Optimization Problem

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.

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tiny-tim
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

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.

tiny-tim
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 …

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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tiny-tim
Homework Helper
No, 0.99 - 0.992 = 0.99 - 0.9801 = 0.0099.

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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tiny-tim