Use Lagrange multipliers to find the shortest distance, d, from the point (1, 0, −2) to the plane x y z = 1.?

Question

Use Lagrange multipliers to find the shortest distance, d, from the point
(1, 0, −2)
to the plane
x y z = 1.


Can someone help me out with this?
I got sqrt(8) for my answer.

Can someone please help me out with this one? And please show your steps.

my x was (lambda 2)/2
my y was (lambda)/2
my z was (lambda-4)/2

Then I got lambda = 4/3
x = 4/3
y=2/3
z=-4/3

Answer 1

It suffices to minimize the square of the distance

D = d^2 = (x - 1)^2 (y - 0)^2 (z 2)^2, subject to g = x y z = 1.



By Lagrange Multipliers, ∇D = λ∇g.

==> <2(x - 1), 2y, 2(z 2)> = λ<1, 1, 1>.



Equate like entries:

2(x - 1) = λ, 2y = λ, 2(z 2) = λ.



So, λ = 2(x - 1) = 2y = 2(z 2)

==> x - 1 = y = z 2.



So, x = y 1 and z = y - 2.

Substitute this into g: (y 1) y (y - 2) = 1 ==> y = 2/3.



Hence, (x, y, z) = (5/3, 2/3, -4/3),

and the minimal distance is d(5/3, 2/3, -4/3) = (2/3)√3.



I hope this helps!

Answer 2

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

g(x) = xyz-1.

You need to solve

grad f = l grad g. l is the Lagrange undetermined multiplier, to be determined.

Let us fine the value

2(x-1) = l yz,

2y = l xz,

2(z+1) = l yx

Now solve for l. Have you written the equation for the plane. Is it x+y+z=1?

Answer 3

F = Distance squared from point (x,y,z) to (1,0,-2) = (x-1)²+y²+(z+2)²

∇F = 2(x-1) i + 2y j + 2(z+2) k (i,j,k are unit vectors)



Equation of plane G = x+y+z-1=0

λ∇G = λ i + λ j + λ k



System of four equations

2x-2 = λ

2y = λ

2z+4 = λ

x+y+z=1



System of three equations (eliminating λ)

y=x-1 or x=y+1

y=z+2 or z=y-2

x+y+z=1



Substituting into third equation

y+1+y+y-2=1

3y=2

y=2/3



x=5/3

z=-4/3



F(5/3,2/3,-4/3) = (5/3-1)²+(2/3)²+(-4/3+2)² = 4/9+4/9+4/9=4/3

distance=√(4/3) = 2√3/3



Please check my algebra and compare it with yours.

Answer 4

It suffices to minimize the square of the distance

D = (x - 1)^2 + y^2 + (z + 2)^2, subject to G = x + y + z = 1.



By Lagrange Multipliers, ∇D = λ∇G.

==> <2(x - 1), 2y, 2(z + 2)> = λ<1, 1, 1>.

==> λ = 2(x - 1) = 2y = 2(z + 2)

==> x - 1 = y = z + 2

==> x = y + 1, and z = y - 2.



Substituting this into G yields (y + 1) + y + (y - 2) = 1.

==> y = 2/3.



So, the point in question is (x, y, z) = (5/3, 2/3, -4/3).

Then, the minimal distance is √(D(5/3, 2/3, -4/3)) = (2/3)√3.



I hope this helps!