# Calculating Extrema on Surface of Sphere

1. Mar 12, 2010

### Hart

1. The problem statement, all variables and given/known data

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.]

2. Relevant equations

Within the problem question statement and solutions.

3. The attempt at a solution

.. not really sure how to start this

2. Mar 12, 2010

### nickjer

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. Mar 12, 2010

### 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.

4. Mar 12, 2010

### Hart

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?

Last edited: Mar 12, 2010
5. Mar 12, 2010

### 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.

6. Mar 12, 2010

### Hart

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?

7. Mar 12, 2010

### 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.

8. Mar 12, 2010

### Hart

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

9. Mar 12, 2010

### nickjer

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.

Last edited: Mar 12, 2010
10. Mar 13, 2010

### Hart

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

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

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

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

11. Mar 13, 2010

### 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.

12. Mar 13, 2010

### Hart

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?

13. Mar 13, 2010

### 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.

14. Mar 13, 2010

### Hart

so:

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

?

15. Mar 13, 2010

### phyzguy

Lookin' good!

16. Mar 13, 2010

### Hart

.. 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

Last edited: Mar 13, 2010
17. Mar 13, 2010

### 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.

18. Mar 13, 2010

### Hart

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?

19. Mar 13, 2010

### Hart

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

20. Mar 13, 2010

### nickjer

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.