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: 6


$f(x,y)=3x^{2}+4y^{2}-xy-2$ has a minimum value subject to the constraint 2x+y=21, it is at the point $(4,\frac{17}{2})$. The value of $(4,\frac{17}{2})$ is 301.

Work Step by Step

$f(x,y)=3x^{2}+4y^{2}-xy-2$ The constraint becomes $2x+y-21=0$ with $g(x,y)=2x+y-21$ $F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=3x^{2}+4y^{2}-xy-2-\lambda(2x+y-21)=3x^{2}+4y^{2}-xy-2-2\lambda x -\lambda y + 21\lambda$ $F_{x}(x,y,\lambda)=6x-y-2\lambda$ $F_{y}(x,y,\lambda)=8y-x-\lambda$ $F_{\lambda}(x,y,\lambda)=-2x-y+21$ $(1) 6x-y-2\lambda =0 \rightarrow \lambda =\frac{6x-y}{2}$ $(2) 8y-x-\lambda =0 \rightarrow \lambda =8y-x$ $(3) -2x-y+21=0$ Set the expressions for equal $\frac{6x-y}{2}=8y-x$ $6x-y=16y-2x$ $x=\frac{17}{8}y$ Substitute for $x=\frac{17}{8}y$ in Equation (3) $-2(\frac{17}{8}y)-y+21=0$ $y=4 \rightarrow x=\frac{17}{2}$ The value of f is $f(4,\frac{17}{2})=301$ Let y=9 and x=6 then f(6,9)=376 which is greater than 301. Because a nearby point has a value greater than 301, the value 301 is probably not a maximum.
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