Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 288: 74

Answer

$85$ dollars per unit.

Work Step by Step

Step 1. Given the marginal cost $\frac{dc}{dx}=0.001x^2-0.5x+115$, we have $c(x)=\frac{10^{-3}}{3}x^3-0.25x^2+115x+D$ where $D$ is a constant. As $c(0)=0$, we have $D=0$. Step 2. For $x=600$, we have $c(600)=\frac{10^{-3}}{3}(600)^3-0.25(600)^2+115(600)$ Steo 3. The production cost per unit is then $\frac{c(600)}{600}=\frac{10^{-3}}{3}(600)^2-0.25(600)+115=120-150+115=85$ dollars per unit.
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