Question

Positive numbers that minimize x+48/x

Answer 1

I was only able to find one.

Answer 2

Ok, sorry it took awhile, had to clean it up a little.. hope it's clear.

First, let's find the critical points of the functions
\[
v(x) = x + \frac{48}{x} = x + 48 \cdot x^{-1}\\
v'(x) = 1 - 48 \cdot x^{-2} \\
v'(x) = 0 \quad \implies \quad 1 - 48\cdot x^{-2} = 0 \quad \implies \quad x^2 = 48 \\
x = \sqrt{48} = \pm \sqrt{16 \cdot 3} = \pm 4 \sqrt{3} \\
\]
Ok, we have critical pionts at \(x= \pm 4\sqrt{3}\)
We only care about the positive one.
Let's check that it's minimum with second derivative
\[
y''(x) = (-2) \cdot (-48) \cdot x^{-3} = 96x^{-3}\\
x > 0 \implies x = 4 \sqrt{3} = 6.92 \\
x > 0 \quad \implies \quad 96 \cdot x^{-3} > 0 \quad \implies \quad y''(x) > 0
\]
Since the second derivative is positive when x>0 it means that it's also true for the critical point we found, and that means it's a minimum point (if you need explanation about this let me know).
And therefore, that critical point is what we're looking for.
Supports:
http://www.wolframalpha.com/input/?i=x+%2B+%2848%2Fx%29

Answer 3

Alright, I understand that. Makes a lot of sense, but the problem is asking for two solutions, both of which need to be positive. Do you have any idea how to find the second?

Answer 4

I should have wrote \(x = \pm \sqrt{48} \) sry.

Answer 5

Thanks so much for the time and effort by the way.

Answer 6

Eh np hehe. as long as you don't leave in the middle i don't mind. lemme read sec

Answer 7

what is dne?

Answer 8

The abbreviation for "Does not exist"

neither it nor -\[\sqrt{48}\] worked.

Answer 9

maybe try opposite order? DNE for smaller one idk..

Answer 10

This is my last attempt at the problem, and is gunna be the final answer. If I do get it backwards then I could be out half the points =/.

Answer 11

But the only answer is definitely \[\sqrt{48}\] then? The only issue is how I'm supposed to enter the other?

Answer 12

Problem is... even a limit doesn't work
\[
\lim_{x \to \infty} x + \frac{48}{x} = \infty + \frac{48}{\infty} = \infty + 0 = \infty \\
\lim_{x \to 0^+} x + \frac{48}{x} = 0^+ + \frac{48}{0^+} = 0 + \infty = \infty
\]

Answer 13

hmmmm.... that's what i'm trying to understand. The only answer, even wolfram says... is a minimum at \(\sqrt{48}\).. how to type it? no idea

Answer 14

Alright then. I really apprecaite the help dude. Thanks so much.

Answer 15

sure np, good luck there =|