How can I show that the global minimum of $4x^2-12x+y^2+2y+10$ is $0$ at $(3/2,-1)$?
I know that at $(3/2,-1)$ is a critical point and I can display that it's not a maximum. But I cannot figure out whether this is enough to conclude it being a minimum or whether I need to show some inequalities of the form $f(x,y)>f(3/2,-1)$ (which I have difficulties discovering).
On
Let $$f(x,y) = 4\bigg[x^2-3x+\left(\frac{3}{2}\right)^2\bigg]+(y^2+2y+1)+10-4\cdot \left(\frac{3}{2}\right)^2-1$$
So $$f(x,y) = 4\left(x-\frac{3}{2}\right)^2+(y+1)^2+10-10\geq 0$$
and equality hold when $$x=\frac{3}{2}\;\;,y=-1$$
On
The function is strictly-convex which you can prove using the definition:
$f(\lambda x + (1-\lambda)y) < \lambda f(x) + (1-\lambda)f(y)$. Such functions have only a unique minimum, and indeed the point you have is a unique minimum.
In general distinguishing minima/maxima and saddle points can be done by looking at first and second derivatives. If $f'(x,y) = 0$, and $f''(x,y)$ has positive eigenvalues it is a minimum, where prime represent derivative w.r.t all variables e.g. (x,y). And double prime stands for Hessian matrix whose elements are second derivatives w.r.t. to variables. If it has negative eigenvalues is a maximum, and everything else is a saddle.
On
If you factorise 4X^2 − 12X you get
(2X−3)^2.
And If you factorise Y^2+2y+1 you get
(Y+1)^2.
Since the highest power is 2 we know that each one is a quadratic equation. Therefore you can make each equation equal to 0 and solve for X and Y.
(2X-3)^2 = 0 Therefore, X = 3/2.
(Y+1)^2 = 0 Therefore, Y = -1.
Therefore (3/2, -1)
Note that $4x^2-12x+y^2+2y+10=(2x-3)^2+(y+1)^2$.