# minimum value

• Oct 27th 2008, 08:56 PM
minimum value
Find the value of p which makes the value of p + 1/p a minimum
• Oct 27th 2008, 09:04 PM
U-God
Quote:

Originally Posted by mathaddict
Find the value of p which makes the value of p + 1/p a minimum

I would attempt it like this:

let $\displaystyle f(p) = p + \frac{1}{p}$

$\displaystyle f'(p) = 1 - \frac{1}{p^2}$

for stationary point let $\displaystyle f'(p) = 0$

this implies that $\displaystyle \frac{1}{p^2} = 1$

so $\displaystyle p = \pm 1$
• Oct 27th 2008, 09:12 PM
Chris L T521
Quote:

Originally Posted by U-God
I would attempt it like this:

let $\displaystyle f(p) = p + \frac{1}{p}$

$\displaystyle f'(p) = 1 - \frac{1}{p^2}$

for stationary point let $\displaystyle f'(p) = 0$

this implies that $\displaystyle \frac{1}{p^2} = 1$

so $\displaystyle p = +/- 1$

Just a friendly suggestion. :)

You can generate the plus or minus sign in LaTeX, using the command \pm. It yields $\displaystyle \pm$

You can also generate a minus or plus sign in LaTeX, use the command \mp. It yields $\displaystyle \mp$.

--Chris
• Oct 29th 2008, 04:52 PM
ThePerfectHacker
Quote:

Originally Posted by mathaddict
Find the value of p which makes the value of p + 1/p a minimum

If $\displaystyle p>0$ then:
$\displaystyle p+\frac{1}{p} \geq 2\left( p \right)\left( \frac{1}{p} \right) = 2$ by AM-GM.
Now for $\displaystyle p=1$ we have a minimum.