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: 1


$f(x,y) =4xy$ has a maximum value subject to the constraint $x + y = 16$ at the point (8,8)

Work Step by Step

$f(x,y)=4xy$ The constraint becomes $x+y-16=0$ with $g(x,y)=x+y-16$ $F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=4xy-\lambda(x+y-16)=4xy-\lambda x -\lambda y + 16\lambda$ $F_{x}(x,y,\lambda)=4y-\lambda$ $F_{y}(x,y,\lambda)=4x-\lambda$ $F_{\lambda}(x,y,\lambda)=-x-y+16$ $(1) 4y-\lambda =0 \rightarrow \lambda =4y$ $(2) 4x-\lambda =0 \rightarrow \lambda =4x$ $(3) -x-y+16=0$ Set the expressions equal: $4y=4x$ $x=y$ Substitute for x=y in Equation (3) $-y-y+16=0$ $y=8 \rightarrow x=8$ The value of is f(8,8)=256 Let x=8.1 and y=7.9; then f(8.1;7.9)=255.96 Because a nearby point has a value smaller than 256, the value 256 is probably not a minimum.
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