application of partial derivat

Let distance d = √(x^2 + y^2 + z^2). We want to minimize f(x,y,z) = x^2 + y^2 + z^2 with constraint g(x,y,z) = z^2 - xy - 1. Using Lagrange multiplier mwthod with L = f(x,y,z) - λg(x,y,z),∂L/∂x = 2x - λy...(1)∂L/∂y = 2y - λx...(2)∂L/∂z = 2z - 2λz...(3)∂L/∂λ = z^2 - xy - 1...(4)Set (1) - (4) equal to 0. From (3) z = (1 - λ). So, either z = 0 or λ = 1. If z = 0, then we have xy = -1 and so x = -1/y. From (1) -2/y = -λy => y^2 = 2/λ => x^2 =2/λ. So x^2 = y^2 => x = ±y. By (4) x = -y where y = ±1. So the critical points are (1, -1, 0) or (-1, 1 , 0) with distance √2. If λ = 1, then (1) and (2) implie that x = (1/4)x and so x = y = 0. Hence z = -1 or 1. That is, the critical points are (0, 0, -1) or (0, 0 , 1) with distance 1. We conclude that (0, 0, -1) or (0, 0 , 1) are the global minima with minimum distance 1.

The distance from any point (x, y, z) to the origin is given by:

d = square root of (x^2 + y^2 + z^2)

So with z^2 = xy + 1, we have:

d = square root of (x^2 + y^2 + xy + 1)

Then consider f(x, y) = x^2 + y^2 + xy + 1:

2011-01-17 15:04:26 補充：
fx = 2x + y

fy = x + 2y

fxx = fyy = 2

fxy = 1

So with fx = fy = 0, we have x = y = 0. Also fxx fyy - (fxy)^2 = 3 > 0

Thus (0, 0) gives a min. value of f(x, y) and hence the points are (0, 0, 1) and (0, 0, -1).