Local minimum and maximum and saddle points problem

**noppawit** · Jan 28th 2009, 02:38 AM

I'm wondering that I did this problem correct, or not.

Find the local maximum and minimum values and saddle point(s) of the function $f(x,y) = 1+2xy-x^2-y^2$
$\nabla f = 0$
$\nabla f =<\frac{\partial f }{\partial x },\frac{\partial f }{\partial y }>=<2y-2x,2x-2y>$

Then, I got x=y. After that, I tried to find $D = fxx*fyy-f^2xy$
I've got 0 from D. Therefore, I concluded D=0, no conclusion.

Am I right? Is there anything wrong?

Thank you.

**Mush** · Jan 28th 2009, 03:03 AM

Originally Posted by noppawit

I'm wondering that I did this problem correct, or not.

Find the local maximum and minimum values and saddle point(s) of the function [tex]f(x,y) = 1-x^2-y^2[/+2xymath]
$\nabla f = 0$
$\nabla f =<\frac{\partial f }{\partial x },\frac{\partial f }{\partial y }>=<2y-2x,2x-2y>$

Then, I got x=y. After that, I tried to find $D = fxx*fyy-f^2xy$
I've got 0 from D. Therefore, I concluded D=0, no conclusion.

Am I right? Is there anything wrong?

Thank you.

If $D = 0$ , then examine the function:

$\Delta(h,k) = f(a+h,b+k)-f(a,b)$

If $\Delta(h,k) > 0 \forall h,k$ , then you have a local minimum.

If $\Delta(h,k) < 0 \forall h,k$ , then you have a local maximum.

If $\Delta(h,k) < 0$ for SOME h,k, and $\Delta(h,k) > 0$ , for OTHER h,k then you have a saddle point.

$\Delta(h,k) = 1+2(a+h)(b+k) -(a+h)^2-(b+k)^2 - (1-a^2-b^2)$

$\Delta(h,k) = 1+2(ab+ak+hb+hk)-(a^2 + 2ah +h^2)-(b^2+2bk+k^2) - 1+a^2+b^2$

$\Delta(h,k) = 2ab+2ak+2hb+2hk - 2ah -h^2-2bk-k^2$

$\Delta(h,k) = 2(ab+ak+ah+hk+hb-bk-ah)-h^2-k^2$

$\Delta(h,k) = 2(ab+ak+hk+hb-bk)-h^2-k^2$

You must plug in a and b. These are the points where you find stationary points to be. After that you should be able to conclude that:

$\Delta(h,k)$ is > 0 for some h k, and < 0 for other. This indicates stationary points.

**Jester** · Jan 28th 2009, 05:32 AM

Originally Posted by noppawit

I'm wondering that I did this problem correct, or not.

Find the local maximum and minimum values and saddle point(s) of the function $f(x,y) = 1+2xy-x^2-y^2$
$\nabla f = 0$
$\nabla f =<\frac{\partial f }{\partial x },\frac{\partial f }{\partial y }>=<2y-2x,2x-2y>$

Then, I got x=y. After that, I tried to find $D = fxx*fyy-f^2xy$
I've got 0 from D. Therefore, I concluded D=0, no conclusion.

Am I right? Is there anything wrong?

Thank you.

You'll notice that your function can be written as

$f(x,y) = 1 - (x-y)^2$

and since the second term is always negative then

$f(x,y) = 1 - (x-y)^2 \le 1$

so the max if f is located on the line $y = x$ and is 1.

**Soroban** · Jan 28th 2009, 06:52 AM

Hello, noppawit!

I'd say your work and conclusion are correct.

Note that if $x = y$ , the function is: . $f(x,x) \:=\:1 + 2x^2 - x^2 - x^2 \:=\:1$

The function is constant above the line $y = x.$

So there is a horizontal tangent plane: $z = 1$

Evidently none of those points $(x,x,1)$ is a maximum/minimum/saddle point.
(Well, they could be ... but I'm not sure.)
.

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