Help with complex systems of equations

**DanielLiwicki** · September 20th 2011, 07:48 PM

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

**CaptainBlack** · September 20th 2011, 08:30 PM

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

**DanielLiwicki** · September 20th 2011, 08:32 PM

pi not hi ....sorry

**HallsofIvy** · September 21st 2011, 05:37 AM

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.)

**DanielLiwicki** · September 21st 2011, 05:58 AM

Your Right!!! Not hi...not pi... but just h.

**HallsofIvy** · September 21st 2011, 06:21 AM

So you have 7r+ h= 4h.

**DanielLiwicki** · September 21st 2011, 06:56 AM

It worked out! Thank you!

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