Answer
$f_{max}(25,50)=62500$
$x=25,y=50$
Work Step by Step
Let our positive numbers be x, y. We know that
$x + y =75 \rightarrow g(x)=x+y-75$
Our objective function is $f(x,y)=xy^{2}$
$F(x,Y,λ)=f(x,y)-\lambda.g(x,y)=xy^{2}-\lambda (x+y-75)=xy^{2} -\lambda x -\lambda y +75\lambda$
$F_{x}(x,y,\lambda)=y^{2}-\lambda$
$F_{y}(x,y,\lambda)=2xy-\lambda$
$F_{\lambda}(x,y,\lambda)=-x-y+75$
$(1) y^{2}-\lambda=0 \rightarrow \lambda =y^{2}$
$(2) 2xy-\lambda =0 \rightarrow \lambda =2xy$
$(3) -x-y+75=0$
Set the expressions equal:
$y^{2}=2xy$
$y=2x$
Substitute for y=2x in Equation (3)
$-x-2x+75=0$
$x=25 \rightarrow y=50$
The value of f is $f(25,50)=62500$
Let x=24 and y=51 then $f(24,51)= 62424$. Because a nearby point has a value smaller than 62500, 62500 is probably not a minimum.
Therefore $f_{max}(25,50)=62500$
