Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.7 - Extreme Values and saddle Points - Exercises 14.7 - Page 843: 11


Local maximum at $f(\dfrac{16}{7},0)=\dfrac{-16}{7}$

Work Step by Step

Given: $f_x(x,y)=\dfrac{112x-8x}{\sqrt{56x^2-8y^2-16x-31}}=0, f_y(x,y)=\dfrac{-8y}{\sqrt{56x^2-8y^2-16x-31}}=0$ Simplify the given two equations. This implies that $x=\dfrac{16}{7},y=0$ Critical point: $(\dfrac{16}{7},0)$ In order to solve this problem we will have to apply Second derivative test that suggests the following conditions to calculate the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$. 1. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\gt 0$ , then $f(a,b)$ is a local minimum. 2. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\lt 0$ , then $f(a,b)$ is a local maximum. 3. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \lt 0$ , then $f(a,b)$ is a not a local minimum or local maximum but a saddle point. $D=f_{xx}f_{yy}-f^2_{xy}=\dfrac{64}{225}\implies D=\dfrac{64}{225} \gt 0$ and $f_{xx}=-\dfrac{-8}{15} \lt 0$ Hence, Local maximum at $f(\dfrac{16}{7},0)=\dfrac{-16}{7}$
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