(x)=−1<0P''(x) = -1 < 0 Hence, profit is maximized at x=30x = 30.

(c) Maximum Profit:

P(30)=−0.5(30)2+30(30)−100=−450+900−100=350P(30) = -0.5(30)^2 + 30(30) - 100 = -450 + 900 - 100 = 350

Maximum profit occurs at x=30 units, Pmax=350 thousand dollars\boxed{ \text{Maximum profit occurs at } x = 30 \text{ units, } P_{\max} = 350 \text{ thousand dollars} }

Original worksheet page 2: question and worked solution for 4-14-002

Original worksheet layout. Use Enlarge or open the PDF for a closer view.

← Section questionsChapter contents