Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 13 - Partial Derivatives - 13.8 Maxima And Minima Of Functions Of Two Variables - Exercises Set 13.8 - Page 987: 46

Answer

$l/4$

Work Step by Step

Let $\mathrm{ABCD}$ be a parallelogram with perimeter $l=p$. From the constraint we know $b=\frac{1}{2}-a$. Using this in the area function: \[ A(a, \theta)=a(l / 2-2 a) \sin \theta=A \] Using the first derivative test, we can determine the critical points \[ A_{a}=\frac{\partial A}{\partial a}=\frac{\partial}{\partial a}\left[\left(\frac{a l}{2}-a^{2}\right) \sin \theta\right]=\sin \theta(l / 2-2 a) \] \[ \begin{aligned} A_{0} &=\frac{\partial A}{\partial \theta}=0 \Rightarrow\left(\frac{a l}{2}-a^{2}\right) \cos \theta=0 \\ & \Rightarrow \theta=\pi / 2 \text { or }(a=l / 2, / b=0) \end{aligned} \] We note that for $b=0, A=0;$ hence $P(a, b, \theta)=P(l / 4, l / 4, \pi / 2)$ is the only critical point such that $A \neq 0$. Using the second derivative test \[ \begin{array}{r} D=A_{a a} A_{l b}-\left.A_{o b}\right|_{(U / 4, l / 4, \pi / 2)}=2\left(f^{2} / 8-P^{2} / 16\right)>0 \text { and } \\ \left.A_{a n}\right|_{(0 / 4 / 4 / 4, \pi / 2)}=-2 \sin (\pi / 2)=-2<0 \end{array} \] This parallelogram is a square with sides $a=l / 4$ and $\ A=(l / 4)^{2}=l^{2} / 16$
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