
#1
Jan1410, 07:55 AM

P: 211

1. The problem statement, all variables and given/known data
Find the points on the surface: zēxy=1 nearest to the origin 2. Relevant equations grad(f)= lambda grad(g) (?) f(x,y,z)=zēxy (?) g=(0,0,0) (?) 3. The attempt at a solution grad(f)=(y, x, 2z)= lambda (0,0,0) = (0,0,0) y=0 => y=0 x=0 => x=0 2z=0 => z=0 But (x,y,z)=(0,0,0) doesn't satisfy f(0,0,0)=1. 



#2
Jan1410, 08:37 AM

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No. You want to minimize distance to the origin. Minimize f=x^2+y^2+z^2. Your constraint g is z^2xy1.




#3
Jan1410, 08:45 AM

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The "f" you are using is the left side of [itex]z^2 xy= 1[/itex], the equation of the surface. But where have you used the function you want to minimize, the distance from the point (x,y,z) to the origin? "g= (0,0,0)" makes no sense because that is a point, not a function!
The function you want to minimize is the distance from (0,0,0). That is [itex]\sqrt{x^2+ y^2+ z^2}[/itex]. But because you know distance is always positive, that is the same as minimizing the square of distance, [itex]x^2+ y^2+ z^2[/itex]. Your "f" is that: [itex]f(x,y,z)= x^2+ y^2+ z^2[/itex]. The "constraint" you are given is that [itex]g(x,y,z)= z^2 xy= 1[/itex]. Geometrically, the shortest line from (0,0,0) to that surface will be perpendicular to that surface. And, if we think of g(x,y,z)= 1 as a "level surface" of the function g(x,y,z), its gradient will be perpendicular to the surface. That is, the shortest distance will be when grad f and grad g are parallel vectors which means one is a multiple of the other: [itex] grad f= \lambda grad g[/itex]. As I said, use [itex]f(x,y,z)= x^2+ y^2+ z^2[/itex] and [itex]g(x,y,z)= x^2 xy[/itex]. 



#4
Jan1410, 09:08 AM

P: 211

Find points on surface nearest to origin
Thanks a lot, I get it now :) !
I get the following system of equations: 2xly = 0 2ylx = 0 z(2+2l) = 0 zēxy = 1 And when I solve this I get (x,y,z)=(0,0,+/1). 



#5
Apr3011, 05:22 AM

P: 5

While searching for finding a solution to my problem, I came across with this thread. It would be nice if someone could explain to me geometrically what kind of vector is the [itex]grad f=(2x,2y,2z)[/itex].
I mean, I can understand that [itex]grad g[/itex] is perpendicular to the surface, but I can't get why [itex]grad f[/itex] and [itex]grad g[/itex] are parallel vectors, since the distance [itex] x^2 + y^2 + z^2[/itex] is already a vector that passes through that point of the surface and the origin. 



#6
Apr3011, 11:50 AM

P: 5

My mistake... [itex]x^2 +y^2 + z^2[/itex] is actually a number and not a vector.
But I still can get this problem solved. I have a surface [itex]x^2 + y^2 z =2[/itex] and I need to find the shortest distance from the origin. The gradient of the distance is [itex]grad(d)=(2x,2y,2z)[/itex] and the gradient of the the surface is [itex]gradf(x,y,z)=(2x,2y,1)[/itex]. Shouldn't I use lagrange multipliers? But then I have [itex]2x= \lambda 2x[/itex] and [itex]2y= \lambda 2y[/itex] and if [itex] \lambda =1[/itex] then I have only two equations with 3 variables which means infinite solutions. Could anyone help me out with the solution? 



#7
Apr3011, 12:46 PM

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#8
Apr3011, 09:40 PM

P: 5

The other case is to have [itex]x=0[/itex] and [itex]y=0[/itex] and I get z=2 from the equation of the surface. Since the length of the vector (0,0,2) is 2 then this is the shortest distance from the surface. Is that right? 



#9
Apr3011, 09:48 PM

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#10
Apr3011, 09:50 PM

P: 5

z=1/2 if I am right since λ=1.




#11
Apr3011, 10:22 PM

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#12
May111, 04:12 AM

P: 5

My fault... It's 3/2 then :)
Have I dealt with the other case correctly? 



#13
May111, 06:59 AM

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Here, the objective is to minimize the the distance to (0, 0, 0). The gradient of the square of the distance is <2x, 2y, 2z>= 2<x, y, z>. If you want to maximize distance, move in the direction of <x, y, z>, directly away from <0, 0, 0>. If you want to minimize distance, move in the direction of <x, y, z>, directly toward <0, 0, 0>. That makes sense! You keep doing that until there is no "direction" to [itex]\nabla f[/itex] until is is 0 which happens when you are at (0, 0, 0). But if you are restricted to a given surface, you cannot alway move parallel to [itex]\nabla f[/itex]. The best you can do look at the projection of [itex]\nabla f[/itex] onto the surface and follow that if [itex]\nabla f[/itex] is a little to the right of perpendicular, go that way. You can do that until [itex]\nabla f[/itex] does NOT have a "projection" onto the surface until it is perpendicular to the surface. But if the surface is given by G(x, y, z)= constant, then [itex]\nabla G[/itex] is perpendicular to the surface. That is, you can go no further on that surface toward your maximum of minimum of f(x,y,z) when [itex]\nabla f[/itex] is perpendicular to the surface which means it is in the same direction as [itex]\nabal G[/itex] [itex]\nabla f[/itex] and [itex]\nabla G[/itex] are parallel there which means one is a multiple of the other: [iitex]\nabla f= \lambda\nabla G[/itex]. 


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