max/min distance from surface to origin?

Question

can you help me find the points on the surface x^2 * y^2 * z = 1 that are closest to the origin?

Answer 1

z = 1/(x^2 y^2)

Because of the symettry let's use cylindrical coordinates.
x = rcosθ
y = rsinθ
z = 4/(r^4 sin^2 (2θ))

R = distance between any point and the origin
R^2 = r^2 + z^2
R^2 = r^2 + 16 / (r^8 sin^4 (2θ))

Now we want to set d(R^2)/dθ and d(R^2)/dr (partial derivatives) equal to zero

d(R^2)/dθ = -16*8*cos(2θ) / r^8 sin^5 (2θ)
= 0 means that cos(2θ) = 0
θ = π/4 (as well as other angles).
This means sin^4 (2θ) = 1

d(R^2)/dr = 2r - 8*16/r^9 sin^4 (2θ)
= 0 means
r^10 = 64
r = 2^(1/10) * sqrt(2)

x = 2^(1/10), y = 2^(1/10), z = 2^(-2/5)

Because of the other angles, all the points are
[ +/- 2^(1/10), +/- 2^(1/10), 2^(-2/5 ]

Answer 2

Let's minimize the square of the distance, which will accomplish the same thing (and avoid derivatives of square roots!) I will denote this square-of-the-distance by D.

D = x² + y² + z²

Surface: z = 1/(x² y²) = x^(-2) y^(-2) so z² = x^(-4) y^(-4)

D = x² + y² + x^(-4) y^(-4)

∂D/∂x = 2x - 4x^(-5) y^(-4)
∂D/∂y = 2y - 4x^(-4) y^(-5)

∂D/∂x = 0 -> x = 2x^(-5) y^(-4), or x^6 y^4 = 2
∂D/∂y = 0 -> y = 2x^(-4) y^(-5), or x^4 y^6 = 2

Therefore x^6 y^4 = x^4 y^6, so x² = y², so x = ±y
(I could legitimately divide by x^4 y^4 here since no point on x²y²z=1 can have x=0 or y=0).

Hence, x^6 y^4 = 2 -> x^10 = 2, so x = ±2^(1/10)

There are four solution points, all the same distance from the origin. There's a test to confirm that these are minima.