The least possible value for $x^2-2xy+2y^2-6y$ If $x,y$ are real numbers .How to find the least possible value for $x^2-2xy+2y^2-6y$
 A: Hint: continue so that it stays true:
$$x^2-2xy+2y^2-6y=(x-y)^2+(y-3)^2-\dots$$
A: This method works for differentiable functions of one or more real variables:


*

*Check whether the function has a minimum. In this case it does,
but if instead of $-2xy$ you have $-3xy$, the formula doesn't have a minimum.
(someone improve this answer by telling how to do this please)

*Find the derivates:

*

*$\cfrac{\mathrm{d}f}{\mathrm{d}x} = 2 x - 2 y$

*$\cfrac{\mathrm{d}f}{\mathrm{d}y} = -6 - 2 x + 4 y$


*Set the derivatives equal to zero and solve the system of equations:


*

*$2x - 2y = 0 \quad\implies\quad x = y$

*$-6 - 2 x + 4 y = 0 \quad\implies\quad -6 + 2x = 0 \quad\implies\quad x = 3$


In this case only one solution is found, but in the general case there could
be more solutions, even infinitely many. The solutions are called critical points.

*For all solutions found, check which one yields the smallest number.
Since we have only one solution, this step can be skipped.


So the minimum of $x^2 -2xy+2y^2-6y$ is $-9$ at $x = 3, y=3$.
