## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 9 - Multiveriable Calculus - 9.4 Lagrange Multipliers - 9.4 Exercises - Page 498: 26

#### Answer

Function $f(x,y)=x^{3}y^{4}$ subject to constraint $g(x,y)=3x+3y-42$ has a maximum value of 884736 when x=6, y=8.

#### Work Step by Step

We are given $f(x,y)=x^{3}y^{4}$ where a unit of x cost is $\$3$, a unit of y cost is$\$3$ and $\$42$is available, so that the combination of x and y is represented by the point$(x,y)=3x+3y=42 \rightarrow g(x,y)=3x+3y-42F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=x^{3}y^{4}-\lambda(3x+3y-42)=x^{3}y^{4}-3\lambda x - 3\lambda y + 42\lambdaF_{x}(x,y,\lambda)=3x^{2}y^{4}-3\lambdaF_{y}(x,y,\lambda)=4x^{3}y^{3}-3\lambdaF_{\lambda}(x,y,\lambda)=-3x-3y+42(1) 3x^{2}y^{4}-3\lambda=0 \rightarrow \lambda =x^{2}y^{4}(2) 4x^{3}y^{3}-3\lambda =0 \rightarrow \lambda =\frac{4x^{3}y^{3}}{3}(3) -3x-3y+42=0$Set the expressions equal:$x^{2}y^{4}=\frac{4x^{3}y^{3}}{3}4x^{3}y^{3}=3x^{2}y^{4}x=\frac{3}{4}y$Substitute for$x=\frac{3}{4}y$in Equation (3)$-3(\frac{3}{4}y)-3y+42=0y=8 \rightarrow x=6$The value of is$f(6,8)=884736$Let x=5 and y=9 then$f(5;9)=820125$. Because a nearby point has a value smaller than 884736, the value 884736 is probably not a minimum. Function$f(x,y)=x^{3}y^{4}$subject to constraint$g(x,y)=3x+3y-42\$ has a maximum value of 884736 when x=6, y=8.

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