# Homework Help: Extrema Points

1. Sep 10, 2009

### manenbu

1. The problem statement, all variables and given/known data

Find exterma points of:
$$z=1-\sqrt{x^2+y^2}$$

2. Relevant equations

Second derivative test.

3. The attempt at a solution

I find that (0,0,1) is a point where an extremum exist. To determine whether it's a maximum or minimum I need to use the second derivative test, but my second derivatives are:
$$z^{''}_{xx} = -\frac{y^2}{\sqrt{(x^2+y^2)^3}}\\ z^{''}_{yy} = -\frac{x^2}{\sqrt{(x^2+y^2)^3}}\\ z^{''}_{xy} = \frac{xy}{\sqrt{(x^2+y^2)^3}}$$
All of those functions are undefined at at x=0 and y=0.
What should I do then?

(edit - any idea why won't the "\\" make a new line in the tex code?)

2. Sep 10, 2009

### Staff: Mentor

You can also look at the original equation. Clearly, the largest value of z will be 1, which occurs when x = 0 and y = 0. For all other values of x and y, a positive number is being subtracted from 1, giving a smaller value of z.

3. Sep 10, 2009

### manenbu

Yes, this is true. But I was wondering if there is a formal way of knowing this. For this particular function it's easy to just look at the function, but for another one it might not be.

4. Sep 10, 2009

### Staff: Mentor

Maybe you noticed that zx and zy are both undefined at (0, 0). This is a point in the domain of your function, which makes it a candidate for an extremum (in fact, the only candidate).

Inasmuch as there are no points where zx and zy are simultaneously zero, you have to use another approach to confirm that you have a maximum at (0, 0). You can do that with inequalities.
$$1 - \sqrt{x^2 + y^2} \le 1~for~all~x~and~y$$
and equals 1 only at (0, 0).
Do you need to say anything more that this?

5. Sep 10, 2009

### manenbu

ok, got it.

6. Sep 10, 2009

### manenbu

I got another question now:
I got this function:
z=-x2 -xy -y2 + 4lnx + 10lny

So the derivatives are:
$$z^{'}_{x} = -2x -y + \frac{4}{x}$$

$$z^{'}_{y} = -x -2y + \frac{10}{y}$$
when solving this system of equations i get the following points:
$$(\pm1,\pm2) and \left(\pm\frac{4}{\sqrt{3}},\mp\frac{5}{\sqrt{3}}\right)$$
all those points satisfy the derivative equation.
in the given answers, only (1,2) is a maximum point. also, when plotting this function i only see one maximum point at (1,2).
so where are the other points coming from?

7. Sep 10, 2009

### Staff: Mentor

For a new problem, you really should start a new thread.

You can't possibly have all of the points you listed. Because of the ln terms, The domain is {(x, y) | x > 0 and y > 0}. That eliminates all but one of the points you show.

I'll bet that you introduced a bunch of extraneous solutions when you solved for x in one of your equations (getting a radical) and substituting it in the other equation.

8. Sep 10, 2009

### manenbu

oh. seems so obvious now.
thanks for helping me, again.