Lagrange optimization: cylinder and plane intersects,

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SUMMARY

The discussion focuses on solving a Lagrange optimization problem involving the intersection of a cylinder defined by the equation x² + y² = 1 and a plane given by x + z = 1. The goal is to find the point on the resulting ellipse that is furthest from the origin. The initial conclusion mistakenly identifies the point as (1,0,0), but the correct solution is (-1,0,2). Additionally, participants emphasize the importance of proper LaTeX formatting for clarity in mathematical expressions.

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  • Familiarity with the equations of cylinders and planes
  • Proficiency in LaTeX for mathematical notation
  • Basic knowledge of optimization techniques
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Students and professionals in mathematics, engineering, and physics who are working on optimization problems involving geometric shapes and require clarity in mathematical communication.

a255c
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Homework Statement



The cylinder x^2 + y^2 = 1 intersects the plane x + z = 1 in an ellipse. Find the point on the ellipse furthest from the origin.

Homework Equations



$f(x) = x^2 + y^2 + z^2$

$h(x) = x^2 + y^2 = 1$

$g(x) = x + z = 1$

The Attempt at a Solution



$\langle 2x, 2y, 2z \rangle = \lambda\langle2x, 2y,0\rangle + \mu\langle1,0,1\rangle$

This results in the equations:

$2x = 2x\lambda + \mu$

$2y = 2y\lambda$

$2z = \mu$

Then $\lambda = 1$, then $2x = 2x + \mu$, then $0 = \mu$, so then $z = 0$.

Then $x + z = 1$, so $x = 1$.

And then $1^2 + y^2 = 1$, so $y = 0$, so then I conclude that the point where this ellipse is furthest from the origin is $(1,0,0)$.

This is wrong. The answer should be $(-1,0,2)$
 
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You'll have a much better chance of replies if you fix up the Latex. You need to double the dollar signs, or if you don't want an equation on a line by itself replace the dollar sign with a double hash(#).
 
a255c said:

Homework Statement



The cylinder x^2 + y^2 = 1 intersects the plane x + z = 1 in an ellipse. Find the point on the ellipse furthest from the origin.

Homework Equations



$f(x) = x^2 + y^2 + z^2$

$h(x) = x^2 + y^2 = 1$

$g(x) = x + z = 1$

The Attempt at a Solution



$\langle 2x, 2y, 2z \rangle = \lambda\langle2x, 2y,0\rangle + \mu\langle1,0,1\rangle$

This results in the equations:

$2x = 2x\lambda + \mu$

$2y = 2y\lambda$

$2z = \mu$

Then $\lambda = 1$, then $2x = 2x + \mu$, then $0 = \mu$, so then $z = 0$.

Then $x + z = 1$, so $x = 1$.

And then $1^2 + y^2 = 1$, so $y = 0$, so then I conclude that the point where this ellipse is furthest from the origin is $(1,0,0)$.

This is wrong. The answer should be $(-1,0,2)$

The equation ##2y = 2y \lambda## implies either ##\lambda = 1## or ##y = 0##.

Note how my LaTeX comes out properly, unlike yours. That is because I used "# # ... # #", but with no space between the two #'s at the start and the end. Had I used $...$ instead it would have come out a mess, like yours. LaTeX/TeX works a bit differently on this Forum than it does in native form.
 

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