## Calculus with Applications (10th Edition)

$f(x,y)=xy^{2}$ 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.
$f(x,y)=xy^{2}$ 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.