## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 3 - Review Exercises - Page 300: 22c

#### Answer

$x=50~m$ $y=50~m$ $A=2500~m^2$

#### Work Step by Step

We need to find the vertex of $A=100x-x^2$ $A=-x^2+100x~~$ ($a=-1,b=100,c=0$): $-\frac{b}{2a}=-\frac{100}{2(-1)}=50$ $f(50)=-50^2+100(50)=-2500+5000=2500$ Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(50,2500)$ That is, when $x=50~m$ the maximum area is obtained ($A=2500~m^2$). $y=100-x~$ (see item (b)) $y=100-50$ $y=50~m$

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