College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 3 - Mid-Chapter Check Point - Page 390: 22


See below.

Work Step by Step

Let $x$ be the height, then the base is $40-2x$, then the area is: $x(40-2x)/2=-x^2+20x$. The maximum value is at $x=-\frac{b}{2a}=-\frac{20}{2\cdot(-1)}=10.$ Hence the maximum value is $f(10)=-(10)^2+20(10)=100.$
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