The minimum distance between the origin and the plane 3x + y + 2z = 28, using the Lagrange Multipliers is 8.73 units.
In the question, we are asked to use Lagrange multipliers to find the minimum distance between the origin and the plans 3x + y + 2z = 28.
Lagrange multipliers state that if we want to minimize a function, f(x, y, z), subject to a constraint g(x, y, z) = constant, then ∇f = λ∇g.
We want to minimize the distance, and the distance formula says that the distance from the origin to a point (x, y, z), is given as:
D = √(x² + y² + z²), which can also be shown as:
D² = x² + y² + z².
The gradient of this function is: (2x + 2y + 2z), which needs to be equal to the gradient of the constraint equation multiplied by the constant, λ, that is, λ(3, 1, 2), which gives:
2x = 3λ, or, x = (3/2)λ,
2y = λ, or, y = λ/2, and,
2z = 2λ, or, z = λ.
Substituting these in the equation of the plane, 3x + y + 2z = 28, we get:
3(3/2)λ + λ/2 + λ = 28,
or, 6λ = 28,
or, λ = 28/6 = 14/3.
Thus, we get:
x = (3/2)λ = 7,
y = λ/2 = 7/3, and,
z = λ = 14/3.
Substituting these in the distance formula, we get:
D = √(7² + (7/3)² + (14/3)²) = √(49 + 49/9 + 196/9) = √(686/9) = 8.73.
Thus, the minimum distance between the origin and the plane 3x + y + 2z = 28, using the Lagrange Multipliers is 8.73 units.
Learn more about the Lagrange Multipliers at
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