## Calculus with Applications (10th Edition)

$f(x,y)=x^{2}+y^{2}$ has a minimum value subject to the constraint x=y-6, it is at the point $f(-3,3)$. The value of $f(-3,3)$ is $18$.
We are given $f(x,y)=x^{2}+y^{2}$ The constraint becomes $x-y+6=0$ with $g(x,y)=x-y+6$ $F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=x^{2}+y^{2}-\lambda(x-y+6)=x^{2}+y^{2}-\lambda x +\lambda y - 6\lambda$ $F_{x}(x,y,\lambda)=2x-\lambda$ $F_{y}(x,y,\lambda)=2y+\lambda$ $F_{\lambda}(x,y,\lambda)=-x+y-6$ $(1) 2x-\lambda =0 \rightarrow \lambda =2x$ $(2) 2y+\lambda =0 \rightarrow \lambda =-2y$ $(3) -x+y-6=0$ Set the expressions equal $2x=-2y$ $x=-y$ Substitute for $x=-y$ in Equation (3) $y+y-6=0$ $2y=6$ $y=3 \rightarrow x=-3$ The value of is $f(-3,3)=18$ Let y=2.9 and x=-3.1 then f(-3.1,2.9)=18.02 which is greater than $18$. Because a nearby point has a value greater than $18$, the value $18$ is probably not a maximum. $f(x,y)=x^{2}+y^{2}$ has a minimum value subject to the constraint x=y-6, it is at the point $f(-3,3)$. The value of $f(-3,3)$ is $18$.