Answer
$f(x,y)=x^{2}y$ has a maximum value subject to the constraint x+y=4, it is at the point $f(\frac{8}{3} ,\frac{4}{3})$.
The value of $f(\frac{8}{3} ,\frac{4}{3})$ is $\frac{256}{27}$.
Work Step by Step
We are given $f(x,y)=x^{2}y$
The constraint becomes $x+y-4=0$ with $g(x,y)=x+y-4$
$F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=x^{2}y-\lambda(x+y-4)=x^{2}y-\lambda x -\lambda y + 4\lambda$
$F_{x}(x,y,\lambda)=2xy-\lambda$
$F_{y}(x,y,\lambda)=x^2-\lambda$
$F_{\lambda}(x,y,\lambda)=-x-y+4$
$(1) 2xy-\lambda =0 \rightarrow \lambda =2xy$
$(2) x^{2}-\lambda =0 \rightarrow \lambda =x^{2}$
$(3) -x-y+4=0$
Set the expressions equal $2xy=x^{2}$
$x=2y$
Substitute for $x=2y$ in Equation (3)
$-2y-y+4=0$
$y=\frac{4}{3} \rightarrow x=\frac{8}{3}$
The value of is $f(\frac{8}{3} ,\frac{4}{3})=\frac{256}{27}$
Let y=2.5 and x=1.5 then f(1.5,2.5)=5.625 which is smaller than $\frac{256}{27}$. Because a nearby point has a value smaller than $\frac{256}{27}$, the value $\frac{256}{27}$ is probably not a minimum.
$f(x,y)=x^{2}y$ has a maximum value subject to the constraint x+y=4, it is at the point $f(\frac{8}{3} ,\frac{4}{3})$. The value of $f(\frac{8}{3} ,\frac{4}{3})$ is $\frac{256}{27}$.
