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: 56


$f_{max}(\frac{80}{3},\frac{160}{3})=37926$ $x=\frac{80}{3}, y=\frac{160}{3}$

Work Step by Step

Let our positive numbers be x, y. We know that $x + y =80 \rightarrow g(x)=x+y-80$ Our objective function is $f(x,y)=x^{2}y$ $F(x,Y,λ)=f(x,y)-\lambda.g(x,y)=x^{2}y-\lambda (x+y-80)=x^{2}y -\lambda x -\lambda y +80\lambda$ $F_{x}(x,y,\lambda)=2xy-\lambda$ $F_{y}(x,y,\lambda)=x^{2}-\lambda$ $F_{\lambda}(x,y,\lambda)=-x-y+80$ $(1) 2xy-\lambda=0 \rightarrow \lambda =2xy$ $(2) x^{2}-\lambda =0 \rightarrow \lambda =x^{2}$ $(3) -x-y+80=0$ Set the expressions equal: $x^{2}=2xy$ $x=2y$ Substitute for x=2y in Equation (3) $-2y-y+80=0$ $y=\frac{80}{3} \rightarrow x=\frac{160}{3}$ The value of f is $f(\frac{80}{3},\frac{160}{3})\approx37926$ Let x=26 and y=54 then $f(26,54)= 36504$. Because a nearby point has a value smaller than 37926, 37926 is probably not a minimum. Therefore $f_{max}(\frac{80}{3},\frac{160}{3})=37926$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.