Answer
$5$ and $3$.
Work Step by Step
Our equations according to the exercise are: $2x+2y=16, xy=15$.
So $x=8-y$, thus $y(y-8)=15\\y^2-8y-15=0\\(y-3)(y-5)=0$
Thus $y=3,x=5$ or $y=5,x=3$, but these are the same, so the dimensions are $5$ and $3$.
College Algebra (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0321979478
ISBN 13:
978-0-32197-947-6
Chapter 8 - Section 8.6 - Systems of Nonlinear Equations - 8.6 Assess Your Understanding - Page 616: 79
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