Optimizing a Function Inside a Tetrahedron: Finding Maxima and Minima

[Addez123]

Homework Statement

$$f(x,y,z) = xyz + xy$$
Boundaries are set by these 4 points:
(0,0,0), (1,0,0), (0,2,0), (0,0,2)

Relevant Equations

None

First step is to find the derivatives:
$$f_x' = yz + y$$
$$f_y' = xz + x$$
$$f_z' = xy $$

When all three equal zero, then you have a stationary point.
1. Let's start with f_z' = 0, when x = 0
Then $$y \in R$$
2. Knowing this we look at f_y' = 0
Since x = 0, z can be any real number just like y.
3. So we look at f_x' = 0
z = -1 and y can be any real number.
Our point of interest is then:
p = (0, y, -1)

This is my problem. That's a line, not a point.
Meaning any point on that line is a max,min?

Im confused..

 

[BvU]

Hi,

Why do you think all derivatives should be zero at the maximum ?
(if that is the relevant equation you use, but fail to mention )

What is the maximum of $f(x) = x^2$ if $0<x<3$ ?

[edit]:

Our point of interest is then:
p = (0, y, -1)

Very unlikely: line or point: it is outside the tetraeder !

[edit edit]Perhaps you want to post a complete problem statement: now we have to guess from the thread title that you are searching for the maximum value of $xyx+xy$ in the first octant...

$\$

 

[Mark44]

Homework Statement:: $$f(x,y,z) = xyz + xy$$
Boundaries are set by these 4 points:
(0,0,0), (1,0,0), (0,2,0), (0,0,2)
Relevant Equations:: None

First step is to find the derivatives:
$$f_x' = yz + y$$
$$f_y' = xz + x$$
$$f_z' = xy $$

This is really nonstandard notation. $f_x$ (without the prime) is alternate notation for $\frac{\partial f}{\partial x}$, and similar for the other two partial derivatives.

In line with what @BvU said, it's not likely that you're going to find a maximum at a point where all three partials are zero. You're going to have to investigate the behavior of the function along all six edges, as well as on the bounding planes.

 

[Addez123]

Hi,

Why do you think all derivatives should be zero at the maximum ?

I dont. As in your x^2 example its found at infinity. But in a restricted search, like my assignment, it's limited by the sides of the tetraed. I'll need to check both for any critical point inside the tetraed and then ill check the sides. But one thing at a time, I can't check the critical points because I don't understand it.

Right now my calculations tell me, any point on
$$(0, t, -1), t \in R$$
is a critical point. That can't be right?

 

[Mark44]

I dont. As in your x^2 example its found at infinity.

No.
As given, the function was $f(x) = x^2$ for $0 < x < 3$.

Right now my calculations tell me, any point on $(0, t, -1), t \in R$
is a critical point. That can't be right?

No, it's not right. First off, this is a line, not a point, and second, it's not within the boundaries that are given.

Again, you need to check the boundary edges and surfaces.

 

[BvU]

Read out loud the function with $z = -1$ It's just not interesting at all.

 

[Addez123]

Read out loud the function with $z = -1$ It's just not interesting at all.

Yeah I noticed that. So I started working on the sides.
I started with the x-z plane by doing: f(x, 0, z) = x*0*z + x*0 = 0
Then f_x and f_z = 0 meaning we have critical points all over this surface.

Somethings seriously wrong.

 

[Mark44]

Then f_x and f_z = 0 meaning we have critical points all over this surface.

Somethings seriously wrong.

Not necessarily. The function $g(x) = x^3$ has a critical point at (0, 0), but this is neither a maximum or minimum. Just because a derivative (or partial derivatives) happen to be zero somewhere doesn't mean there is a max or min there. As I said before, you have four surfaces and three edges you need to check.

 

[Addez123]

The book doesn't prepare you at all for this! I'm lost at every turn.

Since x-z side had derivative = 0 everywhere, the whole side has the same value. Which was 0.
Same thing for x-z side.

Looking at the bottom (z = 0)
f(x, y, 0) = xy
f_x' = y = 0 => y = 0
f_y' = x = 0 => x = 0
So we have critical point in f(0,0,0) = 0Lastly we have the sloped side of the tetraed.
I have no frkn clue how to solve that so I extracted a normal vector from the top (0, 2, 0) to the x and y points.
$$n = (-4, -2, -2)$$
$$n * (x - p) = 0, p = (0, 2, 0)$$
$$-4x -2(y -2) - 2z = 0$$

Doing the particle derivative of the last equation yields static values such as -4, -2 etc.
That means there are not maximums along this surface.

So there's no maximum literally ANYWHERE, yet f(.5, .5, 0) = .25

I've checked every side, and every point.
All of them are ZERO yet f(0.5, 0.5, 0) IS NOT?

Wasnt it suppose to show up at the derivatives then??
It's 5am and I've been stuck for a good 12 hours on this.

 

[Addez123]

I suppose you don't just have to do the surfaces and the dots but the lines connecting the dots aswell.
Tremendous amounts of work, but when doing the line connecting x and y points I get the f(1/2, 1/2, 0) as maximum.

That's the solution. Still feels like a lucky shot since the examination of the surfaces yielded nothing of value.

 

[Mark44]

Watch the language. We don't take kindly to obscene language... I have edited your post to remove part of what you wrote.

The book doesn't prepare you at all for this! I'm lost at every turn.

I'll bet if you look closely, the book tells you, via a theorem or something similar, the places where an extreme value can occur. Only one of those places is where a derivative is zero.

Since x-z side had derivative = 0 everywhere, the whole side has the same value.

OK

Looking at the bottom (z = 0)
f(x, y, 0) = xy
f_x' = y = 0 => y = 0
f_y' = x = 0 => x = 0

All this means is that $f_x = 0$ and $f_y = 0$ at the point (0, 0, 0). What happens along the boundary line from (1, 0, 0) to (0 2, 0)?
Don't write f_x'. As already mentioned f_x is the partial derivative $\frac{\partial f}{\partial x}$. Don't mix Newton notation (with primes) should not be used when you're talking about partial derivatives.

Means the whole side has same value, which if we look at f(0,0,0) = 0

But there are many other points on the base of that figure where the function values aren't zero.

 

[Mark44]

So there's no maximum literally ANYWHERE, yet f(.5, .5, 0) = .25

You didn't check EVERYWHERE. All you checked, as far as I can tell, are the base and the two vertical sides. There is another side that I don't think you checked, and there are three edges that you didn't check.

I've checked every side, and every point.
All of them are ZERO yet f(0.5, 0.5, 0) IS NOT?

No, you didn't check every point.

Wasnt it suppose to show up at the derivatives then??

Not necessarily.

1. Where does the maximum value occur for $f(x) = x^2, x \in [0, 3]$?
2. Where does the minimum value occur for $g(x) = \sqrt x$?
3. Where does the maximum value occur for $h(x) = -|x| + 2$?

Setting the derivative to zero doesn't tell you the whole story.

 

[Addez123]

Apologize. I edited like 15 times to remove most swear words, its just hard to express yourself without swearing sometimes..

Oh I didnt understand f_x was df/dx

Eitherway, along the path (1, 0, 0) to (0, 2, 0) was the answer
I start by making x = 1- y/2 we get
$$f(1-y/2, y, 0) = (1-y/2)*y = y - y^2/2$$
$$f_y = 1 - y = 0, y = 1$$
$$x = 1 - y/2 = 1 - 1/2 = 1/2$$
f(1/2, 1, 0) = 1/2 which is correct maximum.

I can finally rest..