# Math

Use lagrange multipliers to find the point on the plane x-2y 3z-14=0 that is closet to the origin?(try and minimize the square of the distance of a point (x,y,z) to the origin subject to the constraint that is on the plane) Help me please!

asked by Justin
1. Read up your textbook or lecture notes on Lagrange multipliers. It is not as difficult as it sounds.

However, you need to master you algebra and basic calculus, which I believe should not be a problem.

To put it in the least words possible, we attempt to optimize (maximize or minimize) a function f(x,y,z) subject to the constraint g(x,y,z).

The suggested objective function is the square of the distance from the origin, which therefore is
f(x,y,z)=x²+y²+z²
The constraint is
g(x,y,z)=x-2y+3z-14=0

We now introduce a Lagrange multiplier, λ, to form a new function Λ:
Λ(x,y,z)=f(x)-λg(x)
Λ(x,y,z)=x²+y²+z²-λ(x-2y+3z-14)

Now apply partial differentiation with respect to each of the variables, and equate result to zero:
∂Λ/∂x=2x-λ...(1)
∂Λ/∂y=2y+2λ...(2)
∂Λ/∂z=2z-3λ...(3)
Add the original constraint equation:
x-2y+3z-14=0...(4)

Now solve the system of 4 equations in x,y,z and λ and voilĂ !

x=1, y=-2, z=3, L=2.

So the distance is
D=√(1²+(-2)²+3²)
=√14

How can we tell if this is correct?
This is not too difficult... in this particular case.

The shortest distance from a point (origin 0,0,0) to a plane is the perpendicular distance, given by the well-known formula:
Dmin=(ax0+by0+cz0+d)/√(a²+b*sup2;+c²)
=(0-0+0-14)/√(1²+(-2)²+3²)
=-14/√(14)
=-√(14)
and the square of the minimum distance
Dmin²=14
which checks with our Lagrange multiplier answer.

posted by MathMate

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