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Calculus IChapter 4: Application of DerivativesSection 4-14

Business Applications — Question 6

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Question 6

The demand for a product is modeled by p(x)=150−3x,p(x)=150-3x, 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)=50x+300dollars.C(x)=50x+300\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 4-14-006
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Question 6 - Solution

(a) Revenue and profit

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

R(x)=15000x−300x2,P(x)=−300x2+14950x−300.R(x)=15000x-300x^2,\qquad P(x)=-300x^2+14950x-300.

P′(x)=−600x+14950,P′′(x)=−600<0.P\prime(x)=-600x+14950,\qquad P\prime\prime(x)=-600<0.

(b) Continuous maximum

The continuous maximum occurs at

x=14950600≈24.916667,Pmax≈$185,952.08.\boxed{x=\frac{14950}{600}\approx 24.916667,\quad P_{\max}\approx\$185,952.08.}

(c) Whole units

This represents 2491.6666672491.666667 units. Comparing the adjacent whole-unit counts gives 2492\boxed{2492} units, with profit $185,952.08\boxed{\$185,952.08}. The price is nonnegative at these levels.

Original worksheet page 2: question and worked solution for 4-14-006

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