algebrA

Hi! 

One way: (Using differentiation).

 

Find the minimum of:          \(x^2+2xy+y^2+4y+1\)

Let \(F(x,y)=x^2+2xy+y^2+4y+1\)

\(F_{x}=2x+2y\)             (Partial derivative with respect to x)

\(F_{y}=2x+4y+4\)      (Partial derivative with respect to y)

We set these equal to zero:

\(x+y=0\)

\(x+2y=-2\)

Solve the system of two equations:    (By subtracting the first equation from the second.)

\(y=-2 \implies x=2\)

Therefore, the minimum occurs when: \(x=2,y=-2\)

The minimum is: \((2)^2+2(2)(-2)+2(-2)^2+4(-2)+1=4-8+8-8+1=-3\)

 

So the minimum value is -3.

 

Second way: (Uses completing the square and the fact that \(a^2\ge 0\) for any real a.)

\(x^2+2xy+2y^2+4y+1\)

Consider: \((x^2+2xy+y^2)+(y^2+4y+1)\) , complete the square on each bracket:

\((x+y)^2+y^2+4y+1=(x+y)^2+(y+2)^2-3\)

Notice: \((x+y)^2 \ge 0 , (y+2)^2 \ge 0\) for all x and y.

So,

\((x+y)^2+(y+2)^2\ge 0 \iff (x+y)^2+(y+2)^2-3 \ge -3\)

The L.H.S is what we want to find the minimum of.

Therefore, the minimum of the desired expression is -3.

Hope this helps!