Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 2 - Functions and Their Graphs - 2.1 Functions - 2.1 Assess Your Understanding - Page 59: 111

Answer

a) Profit function $P(x) = -0.05x^3+0.8x^2+155x-500$ b) $P(15) =$\$$1836.25$ c) The profit is $\$1836.25$ when $1500$ cell phones are sold.

Work Step by Step

a) We know we can find the profit function by subtracting the cost function from the revenue function. $P(x) = R(x) -C(x)$ $=(-1.2x^2+220x)-(0.05x^3-2x^2+65x+500)$ $=-1.2x^2+220x-0.05x^3+2x^2-65x-500$ $=-0.05x^3+0.8x^2+155x-500$ b) Given $x=15$, when substituted in profit function $P(x)$ $P(15)= -0.05(15)^3+0.8(15)^2+155(15)-500$ $=-168.75+180+2325-500$ $=$\$$1836.25$ c) When $15$ hundred cell phones are sold, the profit is $\$1836.25$
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