Find the maximum and minimum values of a tri-variable function

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November 3rd 2010, 12:01 PM
Runty

Find the maximum and minimum values of a tri-variable function

I'm not sure if this is another Lagrange Multiplier question or not, but so far this question has me a little confused.

Let $R$ be the region in $R^3$ which is inside the ellipsoid $3x^2+3y^2+z^2=28$ and above the paraboloid $z=x^2+y^2$ . Find the maximum and minimum values of the function $f(x,y,z)=2x^3-2y^3+z^2$ on the region $R$ .
HINT: $R$ is inside the cylinder $x^2+y^2=4$ .

It's probably the wording again (I always trip on word problems), but it probably shouldn't cause too much trouble.
November 4th 2010, 10:19 AM
Runty

EDIT: Asking around, I found a good source to provide me with a means to solve this. You can find it here: http://answers.yahoo.com/question/in...4091520AA7dWAg

I just hope his method is correct.

Through this, I found critical points at
$f(0,0,0)=0$ (undetermined)
$f(\frac{-3}{2},0,\frac{9}{4})=\frac{-27}{16}$ (relative minimum)
$f(0,\frac{3}{2},\frac{9}{4})=\frac{-27}{16}$ (relative minimum)
$f(0,-1,1)=3$ (saddle point)
$f(1,-1,2)=8$ (relative maximum)
$f(1,0,1)=3$ (saddle point)

Can anyone confirm if my answers are right or not? This is one question where making a mistake is pretty easy.
November 4th 2010, 01:42 PM
HallsofIvy

You may be missing an important point- the region, R, is inside the ellipsoid and above the paraboloid. To me that implies that points on the ellipsoid itself or on the paraboloid are not in R. I see only NO critical points in R. Since R is not a closed set there may NO maximum or minimum points of the function in R.
November 5th 2010, 06:35 AM
Runty

Quote:

Originally Posted by HallsofIvy

You may be missing an important point- the region, R, is inside the ellipsoid and above the paraboloid. To me that implies that points on the ellipsoid itself or on the paraboloid are not in R. I see only NO critical points in R. Since R is not a closed set there may NO maximum or minimum points of the function in R.

True, but you might've missed the hint: R is inside the cylinder $x^2+y^2=4$ . That would make it a closed set, wouldn't it? You'd have $-2\leq x,y\leq 2$ .
November 5th 2010, 01:45 PM
Runty

Okay, I found a different method on a website (click here) for Lagrange Multipliers with more than one constrait. This method produces a different answer, yet I'm wondering if I used the method right. It's radically different from my first answers, and I suspect I did something wrong.

Using the method I found on that page (look at the bottom of it), I came up with the following two solutions:

$f(\frac{2+\sqrt{2}}{3},\frac{-2-\sqrt{2}}{3},\pm 2\sqrt{\frac{2}{3} (9-\sqrt{2})}\approx 26.125$ , which is a minimum.
$f(\frac{2-\sqrt{2}}{3},\frac{\sqrt{2}-}{3},\pm 2\sqrt{\frac{2}{3} (9+\sqrt{2})}\approx 27.801$ , which is a maximum.

This question is causing me a lot more trouble than I'd like; I've seen plenty of others for assistance, and even more experienced people are having trouble with it. I think my Prof. made this one too tough even for himself.
November 5th 2010, 10:47 PM
CaptainBlack

Quote:

Originally Posted by Runty

True, but you might've missed the hint: R is inside the cylinder $x^2+y^2=4$ . That would make it a closed set, wouldn't it? You'd have $-2\leq x,y\leq 2$ .

No it does not, R can still be non-closed and be inside a close region, and you are given an open cylinder in the hint anyway!

If I were you I would go and re-examine the exact wording of the original question.

This problem as worded only has a solution if there are calculus type maxima and minima of f in R which are greater than any value of the objective on the boundary. If f has no extrema in (the interior of) R then there is no solution.

CB
November 7th 2010, 10:47 AM
Runty

Man, this question is proving to be so frustrating for me. I've asked all around, and still I haven't solved it.

Alright, starting it over again, we have
$f(x,y,z)=2x^3-2y^3+z^2$
constrained by the following two
$3x^2+3y^2+z^2=28$
$z=x^2+y^2$

I've been told that to solve a two-constraint Lagrange Multiplier problem, we need to find the following:

$\nabla f(x,y,z)=-\lambda\nabla g(x,y,z)-\mu\nabla h(x,y,z)$

$f_x=-\lambda g_x-\mu h_x$
$f_y=-\lambda g_y-\mu h_y$
$f_z=-\lambda g_z-\mu h_z$
$g(x,y,z)=k$
$h(x,y,z)=c$

This part gets frustrating for me, because supposedly, $h(x,y,z)=0=x^2+y^2-z$ , which clearly causes problems when trying to determine $\lambda$ and $\mu$ (which is needed to get some critical points). If it's so, I'd have this: (I'm guessing at this point)

$6x^2=-6\lambda x-2\mu x$
$-6y^2=-6\lambda y-2\mu y$
$2z=-2\lambda z+\mu$ (there's got to be a mistake I've made here)
$g(x,y,z)=28=3x^2+3y^2+z^2$
$h(x,y,z)=0=x^2+y^2-z$

Trying to find $\lambda$ or $\mu$ at this point becomes really tough (if not impossible).

