# Math

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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!

• Math - ,

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)
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.

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