# Partial Derivatives

• Mar 25th 2009, 11:00 AM
Hockey_Guy14
Partial Derivatives
Im having some trouble finding the critical points and relative max/min/saddle points for some questions

1. f(x,y) = 7x^2 + 4y^2
I found the critical point to be (0,0), but Im not sure how to calculate for min/max or saddle point

2. f(x,y) = e^1/2xy
Im having trouble finding the critical points and determining whether the points are max/min or saddle

3. f(x,y) = x^4 - 8xy + 2y^2 - 3
I think the critical points are (0,0), (2,4) and (-2,-4) but im not sure, and again im having trouble determining whether the points are a max/min or saddle point.

Any help or clarification would be much appreciated. Thanks!
• Mar 25th 2009, 11:17 AM
thisisstillme
You need the 2nd partial derivative test. Cant be bothered to type it all out, just look at the wikipedia entry for 'second partial derivative test'.
• Mar 25th 2009, 11:17 AM
Chris L T521
Quote:

Originally Posted by Hockey_Guy14
Im having some trouble finding the critical points and relative max/min/saddle points for some questions

1. f(x,y) = 7x^2 + 4y^2
I found the critical point to be (0,0), but Im not sure how to calculate for min/max or saddle point

You need to apply the second partials test (Note that $\left(x_0,y_0\right)$ is the critical point):

$D=f_{xx}\!\left(x_0,y_0\right)f_{yy}\!\left(x_0,y_ 0\right)-f_{xy}^2\!\left(x_0,y_0\right)$

Now, if:

$D>0$ and $f_{xx}\!\left(x_0,y_0\right)>0$, then there is a minimum.

$D>0$ and $f_{xx}\!\left(x_0,y_0\right)<0$, then there is a maximum.

$D<0$, then there is a saddle point.

$D=0$, then no conclusion can be drawn.

In your case, $f_{xx}\!\left(x,y\right)=14$, $f_{yy}\!\left(x,y\right)=8$ and $f_{xy}\!\left(x,y\right)=0$

So what can you conclude?

Quote:

2. f(x,y) = e^1/2xy
Im having trouble finding the critical points and determining whether the points are max/min or saddle
If I'm interpreting this right, the function is $f\!\left(x,y\right)=e^{\frac{1}{2}xy}$

Thus, $f_x\!\left(x,y\right)=\tfrac{1}{2}ye^{\frac{1}{2}x y}$ and $f_y\!\left(x,y\right)=\tfrac{1}{2}xe^{\frac{1}{2}x y}$

Can you find the critical points now and determine whether there are max/min/saddle pts (refer to my explanation above)?

Quote:

3. f(x,y) = x^4 - 8xy + 2y^2 - 3
I think the critical points are (0,0), (2,4) and (-2,-4) but im not sure, and again im having trouble determining whether the points are a max/min or saddle point.

Any help or clarification would be much appreciated. Thanks!

Now apply the second partials test to see if its a maximum, minimum, or saddle point there.
• Mar 25th 2009, 11:21 AM
Hockey_Guy14
Partial Derivatives
For the first one, would it be a minimum at (0,0), because D >0 and fxx >0?

For the second one, im not sure how to solve those new equations in order to get the critical points.Thats where I am stuck.

The third one was more or less clarification on the critical points, I can figure out the max/min stuff now. Thanks.

So, maybe a little help with how to go about solving the equations in the second question? Thanks alot!
• Mar 25th 2009, 11:24 AM
Chris L T521
Quote:

Originally Posted by Hockey_Guy14
For the first one, would it be a minimum at (0,0), because D >0 and fxx >0?

For the second one, im not sure how to solve those new equations in order to get the critical points.Thats where I am stuck.

The third one was more or less clarification on the critical points, I can figure out the max/min stuff now. Thanks.

So, maybe a little help with how to go about solving the equations in the second question? Thanks alot!

Yes, the first one is a minimum (but what value does it have? plug it into the original equation to come up with this value.)

My hint for number two: $e^{\frac{1}{2}xy}\neq 0\,\forall\, x,y\in\mathbb{R}$
• Mar 25th 2009, 11:26 AM
Hockey_Guy14
Partial Derivatives
What does that upside down A mean..haha..and is it basically saying, x cannot be 0 and y is all real numbers?
• Mar 25th 2009, 11:41 AM
Chris L T521
Quote:

Originally Posted by Hockey_Guy14
What does that upside down A mean..haha..and is it basically saying, x cannot be 0 and y is all real numbers?

Its saying that $e^{\frac{1}{2}xy}$ is never equal to zero for all (that's what $\forall$ means) real numbers x and y.

With this fact, $f_x\!\left(x,y\right)=0\implies \tfrac{1}{2}y=0$ and $f_y\!\left(x,y\right)=0\implies\tfrac{1}{2}x=0$

So what are the critical points?
• Mar 25th 2009, 11:50 AM
Hockey_Guy14
Partial Derivatives
Would it just be (0,0) then? it seems to look like that might be the only one on the contour plot that I have.
• Mar 25th 2009, 01:10 PM
Chris L T521
Quote:

Originally Posted by Hockey_Guy14
Would it just be (0,0) then? it seems to look like that might be the only one on the contour plot that I have.

(Yes)