Question

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 :/

Answer 1

\[4x^{2}+3y ^{2}-9xy\]

Answer 2

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.

Answer 3

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

Answer 4

First, find the derivative with respect to x and equate it to 0. Then do the same with y.
Solve these two equations together.

Answer 5

by solving together, do you mean equating them together?
\[\frac{ df }{ dx } = 8x-9y\] and \[\frac{ df }{ dy } = 6y-9x
where to from here?

Answer 6

\[\frac{ df }{ dy } = 6y-9x\]

Answer 7

Let them both equal to 0. Then treat like a system:
\[\begin{cases}
8x-9y=0\\
6x-9x=0
\end{cases}\]

Answer 8

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?

Answer 9

AFAIK, this is the only way to find stationary points, because that's the definition of stationary point.

Answer 10

so how do I use the system you showed me?

Answer 11

I get the answer when solving the two as; x=0 and y=0

Answer 12

You're right. The only stationary point is (0,0). Now classify that stationary point as a min, max, or saddle point.

Answer 13

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?

Answer 14

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.

Answer 15

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

Answer 16

What are f_xx, f_yy, and f_xy first?

Answer 17

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?

Answer 18

yep

Answer 19

therefore that makes, the stationary point be that of a saddle shape and at position 0,0??

Answer 20

Yep. That's right.

Answer 21

woah, thanks a lot!!! This was the one part of the topic the lecturer didn't go over, so you helped a lot :)

Answer 22

No problem :)