# Thread: Local Maxima or Minima

1. ## Local Maxima or Minima

I need to determine whether the fct has any local minimums and/or maximums.

$f(x,y)=x^2-2xy+y^2$

I optimized and found that the f.o.c gives: x=y (right?). Does that mean that at ALL constants, c, where x=y=c, we have a local minima? Or does this mean that the function doesn't have any optimal values?

tks

2. First of all, note that $x^2 - 2 x y + y^2 = (x-y)^2$. As you should see, the function cannot obtain negative values. Setting $x=y$ gives a function value of 0, which obviously is a minimum for all $x=y$. You might also want to have a look at the second plot here: http://www.wolframalpha.com/input/?i=x^2-2+x+y+%2B+y^2

3. Originally Posted by altave86
I need to determine whether the fct has any local minimums and/or maximums.

$f(x,y)=x^2-2xy+y^2$

I optimized and found that the f.o.c gives: x=y (right?). Does that mean that at ALL constants, c, where x=y=c, we have a local minima? Or does this mean that the function doesn't have any optimal values?

tks
Practically speaking, this problem ist a joke: since $f(x,y)=x^2-2xy+y^2=(x-y)^2\geq 0$ you can immediately say that f assumes its global minimum 0 at all points $(c,c),\;c\in\mathbb{R}$.

Speaking more theoretically: with $x=y$ you have only found a necessary condition for f assuming a locally extreme value at all points $(c,c),\;c\in\mathbb{R}$. Now, in principle, you would have to examine the behavior of the second derivative.

4. Originally Posted by Failure
this problem ist a joke
Got you! Another German speaker here. Can't remember how often that happened to me.