## Calculus with Applications (10th Edition)

$f(x,y)=x^{2}-10y^{2}$ has a maximum value subject to the constraint $x-y=18$; it is at the point (20,2). The value of (20,2) is 360
$f(x,y)=x^{2}-10y^{2}$ The constraint becomes $x-y-18=0$ with $g(x,y)=x-y-18$ $F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=x^{2}-10y^{2}-\lambda(x-y-18)=x^{2}-10y^{2}-\lambda x + \lambda y + 18\lambda$ $F_{x}(x,y,\lambda)=2x-\lambda$ $F_{y}(x,y,\lambda)=-20y+\lambda$ $F_{\lambda}(x,y,\lambda)=-x+y+18$ $(1) 2x-\lambda =0 \rightarrow \lambda =2x$ $(2) -20y+\lambda =0 \rightarrow \lambda =20y$ $(3) -x+y+18=0$ Set the expressions equal: $2x=20y$ $x=10y$ Substitute for x=y in Equation (3) $-10y+y+18=0$ $y=2 \rightarrow x=20$ The value of f is $f(20,2)=360$ Let x=19.9 and y=1.9 then $f(19.9;1.9)=259.91$. Because a nearby point has a value smaller than 360, the value 360 is probably not a minimum. $f(x,y)=x^{2}-10y^{2}$ has a maximum value subject to the constraint x-y=18; it is at the point (20,2). The value of (20,2) is 360.