Ray10
  • Ray10
Determine position and nature of stationary point of function: f(x,y) = (4x^2)+(3y^2)-(9xy) I have no idea how to work this out due to the implicit aspect :/
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Ray10
  • Ray10
\[4x^{2}+3y ^{2}-9xy\]
blockcolder
  • blockcolder
Solve \(\Large \frac{\partial f}{\partial x}=0\) and \(\Large\frac{\partial f}{\partial y}=0\) for the stationary points. To determine the nature, let (a,b) be a stationary point and \(D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-(f_{xy}(a,b))^2\). 1. If D(a,b)>0 and \(f_{xx}(a,b)>0\), then (a,b) is a local minimum. 2. If D(a,b)>0 and \(f_{xx}(a,b)<0\), then (a,b) is a local max. 3. If D(a,b)<0, then (a,b) is a saddle point. 4. If D(a,b)=0, then you can't make a conclusion. You have to determine the nature by some other means.
Ray10
  • Ray10
so do I first work out the derivative with respect to x, then sole for it equaling x? I'm not sure where to go from finding each derivative?? :S

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blockcolder
  • blockcolder
First, find the derivative with respect to x and equate it to 0. Then do the same with y. Solve these two equations together.
Ray10
  • Ray10
by solving together, do you mean equating them together? \[\frac{ df }{ dx } = 8x-9y\] and \[\frac{ df }{ dy } = 6y-9x where to from here?
Ray10
  • Ray10
\[\frac{ df }{ dy } = 6y-9x\]
blockcolder
  • blockcolder
Let them both equal to 0. Then treat like a system: \[\begin{cases} 8x-9y=0\\ 6x-9x=0 \end{cases}\]
Ray10
  • Ray10
I did that and got both, equal to zero..is that correct? I've blanked at seeing that system, is there another way to solve?
blockcolder
  • blockcolder
AFAIK, this is the only way to find stationary points, because that's the definition of stationary point.
Ray10
  • Ray10
so how do I use the system you showed me?
Ray10
  • Ray10
I get the answer when solving the two as; x=0 and y=0
blockcolder
  • blockcolder
You're right. The only stationary point is (0,0). Now classify that stationary point as a min, max, or saddle point.
Ray10
  • Ray10
I believe it is a saddle point due to its shape, would you agree? Now that I've reached this point, how may I progress to acquire the actual values of the stationary point?
blockcolder
  • blockcolder
Use the test I described previously, the one with the D(a,b). I believe there's only a single stationary point, and you've already found it.
Ray10
  • Ray10
I read the test you described, is this the one I should be looking at?; 4. If D(a,b)=0, then you can't make a conclusion. You have to determine the nature by some other means. in this case, it is D(a,b) = D(0,0) ? That makes the position at 0,0
blockcolder
  • blockcolder
What are f_xx, f_yy, and f_xy first?
Ray10
  • Ray10
I think I am lost... f_xx is 8 and f_yy is 6 ? This is fairly new to me, so f_xx is the second derivative with respect to x correct?
blockcolder
  • blockcolder
yep
Ray10
  • Ray10
therefore that makes, the stationary point be that of a saddle shape and at position 0,0??
blockcolder
  • blockcolder
Yep. That's right.
Ray10
  • Ray10
woah, thanks a lot!!! This was the one part of the topic the lecturer didn't go over, so you helped a lot :)
blockcolder
  • blockcolder
No problem :)

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