Finding minimum of the modulus of a 2 variable function

Question

How to find the minimum value of $$|f(x,y)|$$ where $f(x,y)$ is a 2nd degree function in x and y with no 'xy' term. $$f(x,y)=ax^2+by^2+cx+dy+e$$ How is the process different from finding the minimum of a function without the modulus operator? See a related post: Finding minimum of a two variable 2nd degree function under a certain constraint?

Any help would be beneficial

Answer 1

You can write your expression as a sum of squares plus a constant.

$f(x,y) = a(x+\frac c {2a})^2+b(y+\frac d {2b})^2+$ [work it out for yourself]

For a well-defined minimum you need $a$ and $b$ positive - and you should then be able to see the minimum.

Answer 2

Concernig the modulus: We suppose $a,b \ne 0$. There are two possibilities. Either $f$ has a zero and the minimum of $|f|$ is 0 or $f$ doesn't change signs and the minimum of $|f|$ is the minimum resp. maximum of $f$ if $f$ is negative resp. positive.

In your case, if you write $f$ in the form @Mark suggested $$ f(x,y) = a\biggl(x + \frac c{2a}\biggr)^2 + b\biggl(y + \frac{d}{2b}\biggr)^2 + C $$ ($C$ to be computed) we have the following cases:

1. $a$, $b > 0$ and $C \ge 0$. Then $f \ge 0$ and therefore $|f| = f$ attains its minimum at $(-\frac c{2a}, -\frac d{2b})$.
2. $a$, $b < 0$ and $C \le 0$. Then $f \le 0$ and $|f| = -f$ attains its minimum at $(-\frac c{2a}, -\frac d{2b})$.
3. $a > 0$, $b < 0$. Then $f$ attains positive and negative values, as for example $f(x,0) \to \infty$, $x \to \infty$ but $f(0,y) \to -\infty$, $y \to \infty$. Therefore by connectedness of $f(\mathbb R^2)$ $f$ has a zero and $|f|$'s minimum is 0.
4. $a < 0$, $b > 0$. As in case 3.
5. $a, b > 0$, $C < 0$. Then $f(-\frac c{2a}, -\frac d{2b}) = C < 0$ but $f(x,0) \to \infty$, $x \to \infty$. Therefore $f$ has a zero.
6. $a,b < 0$, $C > 0$. As in case 5, $f$ has a zero.