Find the stationary points of the following functions, and
identify the nature of each (ie local maxima, minima, saddle point or point of inflexion):

(a) f(x, y) = x2 + y2 - x - y

(b) g(x, y) = 3x2 - 3xy + 4y2 + 8x - 5y - 4

2. Hello,
Originally Posted by brumby_3
Find the stationary points of the following functions, and identify the nature of each (ie local maxima, minima, saddle point or point of inflexion):
(a) f(x, y) = x2 + y2 - x - y
(b) g(x, y) = 3x2 - 3xy + 4y2 + 8x - 5y - 4
Find $x_0$ and $y_0$ such that $\frac{\partial f}{\partial x}(x_0,y_0)=0$ and $\frac{\partial f}{\partial y}(x_0,y_0)=0$

These are the critical points.

I hope you've done bits of algebra including matrices...

The Hessian matrix is defined at a point as :

$H_f(x_0,y_0)=\begin{pmatrix} \frac{\partial^2 f}{\partial x^2}(x_0,y_0) & \frac{\partial^2 f}{\partial x \partial y}(x_0,y_0) \\ \\ \frac{\partial^2 f}{\partial y \partial x}(x_0,y_0) & \frac{\partial^2 f}{\partial y^2}(x_0,y_0) \end{pmatrix}$

$\frac{\partial^2 f}{\partial x^2}=\frac{\partial}{\partial x} \left(\frac{\partial f}{\partial x}\right) \quad \quad \quad \quad \frac{\partial^2 f}{\partial x \partial y}=\frac{\partial}{\partial x} \left(\frac{\partial f}{\partial y}\right)$

Work with eigenvalues
If the eigenvalues are all positive, then this point $(x_0,y_0)$ is a maximum.
If they are all negative, then this point $(x_0,y_0)$ is a minimum.
If one is positive and one is negative, then this point $(x_0,y_0)$ is a saddle point.

Work with discriminant
If the discriminant is negative, then it's a saddle point.
If the discriminant is positive and $\frac{\partial^2 f}{\partial x^2}(x_0,y_0) >0$, then it's a maximum.
If the discriminant is positive and $\frac{\partial^2 f}{\partial x^2}(x_0,y_0) <0$, then it's a minimum.

If you don't know how to proceed, google for Hessian matrix.

3. Originally Posted by brumby_3
Find the stationary points of the following functions, and
identify the nature of each (ie local maxima, minima, saddle point or point of inflexion):

(a) f(x, y) = x2 + y2 - x - y

(b) g(x, y) = 3x2 - 3xy + 4y2 + 8x - 5y - 4

First find where $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$ [and in part (b), where the partials of g are equal to zero.]

These will give you the critical points. (Be careful on how you pair them together)

Then apply the second partials test to see if its a max, min, or is a saddle point.

$D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f_{xy}^2(x_0,y_0)$

Recall that if:

$D>0~\text{and}~f_{xx}(x_0,y_0)>0$, then $f$ has a relative minimum at $(x_0,y_0)$

$D>0~\text{and}~f_{xx}(x_0,y_0)<0$, then $f$ has a relative maximum at $(x_0,y_0)$

$D<0$, then $f$ has a saddle point at $(x_0,y_0)$

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

You have enough information to help you get started on the problem(s)

--Chris

4. I've never heard of the Hessian matrix and it doesn't seem to be mentioned in my course outline. Can someone please go through this question step by step as an example so that I can try and do the next homework questions myself?

5. Stationary points

6. I'm fairly certain that points of inflection are not considered stationary points. That being said, given f(x, y), you must calculate all points at which df/dx = df/dy = 0, and then you must calculate $f_{xx}, f_{xy}, f_{yx}, f_{yy}$ and evaluate $f_{xx}f_{yy} - f_{xy}f_{yx}$. If that value is positive and both $f_{xx}$ and $f_{yy}$ are positive, it means that the point is a minimum. If that value is positive and both $f_{xx}$ and $f_{yy}$ are negative, it means that the point is a maximum. If $f_{xx}f_{yy} - f_{xy}f_{yx}$ is negative, the point is a saddle point. Edited for correctness.

7. Can you show how to do it using the questions asked please?

8. (a): $\frac{df}{dx} = 2x - 1; \frac{df}{dy} = 2y - 1; f_{xx} = 2; f_{yy} = 2; f_{xy} = 0; f_{yx} = 0$

Hence we have x = 0.5; y = 0.5 and thus our only point of consideration is (0.5, 0.5). What type of point is this? Use the formula:

$f_{xx}f_{yy} - f_{xy}f_{yx} = 2 \cdot 2 - 0 \cdot 0 = 4 - 0 = 4$

Since 4 is positive and both 2 and 2 are positive, this point is a local minimum. In this case, it is also an absolute minimum of the function.

See if you can handle (b).

Edited for clarity and correctness.

9. If you need an example, we need only do one of these, right?

the rules are, if $(a,b)$ is a critical point of a function $f(x,y)$, that is, $f_x(a,b) = f_y(a,b) = 0$ (and of course we assume $f$ is differentiable on some disk containing $(a,b)$ and has continuous partial derivatives on the said disk etc), then we can classify the critical points as follows:

Let $D = D(a,b) = f_{xx}(a,b)f_{yy}(a,b) - [f_{xy}]^2$, then

(a) If $D > 0$ and $f_{xx}>0$, then $f(a,b)$ is a local minimum

(b) If $D > 0$ and $f_{xx}<0$, then $f(a,b)$ is a local maximum

(c) If $D<0$, then $f(a,b)$ is a saddle point.

lets see how this works with your $f(x,y)$

Note that:

$\begin{array}{lcl} f_x = 2x - 1 & & f_y = 2y - 1 \\ & & \\ f_{xx} = 2 & & f_{yy} = 2 \\ & f_{xy} = 0 & \end{array}$

so that $x = y = \frac 12$ are the only values of $x$ and $y$ that make our partials zero. hence, the only critical point is $\bigg( \frac 12 , \frac 12 \bigg)$

Now, $D \bigg( \frac 12 , \frac 12 \bigg) = 4 > 0$ and $f_{xx} \bigg( \frac 12 , \frac 12 \bigg) = 2 > 0$, so that $f \bigg( \frac 12 , \frac 12 \bigg)$ is a local minimum

EDIT: I should have known iceman would pick up the slack

10. Okay I've looked at b) and I'm still not sure how to go around it? Am I suppose to collect like terms first?

11. Originally Posted by brumby_3
Okay I've looked at b) and I'm still not sure how to go around it? Am I suppose to collect like terms first?
collect like terms where? where are you stuck? you found the partial derivatives already, right? what did you get?

12. lol I'm lost altogether

13. Originally Posted by brumby_3
lol I'm lost altogether
you must be able to at least find the partial derivatives. if not, then us going through these examples will not help you.