## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 9 - Multiveriable Calculus - 9.4 Lagrange Multipliers - 9.4 Exercises - Page 498: 2

#### Answer

$f(x,y)=2xy+4$ has a minimum value subject to the constraint $x+y=20$; it is at the point (10,10). The value of (10,10) is 204

#### Work Step by Step

$f(x,y)=2xy+4$ The constraint becomes $x+y-20=0$ with $g(x,y)=x+y-20$ $F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=2xy+4-\lambda(x+y-20)=2xy+4-\lambda x -\lambda y + 20\lambda$ $F_{x}(x,y,\lambda)=2y-\lambda$ $F_{y}(x,y,\lambda)=2x-\lambda$ $F_{\lambda}(x,y,\lambda)=-x-y+20$ $(1) 2y-\lambda =0 \rightarrow \lambda =2y$ $(2) 2x-\lambda =0 \rightarrow \lambda =2x$ $(3) -x-y+20=0$ Set the expressions equal: $2y=2x$ $x=y$ Substitute for x=y in Equation (3): $-y-y+20=0$ $y=108 \rightarrow x=10$ The value of is f(10,10)=204 Let x=10.1 and y=9.9; then, f(10.1;9.9)=203.98 Because a nearby point has a value smaller than 204, the value 204 is probably not a minimum. f(x,y)=2xy+4 has a minimum value subject to the constraint x+y=20; it is at the point (10,10). The value of (10,10) is 204.

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.