# confusing lagrange multiplier question

• Aug 18th 2010, 01:19 AM
tsang
confusing lagrange multiplier question
f(x,y,z)=xy+z^3
subject to x^2+y^2+z^2=1
Find all the absolute maximum and minimum.

I tried to solve the simutaneous equations, and I ended up with 14 critical points! (Headbang)
So I doubt if I did right or wrong.
Can anybody please help me and see how many of them? What are they? I solved that x=o, plus or minus square root 2 on 2, plus or minus 2/3
y=0, plus or minus square root 2 on 2, plus or minus 2/3
z=0, plus or minus 1, plus or minus 1/3
Which rearrange them, I end up with 14 critical points. Am I right? Thanks a lot.
• Aug 18th 2010, 10:50 AM
Vlasev
The equations are

$\displaystyle y=2 x \lambda$
$\displaystyle x=2 y \lambda$
$\displaystyle 3 z^2=2 z \lambda$
$\displaystyle x^2+y^2+z^2=1$

Right?
If $\displaystyle x = 0$, so is $\displaystyle y = 0$. Then from the last equation $\displaystyle z = +1$ or $\displaystyle z = -1$. These are 2 points.
If $\displaystyle z = 0$, then you get $\displaystyle x = \sqrt{2}/2$ or $\displaystyle x = -\sqrt{2}/2$ and $\displaystyle y = \sqrt{2}/2$ or $\displaystyle y = -\sqrt{2}/2$. All four combinations are valid, so you get another 4 points.

The last option is if none of them are 0. Then \lambda is non-zero also because if it was, then we'd get one of $\displaystyle x,y,z = 0$. If you solve this you should get 4 solutions where $\displaystyle x = 2/3$, $\displaystyle y = 2/3$, $\displaystyle z = 1/3$ (where some of the signs change, i.e. $\displaystyle z = 1/3$, $\displaystyle x = 2/3$, $\displaystyle y = -2/3$)

This is a grand total of 10 solutions. The way you did it is incorrect.