Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - 7.1 Antiderivatives - 7.1 Exercises - Page 366: 50

Answer

$$C\left( x \right) = \frac{{{x^2}}}{2} - \frac{1}{x} + 4$$

Work Step by Step

$$\eqalign{ & C'\left( x \right) = x + 1/{x^2};{\text{ 2 units cost }}\$ 5.50 \cr & {\text{The marginal function cost is the derivative of the function cost }}C'\left( x \right) \cr & {\text{then}}{\text{, the function cost }}C\left( x \right){\text{ is }} \cr & C\left( x \right) = \int {C'\left( x \right)} dx \cr & {\text{replacing }}x + 1/{x^2}{\text{ for }}C'\left( x \right) \cr & C\left( x \right) = \int {\left( {x + \frac{1}{{{x^2}}}} \right)} dx \cr & {\text{rewrite the integrand}} \cr & C\left( x \right) = \int {\left( {x + {x^{ - 2}}} \right)} dx \cr & {\text{integrating }} \cr & C\left( x \right) = \frac{{{x^2}}}{2} + \frac{{{x^{ - 1}}}}{{ - 1}} + K \cr & C\left( x \right) = \frac{{{x^2}}}{2} - \frac{1}{x} + K{\text{ }}\left( {\bf{1}} \right) \cr & \cr & {\text{Find }}K,{\text{we know that the 2 units cost }}\$ 5.50{\text{ then }}C\left( 2 \right) = 5.50 \cr & 5.50 = \frac{{{{\left( 2 \right)}^2}}}{2} - \frac{1}{2} + K \cr & 5.50 = 2 - 0.5 + K \cr & K = 5.50 - 2 + 0.5 \cr & K = 4 \cr & {\text{then substituting }}K = 4{\text{ into the equation }}\left( {\bf{1}} \right){\text{ we obtain}} \cr & C\left( x \right) = \frac{{3{x^{5/3}}}}{5} + 2x + \frac{{114}}{5} \cr & C\left( x \right) = \frac{{{x^2}}}{2} - \frac{1}{x} + 4 \cr} $$
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