# Lagrange optimization: cylinder and plane intersects,

• a255c
In summary, the given cylinder and plane intersect in an ellipse. Using the equations $f(x)$, $h(x)$, and $g(x)$, a system of equations is set up to find the point on the ellipse furthest from the origin. Solving the system results in the point $(-1, 0, 2)$, rather than the incorrect point $(1, 0, 0)$.

## 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 $(-1,0,2)$

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.