Maximization of a Multivariable Function

In summary: The others all seem to lead to undefined values when I use the sin(u)=-sqrt(2)*cot(v) equation, so I will exclude...-cos(u)=1 yields u=-1, pi-sin(u)=0 yields u=-sqrt(2)*pi, -1-cos(u)=-1 yields u=-sqrt(2)*-1, 0-cos(u)=1 yields u=-1, pi-sin(u)=0 yields u=-sqrt(2)*pi, -1-cos(u)=-1 yields u=-sqrt(2)*-1, 0
  • #1
The Head
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Homework Statement
See Picture Attached
Relevant Equations
fx=0, fy=0, fz=0
The beginning is straight forward and I found f=x^2-2yz, which satisfies grad(f)=F. Then I calculated W= f(x,y,z)-f(0,1,1) since it's conservative.

I get stuck when trying to find the max and mins. Given grad(f)=0 at extrema, we can see (0,0,0) is a point. On the boundary, I have to parameterize:
x=sqrt(2)cos(u)sin(v)
y=sqrt(2)sin(u)sin(v)
z=2cos(v)

Then I'm not exactly sure where to go. I feel like the options are either:
1) Substitute my parameters into f=x^2-2yz, and take the derivative. But then I'm not sure what to do because I'm used to just having one variable/parameter now.
2) Create the second partial derivative matrix (e.g., fxx, fxy, fxz in column one, fyx, fyy, fyz in column two, and fzx, fzy, fzz in column three. But I get a weird result there, with fxx>0, but fxx*fyy- fxy*fyx=0.

I usually at this point substitute the parameters and then take the derivative (with respect to a single parameter like t) and set equal to zero. Then I find all of the t's that work. I'm just not sure how to proceed here.
 

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  • #2
The Head said:
grad(f)=0 at extrema, we can see (0,0,0) is a point
But not necessarily a max or min?
The Head said:
Substitute my parameters into f=x^2-2yz, and take the derivative.
Sounds like a good idea. What do you get?
 
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  • #3
haruspex said:
But not necessarily a max or min?

Sounds like a good idea. What do you get?
Thanks for the guidance.

For (0,0,0), I made the second partial derivatives matrix and substituted, and obviously get an R^3 zero matrix. This just means this one is indeterminate?

Also, I went ahead and parameterized and derived, but I want to make sure that my logic is correct.
f-x^2-2yz= 2*(cosu)^2 * (sinu)^2 - 4sqrt(2)*sinu*sinv*cosv

Usually there is just one parameter at this point, but with two, I assume I just find df/du and df/dv and solve a system?
So df/du= -4cosu*sinu*(sinv)^2 - 4sqrt(2)cosu*sinv*cosv = 0
Simplifying gives me: (4sinv)(cosu)(sinu*sinv+sqrt(2)*cosv)=0
So either sinv=0, cosu=0, or sinu=-sqrt(2)*cotv

Then do the same for df/dv and substitute to solve the system, right?

Really appreciate the help
 
  • #4
Also, the weird thing that happens when I do this is I get a lot of undefined values. For example, I'll get sin(v)=0, which implies v=0. So then when I substitute back into the equation sin(u)=-sqrt(2)*cot(v), I get an error, since cot(0) is undefined. So while the z-coordinate is defined, x and y are not, so I have a hard time getting a value for x^2-2yz. Do I just ignore these points? Or perhaps I've gone about this wrong (because I think my computations are correct).
 
  • #5
The Head said:
Then do the same for df/dv and substitute to solve the system, right?
Yes. Looks good so far.
The Head said:
which implies v=0. So then when I substitute back into the equation sin(u)=-sqrt(2)*cot(v), I get an error
As you wrote
The Head said:
So either sinv=0, cosu=0, or sinu=-sqrt(2)*cotv
Only one of these need be true. If sin v=0 there is no equation sin(u)=-sqrt(2)*cot(v).
 
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  • #6
haruspex said:
Yes. Looks good so far.

As you wrote

Only one of these need be true. If sin v=0 there is no equation sin(u)=-sqrt(2)*cot(v).
Thanks so much for your reply. So all the undefined points are just ignored, and I should only consider the ones that have discrete values for u and v?
 
  • #7
The Head said:
Thanks so much for your reply. So all the undefined points are just ignored, and I should only consider the ones that have discrete values for u and v?
From the ##\frac{\partial f}{\partial u}## equation you have three possibilities for solutions. But there is also the ##\frac{\partial f}{\partial v}## equation. Which of the three are consistent with one or more of the solutions of the second equation?
 
