• Support PF! Buy your school textbooks, materials and every day products Here!

Optimization Problem

  • Thread starter Lorek
  • Start date
  • #1
8
0

Homework Statement



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

Homework Equations



The Attempt at a Solution



[tex] 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 [/tex]

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

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
249
Hi Lorek! :smile:
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 ?? :confused:

but if you must, you should use 1/n also, not n :wink:
 
  • #3
8
0
I don't understand why you need to use [tex] \frac{1}{x} [/tex] instead of x.

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

[tex] x^2 = x [/tex]

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
[tex] \frac{1}{x^2} - \frac{1}{x} [/tex] ?

I don't understand why though.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
249
Hi Lorek! :smile:

Sorry, I don't understand any of your reasoning. :redface:

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 :smile:
 
  • #5
8
0
We're looking for the largest number x that is greater than its square.

Setting the problem up as [tex] x^2 -x [/tex] 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).
 
Last edited:
  • #6
tiny-tim
Science Advisor
Homework Helper
25,832
249
No, 0.99 - 0.992 = 0.99 - 0.9801 = 0.0099. :wink:
 
  • #7
8
0
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 [tex] \lim_{x \rightarrow \infty} \frac{1}{1 + \frac{1}{x}} [/tex]

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.
 
Last edited:
  • #8
tiny-tim
Science Advisor
Homework Helper
25,832
249
Find the positive number that exceeds its square by the largest amount.
x^2 = x. We're looking for the largest number that that holds true.
No, you're looking for the largest value of x - x2.
 

Related Threads for: Optimization Problem

  • Last Post
Replies
3
Views
583
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
944
  • Last Post
Replies
0
Views
931
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
2
Replies
36
Views
3K
  • Last Post
Replies
3
Views
2K
Top