# Critical points question

1. Jul 28, 2010

### jegues

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

2. Relevant equations
N/A.

3. The attempt at a solution

Part A:

The volume is,

$$xyz = xy(1 - x^{2} - y^{2})$$

Critical points:
$$f_{x} = y-3x^{2}y-y^{3} = 0$$

$$f_{y} = x -x^{3} - 3xy^{2} = 0$$

Part B:

This is where I get confused, how do I find the critical point in the first quadrant assuming x>0 and y>0 ?

NOTE: I haven't gotten to parts C and D yet, I'll update this thread as I go.

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2. Jul 28, 2010

Set $$f_x = f_y = 0.$$ Since x and y are not zero here, we can divide f_x by y and f_y by x, giving us:

$$f_{x} = 0 = 1 - 3x^2 - y^2 = 0$$

$$f_{y} = 0 = 1 - x^2 - 3y^2 = 0$$

You should be able to find an (x,y) that satisfies these equations easily. [STRIKE]In reality, there are an infinite number of such points that would work.[/STRIKE] Ignore that last bit, idk what I was thinking xD...

Last edited: Jul 28, 2010
3. Jul 28, 2010

### jegues

$$x^{2}=y^{2}= \frac{1}{4}$$

Therefore,

$$(x,y) = (\frac{1}{2},\frac{1}{2})$$

I'll keep updating this thread as I progress through the other parts.

Thanks again!

4. Jul 28, 2010

Yep, that works!

5. Jul 29, 2010

### jegues

Working on Part C:

I'm a little confused with this part. I took a peak at the solutions and I see the following,

$$f_{xx} = -6xy = \frac{-3}{2}$$

$$f_{yy} = -6xy = \frac{-3}{2}$$

$$f_{xy} = 1 - 3x^{2} - 3y^{2} = \frac{-1}{2}$$

The mechanics of all this (i.e. taking the partial of fx wrt x again) makes sense and so do the results. The part I don't understand is WHY we are doing this to obtain information about the nature of the critical point. (in other words how is this helping us?)

Also they throw in this line at the end,

So,

$$f_{xx}f_{yy}-f_{xy}^{2} > 0,$$ and $$f_{xx} < 0$$

Therefore it is a local maximum.

I'm confused as to how he drew this conclusion??? How do all these partial derivatives, and specifically that line above, tell us whether it is a max or a min?

6. Jul 29, 2010

Let M = $$f_{xx}f_{yy}-f_{xy}^{2}.$$

If M > 0, then $$f_{xx}f_{yy} > f_{xy}^{2}.$$ So f_xx and f_yy must have the same sign (positive or negative) and are sufficiently large enough to offset f_xy (this is a bit more complicated, ignore for now).
If positive, then both the x and y cross sections are concave up, i.e., local min.
Similarly, if negative, then concave down, i.e., local max.

If M < 0 and f_xx and f_yy have different signs, then the x and y cross sections have opposite concavity. Thus, we have a saddle point.

If M < 0 and f_xx and f_yy have the same sign, then things are a tiny bit more complicated. I'm too lazy to type it all out, but the result is that the critical point is a saddle point here.

By the way, all this should be on wikipedia, wolframalpha, etc. Just search "second derivative test in multivariable calculus" in Google.

7. Jul 29, 2010