Stationary points of a function

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Bener123

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Hi guys,

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

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Bener123 said:
I'm really struggling with it.
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Hello Bener123. Please share your beginning thoughts/work, or explain the part(s) at issue. Thank you. ?

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[imath]\;[/imath]
 
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.
 

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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.
 
Bener123 said:
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.
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I said they look good so far.
[imath]f_{xy}=f_{yx}[/imath], You can either take the derivative of [imath]6x^2-96 [/imath] with respect to [imath]y[/imath] OR [imath]6y+12[/imath] with respect to [imath]x.[/imath]
 
BigBeachBanana said:
I said they look good so far.
[imath]f_{xy}=f_{yx}[/imath], You can either take the derivative of [imath]6x^2-96 [/imath] with respect to [imath]y[/imath] OR [imath]6y+12[/imath] with respect to [imath]x.[/imath]
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Understood. So if I took the derivative of X or Y, what answer do I get? Is it 6?
 
Bener123 said:
I just went by the only thing the 2 equations have in common. Please can someone tell me what the fxy is.
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When you found [imath]f_x[/imath] you took the derivative as if y were a constant. When you found [imath]f_y[/imath] you took the derivative as if x were a constant. So how do you find [imath]f_{xy}[/imath]?

-Dan
 
topsquark said:
When you found [imath]f_x[/imath] you took the derivative as if y were a constant. When you found [imath]f_y[/imath] you took the derivative as if x were a constant. So how do you find [imath]f_{xy}[/imath]?

-Dan
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If I knew that, I wouldn't be posting it on here ?
 
Bener123 said:
If I knew that, I wouldn't be posting it on here ?
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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.
[imath]f_{xx}[/imath] means to take the derivative with respect to x twice.
[imath]f_{yy}[/imath] means to take the derivative with respect to y twice.
[imath]f_{xy}[/imath] means to take the derivative with respect to x first, then respect to y.
[imath]f_{yx}[/imath] means to take the derivative with respect to y first, then respect to x.

For example, [imath]f(x,y)=xy[/imath]
[imath]f_x=y[/imath]
[imath]f_{xy}=1[/imath]
[imath]f_{xx}=0[/imath]

[imath]f_y=x[/imath]
[imath]f_{yx}=1[/imath]
[imath]f_{yy}=0[/imath]
 
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.
 
Bener123 said:
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.
Click to expand...

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 [imath], 6x^2-96[/imath] 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 [imath]f_y=6y+12[/imath], then [imath]\red{f_{yx}=\frac{d}{dx}(6y+12)}[/imath].
Compute the derivative in red.
 
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