# Help with complex systems of equations

• Sep 20th 2011, 08:48 PM
DanielLiwicki
Help with complex systems of equations
Hi All
These equations are the resulte from a lagrange multiplier optimization problem. I can not get the algebra right! Please help!

280*pi*r + 80*pi*hi - 2*lamda*r*h = 0
80*pi*r - lamda*pi*r^2=0
-pi*r^2*h + 100=0
• Sep 20th 2011, 09:30 PM
CaptainBlack
Re: Help with complex systems of equations
Quote:

Originally Posted by DanielLiwicki
Hi All
These equations are the resulte from a lagrange multiplier optimization problem. I can not get the algebra right! Please help!

280*pi*r + 80*pi*hi - 2*lamda*r*h = 0
80*pi*r - lamda*pi*r^2=0
-pi*r^2*h + 100=0

What is hi?

CB
• Sep 20th 2011, 09:32 PM
DanielLiwicki
Re: Help with complex systems of equations
pi not hi ....sorry
• Sep 21st 2011, 06:37 AM
HallsofIvy
Re: Help with complex systems of equations
Lagrange multiplier problems typically give something like $f(x,y,z)= \lambda g(x,y,z)$, $h(x,y,z)= \lambda k(x,y,z)$, etc. I have found that it is often best to divide one equation by another, immediately eliminating $\lambda$ (which is not part of the solution, any way).

Here, you have $40\pi(7r+ \pi)= 2\lambda r h$ and $80\pi r= \lambda \pi r^2$
Dividing the first equation by the second,
$\frac{7r+ \pi}{2r}= \frac{2h}{r}$
so that $7r+ \pi= 4h$.

You can solve that for either r or h in terms of the other and put into the final equation.

(Are you sure about the "hi= pi"? It seems strange that you would write "pi*pi" rather than "pi^2" and also strange that "pi^2" would show up in such a problem.)
• Sep 21st 2011, 06:58 AM
DanielLiwicki
Re: Help with complex systems of equations
Your Right!!! Not hi...not pi... but just h. (Clapping)
• Sep 21st 2011, 07:21 AM
HallsofIvy
Re: Help with complex systems of equations
So you have 7r+ h= 4h.
• Sep 21st 2011, 07:56 AM
DanielLiwicki
Re: Help with complex systems of equations
It worked out! Thank you!