# Lagrange multipliers Q

1. Feb 20, 2006

### ElDavidas

I'm stuck on the following question

"Find the maximum and minimum values of f(x,y,z) = $$x^2y^2-y^2z^2 + z^2x^2$$ subject to the constraint of $$x^2 + y^2 + z^2 = 1$$ by using the method of lagrange multipliers.

Write the 4 points where the minimum value is achieved and the 8 points where the max is achieved"

So far, I've managed to calculate the gradient for both formulae and have the resulting equations:

$$2xy^2 + 2z^2x = \lambda(2x)$$

$$2x^2y - 2z^2y = \lambda(2y)$$

$$2x^2z - 2y^2z = \lambda(2z)$$

Do I just substitute these equations in turn into the constraint equation and determine what the points are that satisfy it?

Also, how do you find out if a point is a local max/min when you know the critical point when looking at a case like this?

Last edited: Feb 20, 2006
2. Feb 20, 2006

### HallsofIvy

Staff Emeritus
You should be able to simplify those. Obviously, x= 0 makes the first of those equations true. If x is not 0, then the first equation is $y^2+ z^2= \lamba$.

Similarly, y= 0 satisfies the second equation. If y is not 0, then the second equation is $x^2- z^2= \lamda$

z= 0 satisfies the third equation. If z is not 0, then the third equation is $x^2- y^2= \lambda$.

Setting $x^2- z^2= x^2- y^2$ gives $z^2= y^2$ or z= y. Similarly, setting $y^2+ z^2= x^2- y^2$ gives z= x. Can you finish from here?

3. Feb 20, 2006

### ElDavidas

Ok, I can follow you up to a certain point.

For whenever x,y or z = 0, I get the following critical points:
(1,0,0)
(-1,0,0)
(0,1,0)
(0,-1,0)
(0,0,1)
(0,0,-1)

I get $z^2 = \frac{x^2} {3}$ here.

I then get a bit lost from now on. By substitution and a number of steps I get $z = \frac {1} {\sqrt{5}}$ and $z = -\frac {1} {\sqrt{5}}$. Do I just substitute that in and get the critical points?

4. Feb 20, 2006

### arildno

hmm..you've missed out quite a few points:
$$2x(y^{2}+z^{2}-\lambda)=0$$
$$2y(x^{2}-z^{2}-\lambda)=0$$
$$2z(x^{2}-y^{2}-\lambda)=0$$
$$x^{2}+y^{2}+z^{2}=1$$
If you set x=0, you get in addition to your own points the following 4:
$$(0,\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}}), (\lambda=-\frac{1}{2})$$

Having dealt with those, note that from combining the first and fourth equation, we must have:
$$x^{2}=1-\lambda$$
(Replacing the first and eliminating x from further consideration)
$$2y(1-z^{2}-2\lambda)=0$$
$$2z(1-y^{2}-2\lambda)=0$$
$$y^{2}+z^{2}=\lambda$$
You can now proceed further with these three equations.

If I've counted correctly, there are a total of 18 (x,y,z) solutions here, in addition to the 8 previous ones.

Last edited: Feb 20, 2006
5. Feb 20, 2006

### HallsofIvy

Staff Emeritus
Oops, you are right! Now you know y= z and x2= 3z2. Now put those into your constraint:
$x^2+ y^2+ z^2= 3z^2+ z^2+ z^2= 5z^2= 1$
So, yex, $z= \pm\frac{1}{\sqrt{5}}$.

Now use y= z and x2= 3z2 to find corresponding values of x and y.

6. Feb 21, 2006

### arildno

FYI, the max/min is $\pm\frac{1}{4}$

7. Feb 21, 2006

### benorin

Since $$f(x,z,y) = x^2z^2-z^2y^2 + y^2x^2 = x^2y^2-y^2z^2 + z^2x^2 = f(x,y,z)$$

which is to say that f is symmetric in the variables y and z, doesn't it follow that the extrema will have y=z?

8. Feb 21, 2006

### arildno

Nope; what follows is that for any particular extremum with one choice of y and z, there must also be another extremum with their values commuted.
The 26 critical points are:
$$(\pm{1},0,0), (0,\pm{1},0), (0,0,\pm{1}),(\pm\sqrt{\frac{3}{5}},\pm\sqrt{\frac{1}{5}},\pm\sqrt{\frac{1}{5}}),(0,\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}}),(\pm\frac{1}{\sqrt{2}},0,\pm\frac{1}{\sqrt{2}}),(\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}},0)$$
The last 12 are the extrema.