1. ## 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!

2. 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'.

3. 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 $\displaystyle \left(x_0,y_0\right)$ is the critical point):

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

Now, if:

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

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

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

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

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

So what can you conclude?

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 $\displaystyle f\!\left(x,y\right)=e^{\frac{1}{2}xy}$

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

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

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.

4. ## 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!

5. 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: $\displaystyle e^{\frac{1}{2}xy}\neq 0\,\forall\, x,y\in\mathbb{R}$

6. ## 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?

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

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

So what are the critical points?

8. ## 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.

9. 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.