Lagrange multipliers; tricky algebra
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From: sg on 6 Aug 2010 11:15 Hello, I am working out of Taylor and Mann's Advanced Calculus and could use a push in the right direction with the following problem: "Find the maximum and minimum squared distances to the origin among the points on the curve x + y + z = 12, xyz = 54." The Lagrange conditions are the following: 2x + \lambda_1 + yz \lambda 2 = 0 2y + \lambda_1 + zx \lambda 2 = 0 2z + \lambda_1 + xy \lambda 2 = 0 together with the constraint equations x + y + z = 12, xyz = 54. I have eliminated x, y, z by writing p^2 = x^2 + y^2 + z^2, and writing the sum of the three Lagrange conditions as 2(x + y + z) + 3 \lambda_1 + (xy + yz + zx) \lambda_2 = 0 Since x + y + z = 12, this is 24 + 3 \lambda_1 + \lambda_2 (144 - p^2)/2. On the other hand, we can multiply the three conditions by x, y, z, and then add, obtaining 2p^2 + 12 \lambda_1 + 162 \lambda_2 = 0. From here I am not sure how to proceed, I am pushed these equations around and not got much of anywhere. What am I missing? I feel it's staring me in the face. Thanks very much in advance. sg
From: Ray Vickson on 6 Aug 2010 13:42 On Aug 6, 8:15 am, sg <suitablygene...(a)gmail.com> wrote: > Hello, > > I am working out of Taylor and Mann's Advanced Calculus and could use > a push in the right direction with the following problem: "Find the > maximum and minimum squared distances to the origin among the points > on the curve x + y + z = 12, xyz = 54." > > The Lagrange conditions are the following: > > 2x + \lambda_1 + yz \lambda 2 = 0 > 2y + \lambda_1 + zx \lambda 2 = 0 > 2z + \lambda_1 + xy \lambda 2 = 0 If you write 2*54/(yz) + u1 + u2*yz = 0 (using u1 instead of lambda_1, etc) you can solve for yz as -1/2*(u1-sqrt(u1^2-432*u2))/u2 (= a) or as -1/2*(u1+sqrt(u1^2-432*u2))/u2 (= b). By symmetry, the same solutions are obtained for xz and yz, so the choices are limited: either all xy, xz and yz are the same (either a or b) or two of them are equal: xz = xz = c1 and yz = c2 or xz = yz = c1 and yz = c2 or xz = yz = c1 and xz = c2, where c1 is a or b and c2 is b or a. If you look at the "all equal" case {xy = c, zx = c, yz = c} you can solve for x, y, z in terms of c: either x=y=z = sqrt(c) or x =y=z = - sqrt(c). Since they are all equal, they each equal 12/3 = 4, and at this value of (x,y,z) you can reverse the arguments and find u1, u2. If you look at one of the "two equal" cases {xy=c1,xz=c1,yz=c2} you can get y = +-sqrt(c2), z = +- sqrt(c2), x = +- c1/sqrt(c2), where we choose '+' for all or '-' for all. If we choose '+' signs we can x, y and z in terms of c1 and c2, then solve for c1 and c2 from x+y+z=12 and xyz=54. The other sign choice would be the mirror image in the opposite orthant. There are different possible solutions, depending on which of xy, xz or yz is taken to be the odd man out, but they differ just by permuting x, y and z. I think it is clear from symmetry of the original problem that a max/min is either unique or three-fold, so you would need to look further into the different roots to figure out what is happening. I don't know if you have yet taken second-order tests in constrained problems for distinguishing a max from a min or a saddle point. If you have not taken this material yet you can always plug in the solutions you get and see numerically which ones are maxima and which are minima. Good luck. R.G. Vickson > > together with the constraint equations x + y + z = 12, xyz = 54. I > have eliminated x, y, z by writing p^2 = x^2 + y^2 + z^2, and writing > the sum of the three Lagrange conditions as > > 2(x + y + z) + 3 \lambda_1 + (xy + yz + zx) \lambda_2 = 0 > > Since x + y + z = 12, this is > > 24 + 3 \lambda_1 + \lambda_2 (144 - p^2)/2. > > On the other hand, we can multiply the three conditions by x, y, z, > and then add, obtaining > > 2p^2 + 12 \lambda_1 + 162 \lambda_2 = 0. > > From here I am not sure how to proceed, I am pushed these equations > around and not got much of anywhere. What am I missing? I feel it's > staring me in the face. > > Thanks very much in advance. > sg
