Answer
$f(x,y)=x^{2}-10y^{2}$ 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
Work Step by Step
$f(x,y)=x^{2}-10y^{2}$
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.