Factorisation with 2 variables

[GregA]

Is there a way to find the lowest possible value of the following expression:
$$x^2 + 4xy + 5y^2 - 4x - 6y +7$$ without using calculus?

I've just got myself a different textbook and am working from the beginning, trying to use only the tools given to me thus far in the book (within reason).

One way to solve it is to complete the squares for x and y to get
$(x-2)^2+5(y- \frac{3}{5})^2 +4xy + \frac {6}{5}$ and then find the value of 4xy which keeps the sum of both squares and itself as small as possible..ie differentiate

Problem is the book has not not even touched upon calculus yet and so I suspect there is a more elementary approach that I have either missed or am not even aware of and this is bugging me ...The furthest I got was the factorisation above and figuring out that 4xy should be negative and its absolute value being as high as possible so that y < 0 whilst x > 2 ... but I still cannot find a basic method of finding the precise values of x and y. Anyone have any suggestions?

 

[Hurkyl]

Try getting rid of the xy term first... by completing the square!

(Hint: instead of looking at (x-a)², try looking at (x-ay)²)

 

[GregA]

Thanks for the reply Hurkyl

By completing the square to eliminate xy I get $$(x+2y)^2 +y^2-6y-4x+7$$...I can go further and wrap up y to get $$(x+2y)^2 + (y-3)^2 -4x - 2$$ but I cannot see any useful clues...(apart from what I discovered from the other factorisation), nor can I see a way to force exact values upon x or y

 

[Hurkyl]

You can still complete the square to get rid of the x term. (Doing so will introduce another y term, but you already know how to get rid of that)

 

[GregA]

Aha!

$$x^2 + 4xy + 5y^2 -4x -6y + 7 =$$
$$(x +2y -2)^2 + y^2 +2y +3 =$$
$$(x +2y -2)^2 +(y+1)^2 +2$$

Thanks for steering me in the right direction Hurkyl!