Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.7 - Optimization Problems - 4.7 Exercises - Page 338: 30


The length of the third side should be $~~\sqrt{2}~a$

Work Step by Step

Let $b$ be the base of the triangle. Then the height $h$ of the triangle is $~~h = \sqrt{a^2-(\frac{b}{2})^2}$ We can write an expression for the area: $A = \frac{1}{2}bh$ $A = \frac{1}{2}b\sqrt{a^2-(\frac{b}{2})^2}$ We can find $\frac{dA}{db}$: $\frac{dA}{db} = \frac{1}{2}\sqrt{a^2-\frac{b^2}{4}}+(\frac{b}{2})(\frac{-\frac{b}{2}}{2\sqrt{a^2-\frac{b^2}{4}}})$ $\frac{dA}{db} = \frac{4(a^2-\frac{b^2}{4})}{8\sqrt{a^2-\frac{b^2}{4}}}-\frac{b^2}{8\sqrt{a^2-\frac{b^2}{4}}}$ $\frac{dA}{db} = \frac{4a^2-2b^2}{8\sqrt{a^2-\frac{b^2}{4}}} = 0$ $4a^2-2b^2 = 0$ $2b^2 = 4a^2$ $b^2 = 2a^2$ $b = \sqrt{2}~a$ The length of the third side should be $~~\sqrt{2}~a$
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