Lagrange multipliers?

Question

find the shortest distance form the origin to the surface 3x^2+6y^2-4z^2 = 54

what I did is I found the distance d from some point on the surface (x,y,z) to the origin would be d = sqrt(x^2+y^2+z^2) and that this could be stated as a function of f(x,y,z) as:

f(x,y,z) = x^2+y^2+z^2 and I must find the min of this functoin subject ot the constraint 3x^2+6y^2-4z^2 = 54, but the method of Lagrange multipliers does not work here, why???

Answer 1

You got the following equations, right?

2x - 6λx = 0
2y - 12λy = 0
2z + 8λz = 0
3x² + 6y² - 4z² = 54
So

x = 0 or λ = 1/3
y = 0 or λ = 1/6
z = 0 or λ = -1/4

The choice can't be x = y = z = 0 because that doesn't satisfy the constraint. So λ must equal one of 1/3, 1/6, or -1/4, and so the possible solutions are

x = ±3√2, y = 0, z = 0
x = 0, y = ±3, z = 0

λ = -1/4 forces x = y = 0, but then there is no real solution for z in the constraint. Evaluating F at these four possible solutions tells us which produce the minimum distance.

Note that if you start with the derivative equations

2x - 6λx = 0
2y - 12λy = 0
2z + 8λz = 0

Multiply the first equation by x/2, the second by y/2, and the third by z/2 and add the equations, you get

x²+y²+z² - λ(3x² + 6y² - 4z²) = 0

But from the constraint, 3x² + 6y² - 4z² = 54 so, at the solution,

x²+y²+z² - 54λ = 0

so the extreme value of F is equal to 54λ. We have to discard λ = -1/4, so it's between λ = 1/3 and λ = 1/6. λ = 1/6 gives the smaller value of F, so that's the value we want for λ (which we usually don't care about). That leads us to choose x=z=0, and so y = ±3 to satisfy the constraint.

Answer 2

Your working is good. You need to consider the equation:

F(x,y,z,λ) = √[x²+y²+z²] + λ(3x²+6y²-4z²-54)
However the square root is a pain. Note that √[x²+y²+z²] will be a mimimum at the same point that x²+y²+z² is a minimum so lets consider
F(x,y,z,λ) = x²+y²+z² + λ(3x²+6y²-4z²-54)

Differentiating with respect to the four variables give:
∂F/∂x = 2x + 6xλ = 0
∂F/∂y = 2y +12yλ = 0
∂F/∂z = 2z - 8zλ = 0
∂F∂λ = 3x²+6y²-4z²-54 = 0
The first equation implies x = 0 or λ = -3.
The second equation implies y = 0 or λ = -6.
The third equation implies z = 0 or λ = 4.
Note that the fourth equation is just the constraint.
Combining these gives you three cases to consider:
Case 1: λ=-3, x≠0, y=0, z=0
Case 2: λ=-6, x=0, y≠0, z=0
Case 3: λ = 4, x=0, y=0, z≠0
Substituting each of these three cases back into the constraint:
Case 1: 3x^2 = 54 hence x = ±2√3. Hence the distance is 2√3.
Case 2: 6y^2 = 54 hence y = ±3. Hence the distance is 3. Note this is better than case 1.
Case 2: -4z^2 = 54 hence the is no solution.
So we see the base solution is Case 2 or in other words the point: (0,3,0).

Answer 3

why is not working?