Answer
$f(x,y)=xy^{2}$ has a maximum value subject to the constraint x+2y=15; it is at the point (5,5). The value of (5,5) is 125.
Work Step by Step
$f(x,y)=xy^{2}$
The constraint becomes $x+2y-15=0$ with $g(x,y)=x+y-15$
$F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=xy^{2}-\lambda(x+2y-15)=xy^{2}-\lambda x -2 \lambda y + 15\lambda$
$F_{x}(x,y,\lambda)=y^{2}-\lambda$
$F_{y}(x,y,\lambda)=2xy-2\lambda$
$F_{\lambda}(x,y,\lambda)=-x-2y+15$
$(1) y^{2}-\lambda =0 \rightarrow \lambda =y^{2}$
$(2) 2xy-2\lambda =0 \rightarrow \lambda =xy$
$(3) -x-2y+15=0$
Set the expressions equal:
$y^{2}=xy$
$x=y$
Substitute for x=y in Equation (3)
$-y-2y+15=0$
$y=5 \rightarrow x=5$
The value of is f(5,5)=125
Let x=3.9 and y=5.55 then f(3.9;5.55)=120.13
Because a nearby point has a value smaller than 125, the value 125 is probably not a minimum.
$f(x,y)=xy^{2}$ has a maximum value subject to the constraint x+2y=15; it is at the point (5,5). The value of (5,5) is 125.
