Calculating Extrema on Surface of Sphere

In summary: I can plug these into the last equation and solve for x,y,z to get my final answer.In summary, to find the extrema on the sphere with function f(x,y,z), you need to use Lagrange multipliers and take partial derivatives with respect to x, y, and z. You then need to solve for x, y, and z in terms of lambda.
  • #1
Hart
169
0

Homework Statement



Considering the surface of a sphere of radius [itex]1[/itex] with its centre at coordinates
[itex](0,0,0)[/itex].

For the function:

[tex]f(x,y,z) = x^{3} + y^{3} + z^{3}[/tex]

Need to find the following:

(i) All the extrema on the surface which have x, y and z all non-zero simultaneously.

(ii) All the extrema on the surface which have x = 0. [don’t need to say whether the extrema are maxima or minima.]

Homework Equations



Within the problem question statement and solutions.

The Attempt at a Solution



.. not really sure how to start this :frown:
 
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  • #2
You should start by using Lagrange multipliers since you have a function and constraints. Write out the Lagrange function and then take the partial derivatives of it.
 
  • #3
Hart,

This is a perfect application of the method of Lagrange multipliers. Try looking up Lagrange multipliers on Wikipedia - there is an excellent explanation. You want to extremize x3+y3+z3, subject to the constraint that x2+y2+z2 = 1.
 
  • #4
Ermm.. so need to do this:

[tex]f(x,y) = x^{3} + y^{3} + z^{3}[/tex]

[tex]g(x,y) = x^{2} + y^{2} + z^{2} = 1[/tex]

Hence need to then extremise:

[tex]F(x,y,\lambda) = f + \lambda y = x^{3} + y^{3} + z^{3} + \lambda \left(x^{2} + y^{2} + z^{2} - 1\right)[/tex]

So need to then calculate partial derivatives of this equation..

[tex]\frac{\partial F}{\partial x} = 3x^{2} + y^{3} + z^{3} + \lambda \left(2x + y^{2} + z^{2} - 1\right) = 0[/tex]

[tex]\frac{\partial F}{\partial y} = 3y^{2} + x^{3} + z^{3} + \lambda \left(2y + x^{2} + z^{2} - 1\right) = 0[/tex]

[tex]\frac{\partial F}{\partial \lambda} = x^{3} + y^{3} + z^{3} + \left(2y + x^{2} + z^{2} - 1\right) = 0[/tex]

.. any good so far?
 
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  • #5
You've got the idea, except for two things:

(1) You have a function of 4 variables, x,y,z, and lambda, so you need to calculate the partial derivative with respect to z as well.

(2) You made a mistake calculating the partial with respect to lambda in the last equation.

(3) Then you'll have four equations that you can solve for your four unknowns, x,y,z,lambda.
 
  • #6
Ok, good.

Yes, I just missed that, so I should have:

[tex]\frac{\partial F}{\partial x} = 3x^{2} + y^{3} + z^{3} + \lambda \left(2x + y^{2} + z^{2} - 1\right) = 0[/tex]

[tex]\frac{\partial F}{\partial y} = 3y^{2} + x^{3} + z^{3} + \lambda \left(2y + x^{2} + z^{2} - 1\right) = 0[/tex]

[tex]\frac{\partial F}{\partial z} = 3z^{2} + x^{3} + y^{3} + \lambda \left(2z + x^{2} + y^{2} - 1\right) = 0[/tex]

I was unsure of how to calculate the partial derivative with respect to lambda, maybe it should be this instead:

[tex]\frac{\partial F}{\partial \lambda} = x^{3} + y^{3} + z^{3} + \left(x^{2} + y^{2} + z^{2} - 1\right) = 0[/tex]

.. correct now?
 
  • #7
Still not quite right. When you take the partial with respect to lambda, all of the x3 terms disappear, since they are not multiplied by lambda.
 
  • #8
Ah ok, so it should be:

[tex]\frac{\partial F}{\partial \lambda} = x^{2} + y^{2} + z^{2} - 1 = 0[/tex]

yes?

.. and then how to I go about getting values of x, y, and [itex]\lambda[/itex]? Not sure how to equate those 4 simulataneous equations :confused:
 
  • #9
When you take the partial derivative with respect to x, y, or z... any of the terms without one of those variables multiplied to it go to 0. For example, you shouldn't have y^3 or z^3 terms in the partial derivative with respect to x, since there is no x multiplying to y^3 or z^3 in the Lagrange function.
 
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  • #10
[tex]\frac{\partial F}{\partial x} = 3x^{2} + 2x \lambda = 0[/tex]

[tex]\frac{\partial F}{\partial y} = 3y^{2} + 2y \lambda = 0[/tex]

[tex]\frac{\partial F}{\partial z} = 3z^{2} + 2z \lambda = 0[/tex]

[tex]\frac{\partial F}{\partial \lambda} = x^{2} + y^{2} + z^{2} - 1 = 0[/tex]
 
  • #11
Now you're in business! Now solve each of the first three to get x,y,z in terms of lambda, then plug these into the last equation and solve for lambda. Then once you know lambda, you can plug it into each of the first three to find x,y,z. You're almost there.
 
  • #12
I can only see how to solve those first three equations in terms of a quadratic? I.e. that:

[tex]\frac{\partial F}{\partial x} = 3x^{2} + 2x \lambda = 0 \implies x = \frac{-2\lambda \pm \sqrt{(2\lambda)^{2}}}{6} \implies x = -3\lambda \pm 2\lambda[/tex]

.. is this the right way to go?
 
