# Optimization Problem

Lorek

## Homework Statement

Find the positive number that exceeds its square by the largest amount. Obviously this is on the open interval (0,1).

## The Attempt at a Solution

$$F(x) = ( \frac{1}{n} ) ^2 - n \Rightarrow F'(x) = \frac{-2}{n^3} - 1 = 0 \Rightarrow 1 = \frac{-2}{n^3} \Rightarrow n^3 = -2$$

This is where I'm stuck. Taking the Cubic Root of a Negative number only nets
me a complex number. The answer at the back says its 1/2.

## Answers and Replies

Homework Helper
Hi Lorek! Find the positive number that exceeds its square by the largest amount. Obviously this is on the open interval (0,1).

why are you using 1/n2 instead of x2 ?? but if you must, you should use 1/n also, not n Lorek
I don't understand why you need to use $$\frac{1}{x}$$ instead of x.

Its asking for the positive number that is greater than its square at its maximum.

$$x^2 = x$$

Since its number can only be greater than its square as a fraction I used (1/x)^2.

Your saying I should set it up as
$$\frac{1}{x^2} - \frac{1}{x}$$ ?

I don't understand why though.

Homework Helper
Hi Lorek! Sorry, I don't understand any of your reasoning. It's much easier to use the number x itself than its reciprocal, ie to differentiate x2 - x …

if you must use the reciprocal, y = 1/x, then it's 1/y2 - 1/y …

you'll just have to think about this until you see it Lorek
We're looking for the largest number x that is greater than its square.

Setting the problem up as $$x^2 -x$$ and then differentiating says its maximum
is at 1/2.

The problem I see with that setup is if you take 0.99 and square it you get .9801. (original number larger than its square)
0.99 is larger than 1/2 (0.5).
Since we're looking for the largest number x that is greater than its square, and .99 falls into that category seeing as .99^2 = .9801.
0.99 > 0.9801 so it qualifies. Shouldn't the answer be The limit as x -> infinity of 1/1+(1/x).

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Homework Helper
No, 0.99 - 0.992 = 0.99 - 0.9801 = 0.0099. Lorek
x^2 = x. We're looking for the largest number that that holds true.

rearranging we got x^2-x = 0.

Plugging in 0.99 to x nets us:
0.99^2 - 0.99 = -0.0099 (not quite 0 but pretty close)

Since -0.0099 != 0 that's a problem.

.5 ^2 - .5 = -.25
-.25 != 0. (no where near 0).

The closest number which gets to x^2 = x is $$\lim_{x \rightarrow \infty} \frac{1}{1 + \frac{1}{x}}$$

Update: I'm heading in to bed. I'll just ask my math teacher when I see him tomorrow. Its a tough problem for being problem #1 in the book.

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