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Setting up Lagrange problem

Yatzy3D

New member
I want to find the point on the ellipse that is the intersection between the plane x+y+z=1 and the cylinder x[SUP]2[/SUP] + y[SUP]2[/SUP] = 1 that is furthest away from (0,0). I'm pretty sure this can be solved by using Lagrange, but I can't figure out how to set up the expression.
 
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Subhotosh Khan

Super Moderator
I want to find the point on the eclipse (ellipse?) that is the intersection between the plane x+y+z=1 and the cylinder x[SUP]2[/SUP] + y[SUP]2[/SUP] = 1 that is furthest away from (0,0). I'm pretty sure this can be solved by using Lagrange, but I can't figure out how to set up the expression.
I would start with calculating the equation of the ellipse (curve of intersection).
 

HallsofIvy

Active member
The function to be minimized is the square of the distance \(\displaystyle f(x, y, z)= x^2+ y^2+ z^2\) subject to the constraints \(\displaystyle g(x, y, z)= x+ y+ z= 1\) and \(\displaystyle h(x, y, z)= x^2+ y^2= 1\).

The first thing I would notice is that, since \(\displaystyle x^2+ y^2= 1\), the function to be minimized reduces to \(\displaystyle f(z)= 1+ z^2\). That is minimized when z= 0. And the condition that x+ y+ z= 1 means that x+ y= 1 so y= 1- x. Using, again, \(\displaystyle x^2+ y^2= 1\), we have \(\displaystyle x^2+ 1- 2x+ x^2= 2x^2- 2x+ 1= 1\). So \(\displaystyle 2x^2- 2x= 2x(x- 1)= 0\). \(\displaystyle x= 0\), in which case \(\displaystyle y= 1\) or \(\displaystyle x= 1\), in which case \(\displaystyle y= 0\).

The solutions are \(\displaystyle (0, 1, 0)\) or \(\displaystyle (1, 0, 0)\).
 

Yatzy3D

New member
The function to be minimized is the square of the distance \(\displaystyle f(x, y, z)= x^2+ y^2+ z^2\) subject to the constraints \(\displaystyle g(x, y, z)= x+ y+ z= 1\) and \(\displaystyle h(x, y, z)= x^2+ y^2= 1\).

The first thing I would notice is that, since \(\displaystyle x^2+ y^2= 1\), the function to be minimized reduces to \(\displaystyle f(z)= 1+ z^2\). That is minimized when z= 0. And the condition that x+ y+ z= 1 means that x+ y= 1 so y= 1- x. Using, again, \(\displaystyle x^2+ y^2= 1\), we have \(\displaystyle x^2+ 1- 2x+ x^2= 2x^2- 2x+ 1= 1\). So \(\displaystyle 2x^2- 2x= 2x(x- 1)= 0\). \(\displaystyle x= 0\), in which case \(\displaystyle y= 1\) or \(\displaystyle x= 1\), in which case \(\displaystyle y= 0\).

The solutions are \(\displaystyle (0, 1, 0)\) or \(\displaystyle (1, 0, 0)\).
I'm pretty sure these are the points closest to (0,0). As you can see here:edited imgur site
 
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Subhotosh Khan

Super Moderator
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