## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 3 - Review Exercises - Page 300: 21e

#### Answer

$A=-\frac{(x-4)^2}{2}+8$ The maximum value for $A$ is obtained when $x=4$: $A=8$

#### Work Step by Step

$A=-\frac{x^2}{2}+4x$ (see item (a)) $A=-\frac{1}{2}(x^2-8x)=-\frac{1}{2}(x^2-8x+16-16)$ $A=-\frac{1}{2}[x^2-2(4)x+4^2-16]$ $A=-\frac{1}{2}[(x-4)^2-16]=-\frac{(x-4)^2}{2}+8$ The maximum value for $A$ is obtained when $x-4=0$, that is, when $x=4$: $A=-\frac{(4-4)^2}{2}+8=8$

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