Calculus with Applications (10th Edition)

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

Chapter 9 - Multiveriable Calculus - Chapter Review - Review Exercises - Page 519: 54

Answer

$f(x,y)=x^{2}y$ has a maximum value subject to the constraint x+y=4, it is at the point $f(\frac{8}{3} ,\frac{4}{3})$. The value of $f(\frac{8}{3} ,\frac{4}{3})$ is $\frac{256}{27}$.

Work Step by Step

We are given $f(x,y)=x^{2}y$ The constraint becomes $x+y-4=0$ with $g(x,y)=x+y-4$ $F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=x^{2}y-\lambda(x+y-4)=x^{2}y-\lambda x -\lambda y + 4\lambda$ $F_{x}(x,y,\lambda)=2xy-\lambda$ $F_{y}(x,y,\lambda)=x^2-\lambda$ $F_{\lambda}(x,y,\lambda)=-x-y+4$ $(1) 2xy-\lambda =0 \rightarrow \lambda =2xy$ $(2) x^{2}-\lambda =0 \rightarrow \lambda =x^{2}$ $(3) -x-y+4=0$ Set the expressions equal $2xy=x^{2}$ $x=2y$ Substitute for $x=2y$ in Equation (3) $-2y-y+4=0$ $y=\frac{4}{3} \rightarrow x=\frac{8}{3}$ The value of is $f(\frac{8}{3} ,\frac{4}{3})=\frac{256}{27}$ Let y=2.5 and x=1.5 then f(1.5,2.5)=5.625 which is smaller than $\frac{256}{27}$. Because a nearby point has a value smaller than $\frac{256}{27}$, the value $\frac{256}{27}$ is probably not a minimum. $f(x,y)=x^{2}y$ has a maximum value subject to the constraint x+y=4, it is at the point $f(\frac{8}{3} ,\frac{4}{3})$. The value of $f(\frac{8}{3} ,\frac{4}{3})$ is $\frac{256}{27}$.
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