Answer
$f(x,y)=x^{2}+2y^{2}-xy$ has a minimum value subject to the constraint x+y=8, it is at the point (5,3). The value of (5,3) is 28.
Work Step by Step
$f(x,y)=x^{2}+2y^{2}-xy$
The constraint becomes $x+y-8=0$ with $g(x,y)=x+y-8$
$F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=x^{2}+2y^{2}-xy-\lambda(x+y-8)=x^{2}+2y^{2}-xy-\lambda x -\lambda y + 8\lambda$
$F_{x}(x,y,\lambda)=2x-y-\lambda$
$F_{y}(x,y,\lambda)=4y-x-\lambda$
$F_{\lambda}(x,y,\lambda)=-x-y+8$
$(1) 2x-y-\lambda =0 \rightarrow \lambda =2x-y$
$(2) 4y-x-\lambda =0 \rightarrow \lambda =4y-x$
$(3) -x-y+8=0$
Set the expressions equal
$2x-y=4y-x$
$x=\frac{5}{3}y$
Substitute for $x=\frac{5}{3}y$ in Equation (3)
$-\frac{5}{3}y-y+8=0$
$y=3 \rightarrow x=5$
The value of f is f(5,3)=28
Let y=3.1 and x=4.9 then f(4.9,3.1)=28.04 which is greater than 28.
Because a nearby point has a value greater than 28, the value 28 is probably not a maximum.
