Setting up Lagrange problem

[Yatzy3D]

I want to find the point on the ellipse that is the intersection between the plane x+y+z=1 and the cylinder x² + y² = 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.

 

[Deleted member 4993]

I want to find the point on the eclipse (ellipse?) that is the intersection between the plane x+y+z=1 and the cylinder x² + y² = 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]

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]

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

 

[Deleted member 4993]

I'm pretty sure these are the points closest to (0,0). As you can see here:edited imgur site

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https://www.freemathhelp.com/forum/...Image-Hosting-Sites-A-WARNING?highlight=imgur