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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

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2 Answers 2

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.

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Sorry, I missed the modulus. If $a$ and $b$ have different signs you should be able to find a zero. Likewise with $a$ and $b$ positive and the constant negative. –  Mark Bennet Mar 28 '12 at 7:24
    
what if zero does not exist??to proceed what we can do is to first we check the minimum value of f(x,y), if it is positive then we are through, if it is negative then we find the minimum value of -f(x,y)? –  Abhinav Mar 28 '12 at 7:39
    
also $a,b,c,d,e$ are real and are not conditioned to be + or -. Is there any general way to deal with minimum of a function. Also $|f(x,y)|=max(f(x,y),-f(x,y))$..can we use it somehow? –  Abhinav Mar 28 '12 at 7:41

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.
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Thanks @martini. –  Abhinav Mar 28 '12 at 9:18

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