Least value of the expression.

Our given expression is $x^{2}+2xy+2y^{2}+4y+7$ .

We can write it as $x^{2}+2xy+y^{2}+y^{2}+4y+4+3$ .

Now using Completing The Square Method, we get:

$(x+y)^{2}+(y+2)^{2}+3$ which is the expression to be minimized.

As, $x$ and $y$ are Real Numbers, we know that the square of a real number is non-negative and its minimum is $0$ .

So, the strategy we should adopt is to minimize the square terms to $0$ to achieve the minimum as $3$ which is our constant in our expression.

The minimum occurs at $x=2$ and $y=-2$ and our required answer is $\color{#69047E}{\huge{3}}$ .

I have a different approach : Partial Differentiation

Differentiate w.r.t $x$ to get : $2x + 2y = 0 \Rightarrow x = -y$

Differentiate w.r.t $y$ to get : $2x + 4y + 4 = 0 \Rightarrow y = -2 , x = 2$

Substitute these values,

$f_{min} (x,y) = \boxed{3}$

$x^2+2xy+2y^2+4y+7$

$x^2+2xy+y^2+y^2+4y+4+3$

$(x+y)^2+(y+2)^2+3$

We can minimize $(x+y)^2$ and $(y+2)^2$ to $0$ , and so the minimum value will be $3$

x^2+2xy+2y^2+4y+7 = (x+y)^2+(y+2)^2 + 3 As we know, the square of any real number is non-negative so its minimum is 0. By this, we can conclude the minimum of the above expression is 3, when (x+y)^2 and (y+2)^2 are equal 0.

completing the square will get us, (x+y)^2 + (y+2)^2 + 3 =0. since (x+y)^2 and (y+2)^2 can only be positive, to obtain the least we have to equate both to zero. y+2=0 gives y = -2 and x = 2... with both terms zero, the least value then becomes 3

You can rewrite it as (x+y)^2 + (y+2)^2 +3 The min is clearly 3