Let $A,B,C,D \in \mathbb{R}$, and let $A>0$, $B^2 <AC$.
We define $f(x,y)=Ax^2 +2Bxy+Cy^2+D$.
Find local maximum and minimum of $f$.
Is there a global minimum of $f$ in the plane?
Note: I am not sure if the "in the plane" was a mistake in the question or not, since we're talking about a function in 3d space, but I wrote it here because it might mean something like finding a minimum on the $x,y$ plane and I'm just not getting it.
My Attempt:
$f_x=2Ax+2By=0 \Longrightarrow 2Ax=-2By \Longrightarrow (A>0) \Longrightarrow x = -\frac{By}{A}$.
$f_y= 2Cy+2Bx=0$
I substituted $x$ into the second equation:
$2Cy+2B(-\frac{By}{A})=0 \Longrightarrow 2Cy – \frac{2B^2y}{A}=0\Longrightarrow 2y(C-\frac{B^2}{A})=0$.
Since $AC>B^2$, $C=\frac{B^2}{A}$ is not possible, so I got a point $(0,0)$.
To check if it's a Local Min or Max, I computed the second partial derivatives:
$f_{xx}=2A$
$f_{yy}=2C$
$f_{xy}=2B$
Now since $f_{xx}*f_{yy}-f_{xy}^2=4AC-4B^2=4(AC-B^2) > 0 \Longrightarrow $ Then $(0,0)$ is the local minimum of the function.
Now I'm not sure how to deal with the second part, how do I check if the function has a global minimum or not?
Any feedback and help is really appreciated, Thanks in advance.
Best Answer
This function is convex, all local minimum is a global minimum. $(0,0)$ is the global minimum.
We can also prove it by completing the squares.
\begin{align}Ax^2+2Bxy+Cy^2 +D &= A \left(x+\frac{By}{A} \right)^2+Cy^2-A\left(\frac{By}{A} \right)^2+D \\ &=A \left(x+\frac{By}{A} \right)^2+\left(C-\frac{B^2}{A} \right)y^2+D \\&=A \left(x+\frac{By}{A} \right)^2+\left(\frac{AC-B^2}{A} \right)y^2+D \\ &\ge D\end{align}
Hence $D$ is the lower bound which can only be attained when $y=0$ and $x=0$.