New Year Countdown - 5

In this question, $x$ and $y$ are independent.

So we can partially differentiate it.

Let $f(x,y)=x^2+2xy+3y^2-6x-2y$

$\dfrac{\partial f(x,y)}{\partial x}=2x+2y-6=0\Rightarrow x+y=3$

$\dfrac{\partial f(x,y)}{\partial y}=2x+6y-2=0\Rightarrow x+3y=1$

Solving both equations, $x=4$ and $y=-1$ .

Substitute $x$ and $y$ to get minima which is $\boxed{-11}$

Solved by -b/2a :D

The given expression can be rewritten as $(x+y-3)^2+2(y+1)^2-11$ And so the minimum value is $-11$ .

min{x^2+2 x y+3 y^2-6 x-2 y} = -11 at (x, y) = (4, -1)