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: 5


$f(x,y)=x^{2}+2y^{2}-xy$ has a minimum value subject to the constraint x+y=8, it is at the point (5,3). The value of (5,3) is 28.

Work Step by Step

$f(x,y)=x^{2}+2y^{2}-xy$ The constraint becomes $x+y-8=0$ with $g(x,y)=x+y-8$ $F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=x^{2}+2y^{2}-xy-\lambda(x+y-8)=x^{2}+2y^{2}-xy-\lambda x -\lambda y + 8\lambda$ $F_{x}(x,y,\lambda)=2x-y-\lambda$ $F_{y}(x,y,\lambda)=4y-x-\lambda$ $F_{\lambda}(x,y,\lambda)=-x-y+8$ $(1) 2x-y-\lambda =0 \rightarrow \lambda =2x-y$ $(2) 4y-x-\lambda =0 \rightarrow \lambda =4y-x$ $(3) -x-y+8=0$ Set the expressions equal $2x-y=4y-x$ $x=\frac{5}{3}y$ Substitute for $x=\frac{5}{3}y$ in Equation (3) $-\frac{5}{3}y-y+8=0$ $y=3 \rightarrow x=5$ The value of f is f(5,3)=28 Let y=3.1 and x=4.9 then f(4.9,3.1)=28.04 which is greater than 28. Because a nearby point has a value greater than 28, the value 28 is probably not a maximum.
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