Lagrange Multipliers II
As I head out for vacation, here is a simple question. What is the distance from the plane
ax + by + cz = d
to the origin? The answer is a simple click away (Mathworld). For fun, suppose we also wanted to know the coordinates of the point on the plane closest to the origin. This can be solved as a constrained optimization problem, serving as another example of the technique of Lagrange multipliers.
Informally, we wish to minimize x2 + y2 + z2 subject to the constraint ax + by + cz – d = 0. Relax the constraint into the objective and form the function,
L(x,y,z) = x2 + y2 + z2 – λ(ax + by + cz – d) .
The stationary points are determined from
δL/δx = 2x – aλ = 0 , or x = aλ/2 
δL/δy = 2y – bλ = 0 , or y = bλ/2 
δL/δz = 2z – cλ = 0 , or z = cλ/2 
δL/δλ = 0 implies the original constraint, ax + by + cz = d 
Subsititue x, y, and z from [1-3] into  to obtain λ = 2d/(a2 + b2 + c2) .
Substitute the λ value from  into [1-3] to obtain the coordinates of the point closest to the origin,
e = a2 + b2 + c2
x*= ad/e, y*= bd/e, z*= cd/e . The distance from (x*, y*, z*) to the origin is
D = [((ad)2 + (bd)2 + (cd)2)/e2]1/2
= d(e/e2)1/2 or D = d/√(a2 + b2 + c2), which matches the result from analytic geometry. Of course, calculus is not necessary (or even convenient) for this derivation, but it is another good illustration of Lagrange multipliers.
Formally proving that the stationary point represents a minimum is well beyond a blog post, but intuitively you can imagine that the distance from the origin to any point on the plane can be made arbitrarily large. The stationary point can not represent a maximum distance.