Answer
Function $f(x,y)=x^{2}y^{3}$ subject to constraint $g(x,y)=2x+y-80$ has a maximum value of 28311552 when x=16, y=48
Work Step by Step
We are given $f(x,y)=x^{2}y^{3}$
where a unit of x cost is $\$2$, a unit of cost y is $\$1$ and $\$80$ is available, so that the combination of x and y are represented by the point $(x,y)=2x+y=80 \rightarrow g(x,y)=2x+y-80$
$F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=x^{2}y^{3}-\lambda(2x+y-80)=x^{2}y^{3}-2\lambda x - \lambda y + 80\lambda$
$F_{x}(x,y,\lambda)=2xy^{3}-2\lambda$
$F_{y}(x,y,\lambda)=3x^{2}y^{2}-\lambda$
$F_{\lambda}(x,y,\lambda)=-2x-y+80$
$(1) 2xy^{3}-2\lambda=0 \rightarrow \lambda =xy^{3}$
$(2) 3x^{2}y^{2}-\lambda =0 \rightarrow \lambda =3x^{2}y^{2}$
$(3) -2x-y+80=0$
Set the expressions equal: $xy^{3}=3x^{2}y^{2}$
$y=3x$
Substitute for x=y in Equation (3) $-2x-3x+80=0$
$x=16 \rightarrow y=48$
The value of is $f(16,48)=28311552$
Let x=15 and y=30 then $f(15,30)=6075000$. Because a nearby point has a value smaller than 28311552, the value 28311552 is probably not a minimum.
Function $f(x,y)=x^{2}y^{3}$ subject to constraint $g(x,y)=2x+y-80$ has a maximum value of 28311552 when x=16, y=48.
