Answer
$f(x,y)=xy^{2}$ subject to constraint $g(x,y)=x+2y-60$ has a maximum value of 8000 when x=20, y=20.
Work Step by Step
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:
$(x,y)=x+2y=60 \rightarrow g(x,y)=x+2y-60$
$F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=xy^{2}-\lambda(x+2y-60)=xy^{2}-\lambda x -2 \lambda y + 60\lambda$
$F_{x}(x,y,\lambda)=y^{2}-\lambda$
$F_{y}(x,y,\lambda)=2xy-2\lambda$
$F_{\lambda}(x,y,\lambda)=-x-2y+60$
$(1) y^{2}-\lambda =0 \rightarrow \lambda =y^{2}$
$(2) 2xy-2\lambda =0 \rightarrow \lambda =xy$
$(3) -x-2y+60=0$
Set the expressions equal:
$y^{2}=xy$
$x=y$
Substitute for x=y in Equation (3)
$-y-2y+60=0$
$y=20 \rightarrow x=20$
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.
