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

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- Sep 20th 2011, 07:48 PMDanielLiwickiHelp 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, 08:30 PMCaptainBlackRe: Help with complex systems of equations
- Sep 20th 2011, 08:32 PMDanielLiwickiRe: Help with complex systems of equations
pi not hi ....sorry

- Sep 21st 2011, 05:37 AMHallsofIvyRe: Help with complex systems of equations
Lagrange multiplier problems typically give something like $\displaystyle f(x,y,z)= \lambda g(x,y,z)$, $\displaystyle h(x,y,z)= \lambda k(x,y,z)$, etc. I have found that it is often best to

**divide**one equation by another, immediately eliminating $\displaystyle \lambda$ (which is not part of the solution, any way).

Here, you have $\displaystyle 40\pi(7r+ \pi)= 2\lambda r h$ and $\displaystyle 80\pi r= \lambda \pi r^2$

Dividing the first equation by the second,

$\displaystyle \frac{7r+ \pi}{2r}= \frac{2h}{r}$

so that $\displaystyle 7r+ \pi= 4h$.

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, 05:58 AMDanielLiwickiRe: Help with complex systems of equations
Your Right!!! Not hi...not pi... but just h. (Clapping)

- Sep 21st 2011, 06:21 AMHallsofIvyRe: Help with complex systems of equations
So you have 7r+ h= 4h.

- Sep 21st 2011, 06:56 AMDanielLiwickiRe: Help with complex systems of equations
It worked out! Thank you!