# Differentiation Problem!

• June 6th 2009, 07:19 AM
puggie
Differentiation Problem!
The variables u and v take positive values such that 1/u + 1/v = 1/20. If s = u + v, find the least value of s as u varies. Thanks
• June 6th 2009, 07:40 AM
Prove It
Quote:

Originally Posted by puggie
The variables u and v take positive values such that 1/u + 1/v = 1/20. If s = u + v, find the least value of s as u varies. Thanks

I don't really see how this is a differentiation problem.

$\frac{1}{u} + \frac{1}{v} = \frac{u + v}{uv}$.

So $\frac{u + v}{uv} = \frac{1}{20}$.

Therefore $u + v = 1$ and $uv = 20$.

Solving these equations simultaneously gives...

$u = 1 - v$

$(1 - v)v = 20$

$v - v^2 = 20$

$0 = v^2 - v + 20$

This value does not have a solution... (check it's discriminant).
• June 6th 2009, 07:44 AM
skeeter
Quote:

Originally Posted by puggie
The variables u and v take positive values such that 1/u + 1/v = 1/20. If s = u + v, find the least value of s as u varies. Thanks

$\frac{1}{u} + \frac{1}{v} = \frac{1}{20}$

$\frac{uv}{u+v} = 20$

$\frac{u(s-u)}{s} = 20$

$s = \frac{u^2}{u - 20}$

find $\frac{ds}{du}$ and minimize
• June 6th 2009, 09:43 AM
HallsofIvy
Quote:

Originally Posted by Prove It
I don't really see how this is a differentiation problem.

$\frac{1}{u} + \frac{1}{v} = \frac{u + v}{uv}$.

So $\frac{u + v}{uv} = \frac{1}{20}$.

Therefore $u + v = 1$ and $uv = 20$.

You are arguing that because a/b= c/d, "therefore" a= c and b= d. That is not true! 2/4= 1/2 but $2\ne 1$ and $4\ne 2$.

Quote:

Solving these equations simultaneously gives...

$u = 1 - v$

$(1 - v)v = 20$

$v - v^2 = 20$

$0 = v^2 - v + 20$

This value does not have a solution... (check it's discriminant).