# extreme values

• Oct 6th 2009, 11:52 AM
alex83
extreme values
hi there

in the shown example (attached ) picture i dont understand the circled phrases in red

• what is fxy ? what did we derive to get 1
• why is it local maximum??
• what is the saddle point in this example

i appreciate your help
• Oct 6th 2009, 12:22 PM
Matt Westwood
$\displaystyle f_{xy}$ is $\displaystyle \frac {\partial^2 f}{\partial x \partial y}$.

That is, you differentiate f partially with respect to x, and then differentiate that partially with respect to y.

When you plug x = -2, y = -2 into the given equation for f, you get 8.
• Oct 6th 2009, 07:17 PM
alex83
Quote:

Originally Posted by Matt Westwood
$\displaystyle f_{xy}$ is $\displaystyle \frac {\partial^2 f}{\partial x \partial y}$.

That is, you differentiate f partially with respect to x, and then differentiate that partially with respect to y.

thank u i appreciate ur help but i still didnt get it how to derive with respect to (x and y) at the same time.
i got all the other parts (with respect to x then with respect to y).
if i consider both x and y constants i should get f(xy)= 4
is it possible to show me how did we get (1) in that part
thanks
• Oct 6th 2009, 09:30 PM
Matt Westwood
Quote:

Originally Posted by alex83
thank u i appreciate ur help but i still didnt get it how to derive with respect to (x and y) at the same time.
i got all the other parts (with respect to x then with respect to y).
if i consider both x and y constants i should get f(xy)= 4
is it possible to show me how did we get (1) in that part
thanks

You have your equation (which I can't see in front of me because you didn't type it in just copied the page in and I can't see it when I'm editing it).

Partially differentiate it with respect to $\displaystyle x$ (partially) to get the expression for $\displaystyle f_x$. That is, $\displaystyle f_x = \frac {\partial f}{\partial x}$. It's just a more compact notation for it.

Then differentiate that partially with respect to $\displaystyle y$ to get $\displaystyle f_{xy}$.

That is, $\displaystyle f_{xy}$ is $\displaystyle (f_x)_y$ or $\displaystyle \frac {\partial}{\partiial y} \left({\frac {\partial f}{\partial x}}\right)$.

And when you work it out (they've done $\displaystyle f_x$ for you) you get $\displaystyle f_{xy} = 1$.