SOLUTION: Find a positive integer for which the sum of its recipricol and four times its square is the smallest possible.

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Question 883627: Find a positive integer for which the sum
of its recipricol and four times its square is the smallest
possible.

Found 2 solutions by stanbon, josgarithmetic:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find a positive integer for which the sum
of its recipricol and four times its square is the smallest
possible.
---------
S = 1/x + 4x^2
------
Take the derivative to get:
S' = -1/x^2 + 8x
------
Solve:: -1/x^2 + 8x = 0
8x^3 = 1
x = 1/2
================
Cheers,
Stan H.
================

Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Integer to find, n. Is this for a Calculus course, or College Algebra/Intermediate Algebra?

y=1%2Fn%2B4n%5E2 and you want y as small as possible; and n must be an integer. Use derivative, dy%2Fdn. This should be 0 at any extreme value for y, including any minimum.

dy%2Fdn=-n%5E%28-2%29%2B8n
8n-1%2Fn%5E2
8n%5E3%2Fn%5E2-1%2Fn%5E2
%288n%5E3-1%29%2F%28n%5E2%29
-
Finding extreme values for y
%288n%5E3-1%29=0 ignoring the denominator because it has no importance for equating the derivative to zero.
Based on the difference of cubes formula, highlight%28%28%282n%29-1%29%28%282n%29%5E2%2B2n%2B1%29=0%29.

This would have three possible solutions for n, but maybe only one of them would be our minimum n for y.

The binomial:
2n-1=0
2n=1
n=1%2F2

The quadratic:
4n%5E2%2B2n%2B1=0
Discriminant, 2%5E2-4%2A4%2A1=4-16=-12%3C0, not a real number.

Only that one, real, extreme value so must be our input for the minimum for y.
We want an INTEGER, so the nearest ones to 1%2F2 are 0 and 1.
-
In fact for smallest POSITIVE value for y, choose n=1.