Maximization of a Multivariable Function

[The Head]

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.

 

[haruspex]

grad(f)=0 at extrema, we can see (0,0,0) is a point

But not necessarily a max or min?

Substitute my parameters into f=x^2-2yz, and take the derivative.

Sounds like a good idea. What do you get?

 

[The Head]

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

 

[The Head]

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).

 

[haruspex]

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

Yes. Looks good so far.

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

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).

 

[The Head]

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?

 

[haruspex]

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?

 

[The Head]

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.

 

[haruspex]

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.

 

[FactChecker]

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 Head]

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.

 

[The Head]

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.

 

[FactChecker]

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.