Answer
$f_{max}(\frac{80}{3},\frac{160}{3})=37926$
$x=\frac{80}{3}, y=\frac{160}{3}$
Work Step by Step
Let our positive numbers be x, y. We know that
$x + y =80 \rightarrow g(x)=x+y-80$
Our objective function is $f(x,y)=x^{2}y$
$F(x,Y,λ)=f(x,y)-\lambda.g(x,y)=x^{2}y-\lambda (x+y-80)=x^{2}y -\lambda x -\lambda y +80\lambda$
$F_{x}(x,y,\lambda)=2xy-\lambda$
$F_{y}(x,y,\lambda)=x^{2}-\lambda$
$F_{\lambda}(x,y,\lambda)=-x-y+80$
$(1) 2xy-\lambda=0 \rightarrow \lambda =2xy$
$(2) x^{2}-\lambda =0 \rightarrow \lambda =x^{2}$
$(3) -x-y+80=0$
Set the expressions equal:
$x^{2}=2xy$
$x=2y$
Substitute for x=2y in Equation (3)
$-2y-y+80=0$
$y=\frac{80}{3} \rightarrow x=\frac{160}{3}$
The value of f is $f(\frac{80}{3},\frac{160}{3})\approx37926$
Let x=26 and y=54 then $f(26,54)= 36504$. Because a nearby point has a value smaller than 37926, 37926 is probably not a minimum.
Therefore $f_{max}(\frac{80}{3},\frac{160}{3})=37926$
