September 6th, 2018, 04:13 PM  #11 
Senior Member Joined: Sep 2016 From: USA Posts: 578 Thanks: 345 Math Focus: Dynamical systems, analytic function theory, numerics 
I suspect the problem here is that if you do a google search for examples of Lagrangian optimization, you find only examples with 0dimensional solution manifolds i.e. the number of variables matches the number of constraints. However, this not what he is asking about I don't think. In general if you have $n$ variables and $m$ constraints containing $n + k$ variables, then you expect (for sufficient smoothness conditions) to have a $n + k  m$ dimensional manifold which satisfies the problem. One of the best resources I know of for this case is the following book. Sadly I don't know of any free options. Hope it helps. https://epubs.siam.org/doi/book/10.1137/1.9781611972573 
September 7th, 2018, 06:59 AM  #12  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550  Quote:
$z = x^2 + y^2.$ $\text {Find extrema of } z \text { subject to } x^2 + 4y^2 + w^2 = 26 \text { and } w \ge 1.$ $L = x^2 + y^2  \lambda (26  x^2  4y^2  w^2)  \kappa (1  w).$ $\dfrac{\delta L}{\delta x} = 0 \text { if } x = \lambda x.$ $\dfrac{\delta L}{\delta y} = 0 \text { if } y = 4 \lambda y.$ $\dfrac{\delta L}{\delta w} = 0 \text { if } 2w \lambda = \ \kappa.$ $\dfrac{\delta L}{\delta \lambda} = 0 \text { if } x^2 + 4y^2 + w^2 = 26.$ $\dfrac{\delta L}{\delta \kappa} = 0 \text { if } w = 1.$ $x = \lambda x \implies x = 0 \text { or } \lambda = 1.$ $\text {CASE I: } x = 0.$ $x = 0, \ w = 1, \text { and } 26 = x^2 + 4y^2 + w^2 = 4y^2 + 1 \implies$ $y = \pm \dfrac{5}{2} \implies z = \dfrac{25}{4}.$ $\text {CASE II: } \lambda = 1.$ $\therefore y = 4 \lambda y \implies y = 4y \implies y = 0.$ $x^2 + 0^2 + 1^2 = 26 \implies x = \pm 5 \implies z = 25.$ $\therefore \text { constrained } z \text { has a minimum of } \dfrac{25}{4} \text { at } \left (0, \ \dfrac{5}{2} \right ) \text { and } \left (0,\ \ \dfrac{5}{2} \right ).$ $\text {And constrained } z \text { has a maximum of } 25 \text { at } (5,\ 0) \text { and } (\ 5,\ 0).$ This is such a simple example that it can be checked by common sense, but it shows how it can be done. Admittedly, this process has all the difficulties associated with Lagrangians: it only works if all relevant functions are differentiable, irrelevant constraints result in an inconsistent system of equations, extrema may be at boundaries, the second derivative test may be a bear, etc. But none of this has anything to do with the number of independent variables in the constraints exceeding the number of independent variables in the objective function. Last edited by JeffM1; September 7th, 2018 at 07:08 AM.  
September 8th, 2018, 06:51 AM  #13  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
x^2=264y^2w^2 z=263y^2w^2 dz=6ydy2wdw dz=0 for arbitrary dy and dw (condition for an extreme) if y=0 and w=0, in which case x^2=26 and z=26 is an extreme at x^2=26, y=0, w=0  I came across last post after doing the following: Nice simple illustration of OP. w=x^2 with constraint x^2+y^2+z^2k=0. Without constraint: dw=2xdx is zero for arbitrary dx (condition for extreme) if x=0. w has a min value of 0. With constraint, two methods: I) Solve constraint for x and put it into object function: w=ky^2z^2 Now y and z are the independent variables and dw=2ydy2zdz=0 for arbitrary dy and dz if y=0 and z=0. Then x^2=k and w=k at the extreme II) Take differentials first: dw=2xdx 2xdx+2ydy+2zdz=0 dw=2x(ydyzdx)/2x=0 dw=ydy+zdz=0 for arbitrary dy and dz if y=0 and z=0. Then x^2=k and w=k at the extreme Method II) saves the trouble of solving constraint for x, which in general might not be that easy. Lagrange multiplier doesn't work here because object equation and constraint equations don't have same number of variables. You need to know why Lagrange multipiers work. EDIT: I thanked JeffM1 for a nice example of why Lagrange multipliers doesn't work if number of variables in object function and constraints are different. Last edited by zylo; September 8th, 2018 at 07:11 AM.  