From: Ray Vickson on 6 Aug 2010 14:03 On Aug 6, 10:42 am, Ray Vickson <RGVick...(a)shaw.ca> wrote: > On Aug 6, 8:15 am, sg <suitablygene...(a)gmail.com> wrote: > > > Hello, > > > I am working out of Taylor and Mann's Advanced Calculus and could use > > a push in the right direction with the following problem: "Find the > > maximum and minimum squared distances to the origin among the points > > on the curve x + y + z = 12, xyz = 54." > > > The Lagrange conditions are the following: > > > 2x + \lambda_1 + yz \lambda 2 = 0 > > 2y + \lambda_1 + zx \lambda 2 = 0 > > 2z + \lambda_1 + xy \lambda 2 = 0 > > If you write 2*54/(yz) + u1 + u2*yz = 0 (using u1 instead of lambda_1, > etc) you can solve for yz as -1/2*(u1-sqrt(u1^2-432*u2))/u2 (= a) or > as -1/2*(u1+sqrt(u1^2-432*u2))/u2 (= b). By symmetry, the same > solutions are obtained for xz and yz, so the choices are limited: > either all xy, xz and yz are the same (either a or b) or two of them > are equal: xz = xz = c1 and yz = c2 or xz = yz = c1 and yz = c2 or xz > = yz = c1 and xz = c2, where c1 is a or b and c2 is b or a. If you > look at the "all equal" case {xy = c, zx = c, yz = c} you can solve > for x, y, z in terms of c: either x=y=z = sqrt(c) or x =y=z = - > sqrt(c). Since they are all equal, they each equal 12/3 = 4, and at > this value of (x,y,z) you can reverse the arguments and find u1, u2. > If you look at one of the "two equal" cases {xy=c1,xz=c1,yz=c2} you > can get y = +-sqrt(c2), z = +- sqrt(c2), x = +- c1/sqrt(c2), where we > choose '+' for all or '-' for all. If we choose '+' signs we can x, y > and z in terms of c1 and c2, then solve for c1 and c2 from x+y+z=12 > and xyz=54. The other sign choice would be the mirror image in the > opposite orthant. Sorry: this is false. Obviously x+y+z=12 cannot be satisfied by the mirror image. If we were to choose all '-' signs instead we would just get different values for c1 and c2, and would get another point on the plane x+y+z=12 and surface xyz=54. R.G. Vickson > > There are different possible solutions, depending on which of xy, xz > or yz is taken to be the odd man out, but they differ just by > permuting x, y and z. I think it is clear from symmetry of the > original problem that a max/min is either unique or three-fold, so you > would need to look further into the different roots to figure out what > is happening. > > I don't know if you have yet taken second-order tests in constrained > problems for distinguishing a max from a min or a saddle point. If you > have not taken this material yet you can always plug in the solutions > you get and see numerically which ones are maxima and which are > minima. > > Good luck. > > R.G. Vickson > > > > > together with the constraint equations x + y + z = 12, xyz = 54. I > > have eliminated x, y, z by writing p^2 = x^2 + y^2 + z^2, and writing > > the sum of the three Lagrange conditions as > > > 2(x + y + z) + 3 \lambda_1 + (xy + yz + zx) \lambda_2 = 0 > > > Since x + y + z = 12, this is > > > 24 + 3 \lambda_1 + \lambda_2 (144 - p^2)/2. > > > On the other hand, we can multiply the three conditions by x, y, z, > > and then add, obtaining > > > 2p^2 + 12 \lambda_1 + 162 \lambda_2 = 0. > > > From here I am not sure how to proceed, I am pushed these equations > > around and not got much of anywhere. What am I missing? I feel it's > > staring me in the face. > > > Thanks very much in advance. > > sg > >
From: Robert H. Lewis on 6 Aug 2010 10:56 I don't see any simple, easy algebraic manipulations. Maybe someone will prove me wrong. Purely algebraically, you have a system of 5 polynomial equations in 5 variables. You could compute the resultants. Granted, the text book probably has not gone into that topic. Geometrically, the first equation is of course just a slanting plane. The second is fairly well known, the z = c/(xy) hyperbolic "curtains". A graph is here: http://www.math.wisc.edu/~miller/old/m223-96/progs/rat0.gif There are 4 sheets to the image, one above/below each xy-quadrant. If you look at that image, picture the slanting plane z = 12 - x - y. When x=y, that is z = 12 - 2x, so that will help visualize it. The intersection with the third quadrant sheet (x<0, y<0) is a sort of slanting hyperbola. It is unbounded. Therefore, there are points on this intersection arbitrarily far from the origin. To get one, let y, say, be some large negative number, like y = -10. Then z = 22-x and also z = -5.4/x. Find an x that makes these equal. It's roughly x = -.24. Repeat for -100, -1000, etc. The intersection with the first quadrant is kind of cool. (I think the 12 and 54 were chosen so that there is an intersection.) It's going to be one of those cubics, a sort of squashed ellipse. So what's the answer? Don't know, but if you let x=y, you get (3,3,6). I'd bet on that, and its permutations. Robert H. Lewis Fordham University
From: Robert H. Lewis on 6 Aug 2010 12:27
Oops, not quite. (3,3,6) is not the answer. It's the other root, (3 + 3 sqrt(5))/2. (4.854, 4.854, 2.292) approx. I computed the resultants. If no one else solves this, I'll post the details later. Robert H. Lewis Fordham University |