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

Business Applications — Question 9

PDF ↗

Question 9

The demand for a product is modeled by p(x)=80−x,p(x)=80-x, where p(x)p(x) is the price per unit in dollars and xx is the number of units sold in hundreds. The total cost of producing 100x100x units is C(x)=20x+100dollars.C(x)=20x+100\quad\text{dollars}. Require x≥0x\geq0 and p(x)≥0p(x)\geq0.

  • (a) Find the revenue and profit functions.

  • (b) Find the production level and maximum profit in the continuous model.

  • (c) Find the best whole-number quantity of units and its profit.

Original worksheet page 1: question and worked solution for 4-14-009
Show solutionHide solution

Question 9 - Solution

(a) Revenue and profit

The number sold is 100x100x, so revenue is 100xp(x)100x\,p(x):

R(x)=8000x−100x2,P(x)=−100x2+7980x−100.R(x)=8000x-100x^2,\qquad P(x)=-100x^2+7980x-100.

P′(x)=−200x+7980,P′′(x)=−200<0.P\prime(x)=-200x+7980,\qquad P\prime\prime(x)=-200<0.

(b) Continuous maximum

The continuous maximum occurs at

x=7980200≈39.900000,Pmax≈$159,101.00.\boxed{x=\frac{7980}{200}\approx 39.900000,\quad P_{\max}\approx\$159,101.00.}

(c) Whole units

This represents 3990.0000003990.000000 units. Comparing the adjacent whole-unit counts gives 3990\boxed{3990} units, with profit $159,101.00\boxed{\$159,101.00}. The price is nonnegative at these levels.

Original worksheet page 2: question and worked solution for 4-14-009

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

← Section questionsChapter contents