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Smallest Value (Posted on 2007-03-27) Difficulty: 3 of 5
Find the smallest possible value of

f =
x4+y4-k

x2-y2

in terms of the constant k, given that xy=k, and x>y>1.

See The Solution Submitted by Dennis    
Rating: 3.3333 (3 votes)

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Solution Correct solution (with hand-waving) | Comment 3 of 6 |
substitute y = x/k and multiply by x^4

Then f = (x^8 +k^4 -kx^4)/(x^2*(x^4-k^2))

          = ((x^8 -2kkxx -k^4) + (2kkx^4-kx^4))/(x^2*(x^4-k^2))

          = (x^4 - k^2)/x^2   + (x^2/(x^4 - k^2))*(2k^2 - k)

Let a = (x^4 - k^2)/x^2

Then f = a + (2k^2 - k)/a

I happen to know (and a little calculus will prove) that the function is minimized when the two terms are equal, i.e. when a = sqrt(2K^2-k).

At that point, f = 2*sqrt(k(2k-1)), which is the solution to the problem.

I have checked this (using Excel) for the case when k = 4.  The function does indeed have a minimum value of 2sqrt(28) = 4*sqrt(7) = roughly 10.583005.
At this point, y is roughly 1.4663 and x is roughly 2.727955.

I have also checked when k = 25, and the minimum f is indeed 70 (this is the only rational pair I found).  At this point,  y is roughly 3.607816 and x is roughly 6.9294

Edited on March 29, 2007, 6:13 am

Edited on March 29, 2007, 6:28 am
  Posted by Steve Herman on 2007-03-28 22:14:43

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