Lagrange multipliers closest point?

[匿名]

2015-03-20 6:40 pm

Use Lagrange multipliers to find the point on the plane
x − 2y + 3z = 6
that is closest to the point
(0, 5, 3).

My professor definitely never taught us how to do this in depth but maybe its something simple that he expected us to figure out. Unfortunately I can't figure out how to solve it.

回答 (2)

2015-03-20 8:48 pm

✔ 最佳答案

It suffices to minimize the square of the distance [to (0, 5, 3)]
D = (x - 0)^2 + (y - 5)^2 + (z - 3)^2, subject to g = x - 2y + 3z = 6.

By Lagrange Multipliers, ∇D = λ∇g.
==> <2x, 2(y - 5), 2(z - 3)> = λ<1, -2, 3>.
==> 2x = λ, 2(y - 5) = -2λ, 2(z - 3) = 3λ

So, 2(y - 5) = -2 * 2x and 2(z - 3) = 3 * 2x
==> y = -2x + 5 and z = 3x + 3.

Substitute this into g:
x - 2(-2x + 5) + 3(3x + 3) = 6
==> x + (4x - 10) + (9x + 9) = 6
==> x = 1/2.

Thus, (x, y, z) = (1/2, 4, 9/2).
(Geometrically, this must give the minimum.)

I hope this helps!

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Andrew

2015-03-20 9:08 pm

Here you go!

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