NMTC 2015

Notice that $x^2+2xy+3y^2+2x+6y+4=(x^2+2xy+y^2)+(2x+2y)+(2y^2+4y)+2+1+1=(x+y)^2+2(x+y)+1+2(y+1)^2+1=(x+y+1)^2+2(y+1)^2+1$

So, minimum is when $x+y+1=y+1=0$ , thus $x=0, y=-1$ , with the expression equals $\boxed{1}$

By the way, do anyone know how to enter the equivalence vertically?

Use partial differentiation.

With respect to $x$ , we get,

$2x + 2y + 2 = 0$

With respect to $y$ , we get,

$2x + 6y + 6 = 0$

Solving both, $x = 0, y = -1$ .

Minimum value is of $f(x,y) = \boxed{1}$

I had a different approach-

* $\underline{WORKING}$ *

Expression is $x^{2} + 2\times{x}\times{(y+1)} + 3y^{2} + 6y + 4$ .

I considered the expression to be a quadratic in x then by applying the formula for min value = $\frac{4ac - b^{2}}{4a}$ ,where a is the coefficient of $x^{2}$ and b is the coefficient of x and c is the constant term, we obtain a quadratic in y .

Applying the formula for minimum value i.e. $\frac{4ac - b^{2}}{4a}$ we obtain that the minimum value of the expression is $\boxed{2y^{2} + 4y + 3}$ ...

Once again applying the formula on the above mentioned quadratic gives that the minimum value of the expression is $\boxed{1}$ ...

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