# Calculating Extrema on Surface of Sphere

## Homework Statement

Considering the surface of a sphere of radius $1$ with its centre at coordinates
$(0,0,0)$.

For the function:

$$f(x,y,z) = x^{3} + y^{3} + z^{3}$$

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

Related Advanced Physics Homework Help News on Phys.org
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.

phyzguy
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.

Ermm.. so need to do this:

$$f(x,y) = x^{3} + y^{3} + z^{3}$$

$$g(x,y) = x^{2} + y^{2} + z^{2} = 1$$

Hence need to then extremise:

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

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

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

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

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

.. any good so far?

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phyzguy
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.

Ok, good.

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

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

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

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

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

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

.. correct now?

phyzguy
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.

Ah ok, so it should be:

$$\frac{\partial F}{\partial \lambda} = x^{2} + y^{2} + z^{2} - 1 = 0$$

yes?

.. and then how to I go about getting values of x, y, and $\lambda$? Not sure how to equate those 4 simulataneous equations

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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$$\frac{\partial F}{\partial x} = 3x^{2} + 2x \lambda = 0$$

$$\frac{\partial F}{\partial y} = 3y^{2} + 2y \lambda = 0$$

$$\frac{\partial F}{\partial z} = 3z^{2} + 2z \lambda = 0$$

$$\frac{\partial F}{\partial \lambda} = x^{2} + y^{2} + z^{2} - 1 = 0$$

phyzguy
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.

I can only see how to solve those first three equations in terms of a quadratic? I.e. that:

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

.. is this the right way to go?

phyzguy
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.

so:

$$3x + 2\lambda = 0 \implies x = \frac{-2\lambda}{3}$$

?

phyzguy
Lookin' good!

.. so from the first three equations I get:

$$x = \frac{-2\lambda}{3}$$

$$y = \frac{-2\lambda}{3}$$

$$z = \frac{-2\lambda}{3}$$

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

$$\frac{4\lambda^{2}}{9} + \frac{4\lambda^{2}}{9} + \frac{4\lambda^{2}}{9} - 1 = 0 \implies \frac{4\lambda^{2}}{3} - 1 = 0$$

.. then rearrange to get in terms of $\lambda$ hence:

$$\lambda = \sqrt{\frac{3}{4}} = \pm\frac{\sqrt{3}}{2}$$

.. correct so far??

So now I know the values of $x , y , z , \lambda$. Great.

.. not too sure what to do now though

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phyzguy
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.

OK, so I now know the values of $x , y , z$ and of $\lambda$ which should be:

$$\lambda = \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$$

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 $\lambda$ again? The values of y and z will still be the same, but will have:

$$\lambda = \sqrt{\frac{9}{8}} = \pm \frac{3\sqrt{2}}{4}$$

.. yes?

.. did you post a reply? I got a notification email but it's not showing up here

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.

So as an answer to the first part of the question, I get that the extrema are:

$$f(x,y,z) = \left(\frac{-2\lambda}{3} , \frac{-2\lambda}{3} , \frac{-2\lambda}{3}\right)$$

since $$\lambda = \frac{\sqrt{3}}{2}$$

.. 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 $\lambda$ to calculate new values for y and z (since x = 0):

$$y = \pm \frac{\sqrt{3}}{3}$$

$$z = \pm \frac{\sqrt{3}}{3}$$

where $$\lambda = \frac{\sqrt{3}}{2}$$

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

$$f(x,y,z) = 0 , \frac{-2\lambda}{3} , \frac{-2\lambda}{3}\right)$$

.. not sure if this is going right?

.. oh wait, for say the first one do I get:

$$f_{1}(x,y,z) = \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3}$$

$$f_{2}(x,y,z) = \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}$$

$$f_{3}(x,y,z) = \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3}$$

$$f_{4}(x,y,z) = \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}$$

$$f_{5}(x,y,z) = -\frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3}$$

$$f_{6}(x,y,z) = -\frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}$$

$$_{7}f(x,y,z) = -\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{3}$$

$$f_{8}(x,y,z) = -\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}$$

.. 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.