Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 9 - Multiveriable Calculus - 9.4 Lagrange Multipliers - 9.4 Exercises - Page 498: 19


f(x) does not have an absolute minimum or maximum

Work Step by Step

$f(x,y)=8x^{2}y$ The constraint becomes $3x-y-9=0$ with $g(x,y)=3x-y-9$ $F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=8x^{2}y-\lambda(3x-y-9)=x^{2}y-3\lambda x - \lambda y + 9\lambda$ $F_{x}(x,y,\lambda)=2xy-\lambda$ $F_{y}(x,y,\lambda)=x^{2}-\lambda$ $F_{\lambda}(x,y,\lambda)=-3x-y+9$ $(1) 2xy-\lambda =0 \rightarrow \lambda =2xy$ $(2) x^{2} -\lambda =0 \rightarrow \lambda =x^{2}$ $(3) -3x-y+9=0$ Set the expressions equal: $x^{2}=2xy$ $x=2y$ Substitute for x=2y in Equation (3) $-6y-y+9=0$ $y=\frac{9}{7} \rightarrow x=\frac{18}{7}$ The value of is $f(\frac{18}{7},\frac{9}{7})\approx68.01$ Let $y=1, x=\frac{10}{3}$ then $f(\frac{10}{3},1)\approx88.89$. Because a nearby point has a value larger than 68.01, the value 68.01 is probably not a maximum. Let's try with a different nearby point $y=1.5,x=3.5$ then $f(3.5,1.5)=147$ Since the two nearby points have a larger value compared to 68.01, f(x) does not have an absolute minimum or maximum.
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