## Calculus with Applications (10th Edition)

$f(x,y)=xy^{2}$ subject to constraint $g(x,y)=x+2y-60$ has a maximum value of 8000 when x=20, y=20.
We are given $f(x,y)=xy^{2}$ where a unit of x cost is $\$1$, a unit of cost y is$\$2$ and $\$60$is available. We will have the combination of x and y:$(x,y)=x+2y=60 \rightarrow g(x,y)=x+2y-60F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=xy^{2}-\lambda(x+2y-60)=xy^{2}-\lambda x -2 \lambda y + 60\lambdaF_{x}(x,y,\lambda)=y^{2}-\lambdaF_{y}(x,y,\lambda)=2xy-2\lambdaF_{\lambda}(x,y,\lambda)=-x-2y+60(1) y^{2}-\lambda =0 \rightarrow \lambda =y^{2}(2) 2xy-2\lambda =0 \rightarrow \lambda =xy(3) -x-2y+60=0$Set the expressions equal:$y^{2}=xyx=y$Substitute for x=y in Equation (3)$-y-2y+60=0y=20 \rightarrow x=20$The value of is$f(20,20)=8000$Let x=19.9 and y=20.5 then$f(19.9;20.05)=7999.85$. Because a nearby point has a value which is smaller than 8000, the value 8000 is probably not a minimum. Function$f(x,y)=xy^{2}$subject to constraint$g(x,y)=x+2y-60\$ has a maximum value of 8000 when x=20, y=20.