# Thread: Lagrange Multipliers and Distance from point to plane

1. ## Lagrange Multipliers and Distance from point to plane

I had a question on my final exam that I can't seem to get my mind off...
Unfortunately I don't remember the equations that were given but the general question was:
How is the maximum/minimum value (found with Lagrange multipliers) related to the distance from a point (the max/min) to a plane?
The only reasoning I've managed to do is this:
the plane is the constraint function, so as the distance between the point and plane goes to zero, you get closer to the max/min value of the objective function.
To me this seems like too basic of reasoning to be on the final.

Any other thoughts?

2. Why would distnce magically decrease? What's a Tangent and what's a Normal?

3. Originally Posted by TKHunny
Why would distnce magically decrease?
If you are trying to find the extreme point of a surface subject to a constraint, the limit of the distance between the extreme point and the constraint surface would approach zero as the constraint approaches the point.

Im not sure how finding the normal to the surface would provide anything... just a vector, not a distance, you dont have any points on the plane.

4. Hello, kateburns!

Without a specific problem, it's difficult to discuss the approach and solution.

Could it be something like this?

Find the minimum distance from the point (3, 1, -2)
. . to the plane 2x + 3y - 6z + 7 .= .0.

Let P(x,y,z) be a point on the plane.

Its distance from Q(3,1,-2) is given by:
. . . . . . . . . ____________________________
. . . d . = . √(x - 3)^2 + (y - 1)^2 + (z + 2)^2

or: .D . = . (x - 3)^2 + (y - 1)^2 + (z + 2)^2

The constraint is: .2x + 3y - 6z + 7 .= .0

The function to minimize is:

. . f(x,y,L) . = . (x - 3)^2 + (y - 1)^2 + (z + 2)^2 + L(2x + 3y - 6z + 7)