I know that hint is supposed to come into play somehow, but I just don't know how. Is it just supposed to mean that the only valid critical points have to have $-2<x,y<2$ ?

Really, the wording is tripping me up way too easily. I don't even know if the method I've been told to use might be completely wrong for this case.
November 8th 2010, 11:14 AM
Runty

It looks like all of us were wrong. I kept thinking this whole thing was a two-constraint Lagrange Multiplier question, but most of the segments were one-constraint; only one part was two-constraint. Talk about heavy misinterpretation; my Prof. needs to work on putting his questions into better context. Everyone in my class had trouble with this one because he didn't word it properly.

HallsOfIvy, I don't think your bit on "No critical points in R" was relevant. I've already handed in my answers, and in return got my Prof.'s answers, which show mine are partly wrong (great...). Since I've got them, I might as well post them up.

Inside
There is only one critical point of $f$ in $R$ at $(0,0,0)$ where $f=0$ .
Top
Let $g=3x^2+3y^2-28$ and $G=f+\lambda g$ .
For critical points of $G$ we get $g=0$ and $6x^2+6\lambda x=0$ , $-6y^2+6\lambda y=0$ , $2z+2\lambda z=0$ ,
so $x=0 \mbox{ or } -\lambda$ ; $y=0 \mbox{ or } \lambda$ ; $z=0 \mbox{ or } -1$ .
$z=0$ gives $x^2+y^2=28/3$ , contrary to $x^2+y^2\leq 4$ (see the hint).
$\lambda =-1:x=y=0$ gives $z=\sqrt{28}$ (as $z\geq 0$ ) and $f=28$ .
$x=0,y=\lambda=-1 \mbox{ and } x=-\lambda=1,y=0$ both give $z=5$ and $f=27$ .
$x=-\lambda=1,y=\lambda=-1$ give $z^2=22$ and $f=26$ .
Bottom
Let $h=z-x^2-y^2$ and $H=f+\lambda h$ .
For critical points of $H$ we get $h=0$ and $6x^2-2\lambda x=0$ , $-6y^2-2\lambda y=0$ , $2z+\lambda=0$ ,
so $x=0 \mbox{ or } \lambda/3$ ; $y=0 \mbox{ or } -\lambda/3$ ; $z=-\lambda/2$ .
$x=y=0$ gives $z=0$ and $f=0$ .
$x=0,y=-\lambda/3,z=-\lambda/2$ give $y=2z/3$ and $z=4z^2/9$ so $z=9/4$ (or 0) and $y=3/2$ so $f=-\frac{27}{16}$ .
$x=\lambda/3,y=0,z=-\lambda/2$ give $x=-2z/3$ and $z=4z^2/9$ so $z=9/4$ and $x=-3/2$ so $f=-\frac{27}{16}$ .
$x=\lambda/3=-y,z=-\lambda/2$ give $y=-x=2z/3$ and $z=8z^2/9$ so $z=9/8$ (or 0) and $y=3/4=-x$ so $f=-\frac{27}{64}$ .
Rim
Let $F=f+\lambda g+\mu h$ .
For critical points of $F$ we get $g=0,h=0$ which give $z=4$ .
Also $6x^2+6\lambda x-2\mu x=0$ , $-6y^2+6\lambda y-2\mu y=0$ , $2z+2\lambda z+\mu=0$ ( $x^2+y^2=4$ ).
Then $x=0$ gives $y=\pm 2$ so $f=\mp 16+16=0 \mbox{ or } 32$ .
$y=0$ gives $x=\pm 2$ so $f=32 \mbox{ or } 0$ .
Otherwise $y=-2(=\lambda-\mu/3)$ gives $x=\pm \sqrt{2}=-y$ so $f=8(2\pm\sqrt{2})$ ( $\cong 27.3$ or $4.7$ )

Therefore, $\max f=32$ at $(2,0,4)$ and $(0,-2,4)$ , while $\min f=-\frac{27}{16}$ at $(-3/2,0,9/4)$ and $(0,3/2,9/4)$ .

So yeah, there was a lot I had made mistakes of in trying to do this. This question definitely was not made for us to properly be able to do it, seeing as our Prof. didn't provide us with the information we would need.
November 8th 2010, 06:25 PM
CaptainBlack

Quote:

Originally Posted by Runty

It looks like all of us were wrong. I kept thinking this whole thing was a two-constraint Lagrange Multiplier question, but most of the segments were one-constraint; only one part was two-constraint. Talk about heavy misinterpretation; my Prof. needs to work on putting his questions into better context. Everyone in my class had trouble with this one because he didn't word it properly.

HallsOfIvy, I don't think your bit on "No critical points in R" was relevant. I've already handed in my answers, and in return got my Prof.'s answers, which show mine are partly wrong (great...). Since I've got them, I might as well post them up.

It was relevant, you either posted a garbled version of the question or your teacher is incompetent.

Having checked the first point claimed as a maximum it is not inside the feasible region but on a boundary which is explicitly not feasible (and I can't be bothered to check the others but the same is probably the case). This is not a surprise to us as it is exactly the point Halls was making.

If you had responded to the point about the feasible region not being closed you might have received some real help (since we do know how to use Lagrange multipliers with inequality constraints for a properly formulated problem)

CB