Find all pairs of positive numbers (x,y) for which f(x,y) has a minimum. Find the smallest possible minimum.?

Question

f(x,y) = ((x^4)/(y^4)) + ((y^4)/(x^4)) - ((x^2)/(y^2)) - ((y^2)/(x^2)) + (x/y) + (y/x)

Answer 1

Let z = x/y + y/x. Then z^2 = x^2/y^2 + y^2/x^2 + 2
and (z^2 - 2)^2 = x^4/y^4 + y^4/x^4 + 2
So f(x, y) = (z^2 - 2)^2 - 2 - (z^2 - 2) + z = g(z)
= z^4 - 4z^2 + 4 - 2 - z^2 + 2 + z
= z^4 - 5z^2 + z + 4
g'(z) = 4z^3 - 10z + 1
= 0 at approximately -1.628949, 0.100405 and 1.528544.
We know from the shape of g'(z) that the first and last of these will be points where g'(z) changes from negative to positive (i.e. minima of g) and the middle will be a maximum of g.
So there are minima for all x, y where x/y + y/x = -1.628949 or x/y + y/x = 1.528544.

However, if we let r = x/y, we know that z = r + 1/r is at least 2 (for r positive) or at most -2 (for r negative)*. So all the variation in g(z) happens in a region z cannot be. So the minimum value will either be at r = 1 (z = 2) or at r = -1 (z = -2).
g(2) = 2^4 - 5.2^2 + 2 + 4 = 2
g(-2) = (-2)^4 - 5.(-2)^2 + (-2) + 4 = -2

So there are minima along the lines x=y (f(x, y) = 2) and x = -y (f(x, y) = -2), and the lowest possible value of f is -2 along the line x = -y.

* this is easy to check; first derivative is 1 - 1/r^2 = 0 for r = ± 1, and second derivative is 2/r^3 so there's a minimum at (1, 2) and a maximum at (-1, -2), with of course a discontinuity at 0.

Answer 2

if xy = 18, then y = 18/x so locate the minimum of 2x + 18/x. Take the derivative and set to 0. 2 + ( (0x - 18)/x^2) = 0 2 - 18/(x^2) = 0 -18/(x^2) = -2 2x^2 = 18 x^2 = 9 x = 3 So y = 6 the respond is (3,6) and the minimum is 12

Answer 3

x = y for all positive values of x and y.

Minimum f = 2

Answer 4

aply diff. and get teh answer