  • #13
No. Since you are given that x is non-zero, you can divide the equation by x and convert it into a linear equation, which is easy to solve for x in terms of lambda.
 
  • #14
so:

[tex]3x + 2\lambda = 0 \implies x = \frac{-2\lambda}{3}[/tex]

?
 
  • #15
Lookin' good!
 
  • #16
.. so from the first three equations I get:

[tex]x = \frac{-2\lambda}{3}[/tex]

[tex]y = \frac{-2\lambda}{3}[/tex]

[tex]z = \frac{-2\lambda}{3}[/tex]

.. then square these values and input into the fourth equation, which gives:

[tex]\frac{4\lambda^{2}}{9} + \frac{4\lambda^{2}}{9} + \frac{4\lambda^{2}}{9} - 1 = 0 \implies \frac{4\lambda^{2}}{3} - 1 = 0[/tex]

.. then rearrange to get in terms of [itex]\lambda[/itex] hence:

[tex]\lambda = \sqrt{\frac{3}{4}} = \pm\frac{\sqrt{3}}{2}[/tex]

.. correct so far??

So now I know the values of [itex] x , y , z , \lambda [/itex]. Great.

.. not too sure what to do now though :frown:
 
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  • #17
The values of x,y,z at the extremal points is what you were asked for, so you're done. Just don't forget that when you take the square root of a number, it has both + and - solutions.
 
  • #18
OK, so I now know the values of [itex]x , y , z[/itex] and of [itex]\lambda[/itex] which should be:

[tex]\lambda = \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}[/tex]

SO that answers part i of the question (where x, y, and z are all non-zero), yep?

Now for part ii (where x = 0), I just do the same but with x = 0? I.e. calculating [itex]\lambda[/itex] again? The values of y and z will still be the same, but will have:

[tex]\lambda = \sqrt{\frac{9}{8}} = \pm \frac{3\sqrt{2}}{4}[/tex]

.. yes?
 
  • #19
.. did you post a reply? I got a notification email but it's not showing up here :confused:
 
  • #20
Yes, but I didn't see the 2nd page since the link in the email took me to the first page. So I deleted my comment since it was taken care of on the 2nd page.

But just to comment, your last solution is correct. You just need to plug it back in for y and z.
 
  • #21
So as an answer to the first part of the question, I get that the extrema are:

[tex] f(x,y,z) = \left(\frac{-2\lambda}{3} , \frac{-2\lambda}{3} , \frac{-2\lambda}{3}\right)[/tex]

since [tex]\lambda = \frac{\sqrt{3}}{2}[/tex]

.. is this sufficient to state just this? I thought I'd get a list of points maybe.

Then for the second part of the question, I have used the value of [itex]\lambda[/itex] to calculate new values for y and z (since x = 0):

[tex]y = \pm \frac{\sqrt{3}}{3}[/tex]

[tex]z = \pm \frac{\sqrt{3}}{3}[/tex]

where [tex]\lambda = \frac{\sqrt{3}}{2}[/tex]

And hence then similarly to the first part, the extrema:

[tex] f(x,y,z) = 0 , \frac{-2\lambda}{3} , \frac{-2\lambda}{3}\right)[/tex]

.. not sure if this is going right?
 
  • #22
.. oh wait, for say the first one do I get:

[tex]f_{1}(x,y,z) = \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3}[/tex]

[tex]f_{2}(x,y,z) = \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}[/tex]

[tex]f_{3}(x,y,z) = \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3}[/tex]

[tex]f_{4}(x,y,z) = \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}[/tex]

[tex]f_{5}(x,y,z) = -\frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3}[/tex]

[tex]f_{6}(x,y,z) = -\frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}[/tex]

[tex]_{7}f(x,y,z) = -\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3}[/tex]

[tex]f_{8}(x,y,z) = -\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}[/tex]

.. not sure if that's actually made things better. I understand what've going on, just having a bit of a confusion on what to actually write to answer the question.
 

What is the formula for calculating extrema on the surface of a sphere?

The formula for calculating extrema on the surface of a sphere is the Lagrange multiplier method. This method involves setting up a system of equations that includes the surface equation of the sphere, the objective function, and the constraint function. The extrema can then be found by solving the system of equations.

What is the significance of calculating extrema on the surface of a sphere?

Calculating extrema on the surface of a sphere can have various applications in fields such as physics, engineering, and mathematics. It can be used to optimize the design of structures, determine the maximum and minimum values of physical quantities, and solve optimization problems involving spherical surfaces.

Can extrema be calculated on any point of a sphere?

No, extrema can only be calculated on critical points of the surface of a sphere. These points are where the gradient of the objective function is parallel to the gradient of the surface equation of the sphere. Other points on the surface may not have extrema or may have saddle points.

How does the radius of the sphere affect the calculation of extrema?

The radius of the sphere affects the calculation of extrema by changing the constraint function in the Lagrange multiplier method. As the radius changes, the constraint function will also change, which will result in different critical points and extrema values.

Are there any limitations to calculating extrema on the surface of a sphere?

One limitation is that the Lagrange multiplier method can only be used for smooth surfaces, so it may not be applicable to spheres with discontinuities or sharp edges. Additionally, the method may become computationally intensive for higher dimensions, making it challenging to calculate extrema on higher-dimensional spheres.

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