Stationary points of a function

[Bener123]

Hi guys,

My last question of this topic and I'm really struggling with it. Would anyone like to help out on this?

 

[mmm4444bot]

I'm really struggling with it.

Hello Bener123. Please share your beginning thoughts/work, or explain the part(s) at issue. Thank you. ?

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[BigBeachBanana]

Would anyone like to help out on this?

We all love to help, but we need to know what you need help with.

 

[Bener123]

Hi, I have added what I have so far. Just need to confirm if I'm right. I'm trying to figure out the second derivatives and the d^2omega/dxdy as you can see from the second page.

I need to know the min the max and saddle points and how to get there if that's doable.

 

[Bener123]

We all love to help, but we need to know what you need help with.

Reference the first image posted on the first question ^

 

[BigBeachBanana]

Those look good so far. Continue with the second partials and use the test below.

 

[Bener123]

I appreciate that, but it's the numerical part I am struggling with. I want to confirm I have the correct xx and yy (on the second page with the second derivatives) and if someone could please give me the numbers for dxdy, I will be able to figure out the rest.

 

[BigBeachBanana]

I appreciate that, but it's the numerical part I am struggling with. I want to confirm I have the correct xx and yy (on the second page with the second derivatives) and if someone could please give me the numbers for dxdy, I will be able to figure out the rest.

I said they look good so far.
$f_{xy}=f_{yx}$, You can either take the derivative of $6x^2-96$ with respect to $y$ OR $6y+12$ with respect to $x.$

 

[Bener123]

I said they look good so far.
$f_{xy}=f_{yx}$, You can either take the derivative of $6x^2-96$ with respect to $y$ OR $6y+12$ with respect to $x.$

 

[Bener123]

I said they look good so far.
$f_{xy}=f_{yx}$, You can either take the derivative of $6x^2-96$ with respect to $y$ OR $6y+12$ with respect to $x.$

Understood. So if I took the derivative of X or Y, what answer do I get? Is it 6?

 

[Deleted member 4993]

Understood. So if I took the derivative of X or Y, what answer do I get? Is it 6?

No

Please share your work showing how you got 6.

 

[Bener123]

No

Please share your work showing how you got 6.

I just went by the only thing the 2 equations have in common. Please can someone tell me what the fxy is.

 

[topsquark]

I just went by the only thing the 2 equations have in common. Please can someone tell me what the fxy is.

When you found $f_x$ you took the derivative as if y were a constant. When you found $f_y$ you took the derivative as if x were a constant. So how do you find $f_{xy}$?

-Dan

 

[Bener123]

When you found $f_x$ you took the derivative as if y were a constant. When you found $f_y$ you took the derivative as if x were a constant. So how do you find $f_{xy}$?

-Dan

If I knew that, I wouldn't be posting it on here ?

 

[BigBeachBanana]

If I knew that, I wouldn't be posting it on here ?

I think you're just unfamiliar with the notation. It has nothing to do with what's in common. You know how to compute derivatives.
$f_{xx}$ means to take the derivative with respect to x twice.
$f_{yy}$ means to take the derivative with respect to y twice.
$f_{xy}$ means to take the derivative with respect to x first, then respect to y.
$f_{yx}$ means to take the derivative with respect to y first, then respect to x.

For example, $f(x,y)=xy$
$f_x=y$
$f_{xy}=1$
$f_{xx}=0$

$f_y=x$
$f_{yx}=1$
$f_{yy}=0$

 

[Bener123]

So for the equations 6x^2-96 and 6y+12 what is the fxy?

 

[BigBeachBanana]

So for the equations 6x^2-96 and 6y+12 what is the fxy?

I told you how to get the answer earlier.

If $f_x=6x^2-96$, then $\red{f_{xy}=\frac{d}{dy}(6x^2-96)}$.
Compute the derivative in red.

 

[Bener123]

I know what the question is asking. With respect to y from x. I don't know how to do this bit. Please show me, answer it too so I can understand how to do it in the future.

 

[BigBeachBanana]

I know what the question is asking. With respect to y from x. I don't know how to do this bit. Please show me, answer it too so I can understand how to do it in the future.

When you differentiate with respect to y, treat all other variables that are not y as a constant, like number 2 for example. So in this case $, 6x^2-96$ only contains x and no y. So you're basically differentiating a constant. What's the derivative of a constant?

Now, try the other way.
If $f_y=6y+12$, then $\red{f_{yx}=\frac{d}{dx}(6y+12)}$.
Compute the derivative in red.

 

[Bener123]

12?