Min. distance from point to parabaloid?

EDIT: I had a mistake in solving the system, below. (something you could've figured out on your own if you followed my work)

Maybe you used the Lagrange multipliers on the wrong function. You shouldn't use it on z = x^2 + y^2, because then you'll be finding the minimum of the function z insteand of the minimum distance of this function with the point (1, 1,14)

Instead, consider the distance between the point (1, 1, 14) and a point of the form which belong to the graph of your function. For now we consider a point (x,y,z) in the plane, and then we will add the fact that this point belong to your function as a constraint!

The distance between (1,1,14) and (x,y,z), by the usual euclidean formula is thus:

d = √[ (x-1)^2 + (y-1)^2 + (z - 14)^2 ]

So we should find the minimum of this function by the Lagrange multiplier. To ease up our work, we may note that this distance is minimum whenever the expression under the square root is minimized, so, we may find the minimum of the following function

f(x, y, z) = (x-1)^2 + (y-1)^2 + (z - 14)^2

(finding the min of d is also fine but more work involving the derivatives)

So ∂f / ∂x = 2(x-1) = 2x - 2

∂f / ∂y = 2(y-1) = 2y - 2

∂f / ∂z = 2(z-14) = 2z - 28

Now what is our constraint? The constaint we have is that the point (x,y,z) belong to the graph of the function mentionned. That is, the coordinates of this point satisfy the equation of the function. So the constraint is simply z =x^2 + y^2. Write this as a function g(x,y,z) = z - x^2 - y^2.

Then the constriant is g(x,y,z) = 0

Note that

∂g / ∂x = -2x

∂g / ∂y = -2y

∂g / ∂z = 1

Now, by the Lagrange multiplier, to minimize, we should solve the system:

∇f = u*∇g

g(x,y,z) = 0

(u is instead of the usual "lamda"--couldn't find the lamda)

where ∇f and ∇g denote the grandients of f and g

2x - 2 = u*(-2x)........(1)

2y - 2 = u*(-2y)........(2)

2z - 28 = u*1...........(3)

z - x^2 - y^2 = 0......(4)

(4) in (3):

2(x^2+y^2) -28 = u.......(5)

(1): x*(2 + 2u) = 2 => (2+2u) = 2/x ....(6) and x = 1/(1+u) ...(7)

(2): y*(2 + 2u) = 2 =>(2+2u) = 2/y

So 2/x = 2/y so x = y

so z = 2x^2 => x^2 = z/2 ....(8)

(7): x = 1/(1+u) = 1/(2z - 28 + 1) = 1/(2z - 27) (from 3)

So x^2 = 1/(4z^2 + 108z - 729) ....(9)

So (8) and (9): z/2 = 1/(4z^2 + 108z - 729) => 4z^2 + 108z - 729 = 2/z

=> 2z^3 + 53z^2 - 364.5z - 1 = 0

Solve this one, find y and x. And see when the distance is minimized