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!
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 voilà!
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.
To find the point on the plane that is closest to the origin, we need to minimize the square of the distance between a point (x, y, z) on the plane and the origin (0, 0, 0), subject to the constraint that the point is on the plane x - 2y + 3z - 14 = 0.
Let's denote the square of the distance between the point (x, y, z) and the origin as D(x, y, z), which can be given by:
D(x, y, z) = x^2 + y^2 + z^2
We will introduce a Lagrange multiplier λ to incorporate the constraint equation. The Lagrangian function L(x, y, z, λ) is defined as:
L(x, y, z, λ) = D(x, y, z) - λ(x - 2y + 3z - 14)
To find the point that minimizes the distance, we need to find the critical points of L(x, y, z, λ), which means finding the solutions to the following system of equations:
1) ∂L/∂x = 2x - λ = 0
2) ∂L/∂y = 2y + 2λ = 0
3) ∂L/∂z = 2z - 3λ = 0
4) x - 2y + 3z - 14 = 0
From equation 2, we can solve for y in terms of λ:
y = -λ/2
Substituting this equation into equations 1, 3, and 4, we get:
x = λ/2
z = 3λ/2
x - 2y + 3z = λ/2 + 2(λ/2) + 3(3λ/2) = 14
Simplifying the equation x - 2y + 3z = 14, we get:
λ + 3λ - 9λ = 14
-5λ = 14
λ = -14/5
Substituting λ = -14/5 back into the equations x = λ/2, y = -λ/2, and z = 3λ/2, we get:
x = -7/5
y = 7/10
z = -21/10
Therefore, the point on the plane x - 2y + 3z - 14 = 0 that is closest to the origin is (-7/5, 7/10, -21/10).
To find the point on the plane that is closest to the origin, we can use the method of Lagrange multipliers. The objective is to minimize the square of the distance between a point (x, y, z) and the origin, which is given by the expression x^2 + y^2 + z^2.
Let's define a function, F(x, y, z), as the square of the distance from a point (x, y, z) to the origin, subject to the constraint that the point lies on the plane x - 2y + 3z - 14 = 0.
Thus, we have the following mathematical expressions:
F(x, y, z) = x^2 + y^2 + z^2
g(x, y, z) = x - 2y + 3z - 14
We can introduce a Lagrange multiplier, λ, to incorporate the constraint into the objective function. The Lagrangian function, L, can be defined as follows:
L(x, y, z, λ) = F(x, y, z) - λg(x, y, z)
Plugging in the expressions for F(x, y, z) and g(x, y, z), we have:
L(x, y, z, λ) = x^2 + y^2 + z^2 - λ(x - 2y + 3z - 14)
To find the point that minimizes F(x, y, z), we need to solve the following system of equations:
∇L/∇x = ∇L/∇y = ∇L/∇z = ∇L/∇λ = 0
Taking the partial derivatives and setting them to zero, we have:
∇L/∇x = 2x - λ = 0
∇L/∇y = 2y + 2λ = 0
∇L/∇z = 2z + 3λ = 0
∇L/∇λ = -x + 2y - 3z + 14 = 0
From the first equation, we get:
2x - λ = 0 => λ = 2x
From the second equation, we get:
2y + 2λ = 0 => 2y + 4x = 0 => y = -2x
From the third equation, we get:
2z + 3λ = 0 => 2z + 6x = 0 => z = -3x
Now, substitute these values of y and z into the equation obtained from the fourth equation:
-x + 2y - 3z + 14 = 0
-x + 2(-2x) - 3(-3x) + 14 = 0
-5x + 14 = 0
x = 14/5
Substituting this value of x back into the expressions we derived for y and z, we get:
y = -2(14/5) = -28/5
z = -3(14/5) = -42/5
So, the point on the plane that is closest to the origin is (x, y, z) = (14/5, -28/5, -42/5).