Find min.

$\begin{aligned} f(x,y) & = x^2-2xy+2y^2 + 2x - 6y & \small \blue{\text{Note that }(x-y+1)^2 = x^2 + y^2 + 1 + 2(-xy-y+x)} \\ & = (x-y+1)^2 + y^2 - 4y - 1 & \small \blue{\text{and }(y-2)^2 = y^2-4y+4} \\ & = ((x-1)-(y-2))^2 + (y-2)^2 - 5 \end{aligned}$

Note that $g(x,y) = ((x-1)-(y-2))^2 + (y-2)^2 \ge 0$ . Therefore, $f(x,y)$ is minimum when $g(x,y)$ is minimum or $g(x,y) = 0$ , $\implies \min(f(x,y) = f(1,2) = 0 - 5 = \boxed{-5}$ .

$f(x,y)=x^2-2xy+2y^2+2x-6y$

The minimum values of $x$ and $y$ will satisfy $f_x=0$ and $f_y=0$ , or $\frac{df}{dx}=0$ and $\frac{df}{dy}=0$ .

$\frac{df}{dx} = 2x-2y+2 = 0$

$\rightarrow 2y=2x+2$

$\rightarrow y=x+1$

Simplifying the above equation gives that $y=x+1$ , so we can now plug that value into $y$ in the second equation to solve for $x$ .

$\frac{df}{dy} = -2x+4y-6 = 0$

$\rightarrow -2x+4(x+1)-6=0$

$\rightarrow 4x+4-2x-6=0$

$\rightarrow 2x=2$

$\rightarrow x=1$

So now that we know that $x=1$ , we can plug that into $y=x+1$ to solve for $y$ .

This gives that $y=2$ and $x=1$ as the minimum values of $x$ and $y$ .

Finally, to solve for the minimum value of $f(x,y)$ , we just need to plug the values above into the function.

$f(x,y)=(1)^2-2(1)(2)+2(2)^2+2(1)-6(2)$

$\rightarrow 1-4+8+2-12 = -5$

So, the answer is -5