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  • #8
haruspex said:
From the ##\frac{\partial f}{\partial u}## equation you have three possibilities for solutions. But there is also the ##\frac{\partial f}{\partial v}## equation. Which of the three are consistent with one or more of the solutions of the second equation?
Thanks, this makes more sense. I got cos(u)=0, sin(v)=0, sinu=-sqrt(2)*cot(v) from the df/du equation. Then for df/dv I get cot(v)=0, cos(v)=1

-The cot(v)=0 is promising, because that works for v=pi/2, 3pi/2. I sub back into then find sin(u)=0, so u= 0, pi.
-Similarly, cos(u)=0 gives u= pi/2, 3pi/2, which means cot(v)=+/- 1/sqrt(2), which yields two solutions.

The others all seem to lead to undefined values when I use the sin(u)=-sqrt(2)*cot(v) equation, so I will exclude those.
 
  • #9
The Head said:
Thanks, this makes more sense. I got cos(u)=0, sin(v)=0, sinu=-sqrt(2)*cot(v) from the df/du equation. Then for df/dv I get cot(v)=0, cos(v)=1

-The cot(v)=0 is promising, because that works for v=pi/2, 3pi/2. I sub back into then find sin(u)=0, so u= 0, pi.
-Similarly, cos(u)=0 gives u= pi/2, 3pi/2, which means cot(v)=+/- 1/sqrt(2), which yields two solutions.

The others all seem to lead to undefined values when I use the sin(u)=-sqrt(2)*cot(v) equation, so I will exclude those.
From the df/dv equation I get ##\tan(2v)\cos^2(u)=2\sqrt 2\sin(u)##. Whence if cos (u)=0 then tan(2v) is infinite, or if sin (v)=0 then sin(u)=0.
 
  • #10
Around the point (0,0,0), consider what happens if you keep x=0 and let y and z vary. The term -2yz can be either positive or negative, depending on which way you change the variables. What does that tell you?
 
  • #11
FactChecker said:
Around the point (0,0,0), consider what happens if you keep x=0 and let y and z vary. The term -2yz can be either positive or negative, depending on which way you change the variables. What does that tell you?

The -2yz term could be positive or negative, depending on which way you go in the y- and z-directions. Which intuitively feels like a saddle point, but I could be wrong there.
 
  • #12
haruspex said:
From the df/dv equation I get tan⁡(2v)cos2⁡(u)=22sin⁡(u). Whence if cos (u)=0 then tan(2v) is infinite, or if sin (v)=0 then sin(u)=0.
Thanks, I will try to look it over. I was pretty careful, but evidently made an error along the way. Perhaps I exchanged a u with a v somewhere. I don't think I could spot the double angle identity, but I'll work through solving the df/du equation again and check my substitutions. Thanks for your help along the way! Conceptually at this point it makes much more sense, which is what I'm going for.
 
  • #13
The Head said:
The -2yz term could be positive or negative, depending on which way you go in the y- and z-directions. Which intuitively feels like a saddle point, but I could be wrong there.
I think you've got it. Suppose you go in the y=z direction. Then suppose you go in the y=-z direction.
 
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Related to Maximization of a Multivariable Function

1. What is the purpose of maximizing a multivariable function?

The purpose of maximizing a multivariable function is to find the maximum value of the function within a given range of input variables. This can be useful in various fields such as economics, engineering, and physics, where finding the optimal solution is crucial.

2. How is the maximum value of a multivariable function determined?

The maximum value of a multivariable function is determined by taking the partial derivatives of the function with respect to each variable and setting them equal to zero. This results in a system of equations that can be solved to find the critical points, which are then evaluated to determine the maximum value.

3. What is the difference between local and global maximum in a multivariable function?

A local maximum is a point where the function has a higher value than all the points immediately surrounding it, while a global maximum is the highest value of the function within the entire range of input variables. In other words, a global maximum is also a local maximum, but a local maximum may not be a global maximum.

4. Can a multivariable function have more than one maximum value?

Yes, a multivariable function can have multiple maximum values. This can occur when there are multiple critical points that have the same maximum value. It can also happen when the function has multiple local maximum points, but only one global maximum.

5. How is the maximum value affected by the range of input variables?

The maximum value of a multivariable function can be affected by the range of input variables. If the range is too small, the function may not have a maximum value within that range. On the other hand, if the range is too large, the maximum value may occur at a point outside of the specified range. It is important to carefully choose the range of input variables to accurately determine the maximum value of the function.

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