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!
!
