Finding Extremum of z=1-sqrt(x^2+y^2)

[manenbu]

Homework Statement

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

Homework Equations

Second derivative test.

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:
$$z^{''}_{xx} = -\frac{y^2}{\sqrt{(x^2+y^2)^3}}\\<br /> z^{''}_{yy} = -\frac{x^2}{\sqrt{(x^2+y^2)^3}}\\<br /> z^{''}_{xy} = \frac{xy}{\sqrt{(x^2+y^2)^3}}$$
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?)

 

[Mark44]

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.

 

[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.

 

[Mark44]

Maybe you noticed that z_x and z_y 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 z_x and z_y 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?

 

[manenbu]

ok, got it.

 

[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?

 

[Mark44]

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.

 

[manenbu]

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