September 10th, 2018, 09:27 AM  #14  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550  Quote:
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$L = x^2  \lambda (x^2 + y^2 + z^2  k = 0)$ Optimize L. $\dfrac{\delta L}{\delta x} = 0 \text { if } 2x = 2 \lambda x.$ Two cases: $\lambda = 1 \text { or } \lambda = 1.$ $\dfrac{\delta L}{\delta y} = 0 \text { if } 2y \lambda = 0.$ $\dfrac{\delta L}{\delta z} = 0 \text { if } 2z \lambda = 0.$ $\dfrac{\delta L}{\delta \lambda} = 0 \text { if } x^2 + y^2 + z^2  k = 0.$ $\text {Case I: } \lambda = 1 \implies y = 0 = z \implies x^2  k = 0 \implies x = \pm \sqrt{k} \text { and } w = k.$ Two points of maximization under the constraint. $\text {Case II: } \lambda = 0 \implies 2x = 2 * 0 * x = 0 \implies x = 0 \implies w = 0.$ One point of minimization under the constraint. {QUOTE]Without constraint: dw=2xdx is zero for arbitrary dx (condition for extreme) if x=0. w has a min value of 0. With constraint, two methods: I) Solve constraint for x and put it into object function: w=ky^2z^2 Now y and z are the independent variables and dw=2ydy2zdz=0 for arbitrary dy and dz if y=0 and z=0. Then x^2=k and w=k at the extreme[/QUOTE] Yes this method sometimes works if the constraint can be restated in a closed form. I never said that Lagrangian multipliers are required to solve all problems in constrained minimization. But it is false that they cannot be applied if the number of independent variables in the constraints exceeds the number of independent variables in the objective function.  
September 10th, 2018, 12:05 PM  #15 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
"Maximize" (really find critical points) f(x,y) subject to constraint g(x,y,z) =0 I) x,y independent. Maximize f(x,y) and plug results into g(x,y,z)=0 to determine z. $\displaystyle df=f_{x}dx+f_{y}dy=o \rightarrow f_{x}=f_{y}=0 $ II) x,z independent a) Solve g(x,y,z) for y=y(x,z) and substitute into f(x,y) to get F(x,z). Maximize F(x,z) to find x,z and plug into g(x,y,z)=0 to find y. b) Use differentials. $\displaystyle df=f_{x}dx+f_{y}dy=o$ $\displaystyle g_{x}dx+g_{y}dy+g_{z}dz=0 $ Solve the latter for dy and plug into former and then equate coefficients of dx and dz to zero to find x,z. Then plug into g(x,y,z) to find y. Example 1: f(x,y)=xy, g(x,y,z)=x+y+zM=0, x,z independent. Method IIa: y=Mxz, xy=x(Mxz), df=Mdx2xdxzdxxdz=0 (M2xz)dxxdz=0 $\displaystyle \rightarrow$ x=0, z=M, y=0 Method IIb: df=ydx+xdy=0 dx+dy+dz=0, dy=dxdz, ydx+x(dxdz)=0, (yx)dxxdz=0, $\displaystyle \rightarrow$ x=0, y=0, z=M Neither result makes much sense but illustrates the procedure. Example 2 $\displaystyle f(x,y)=x^{2}+y^{2}$, $\displaystyle g(x,y,z)=x^{2}+4y^{2}+w^{2}26$, x,z independent Method IIb df=2xdx+2ydy=0 2xdx+8ydy+2wdw=0 $\displaystyle dy=\frac{x}{4y}dx\frac{w}{4y}dw$, then df=(3/2)xdx1/2wdw=0, \rightarrow x=0, w=0, y^{2}=26/4 y and w independent $\displaystyle \rightarrow$ y=0, w=0, x$\displaystyle ^{2}$=26 ========================================== Lagranges Multiplier Max f(x,y) subject to g(x,y)=0 $\displaystyle df=f_{x}dx+f_{y}dy=0$ $\displaystyle \lambda g_{x}dx+\lambda g_{y}dy=0$ $\displaystyle (f_{x}\lambda g_{x})dx+(f_{y}\lambda g_{y})dy=0$ So instead of solving for dy in terms of dx, if $\displaystyle f_{y}\lambda g_{y}=0$ and dx arbitrary, $\displaystyle f_{x}\lambda g_{x}=0 $is condition for extremum. This gives with g(x,y)=0 three conditions for x,y,z. Now try the same method with: Maximize" f(x,y) subject to constraint g(x,y,z) =0 $\displaystyle df=f_{x}dx+f_{y}dy=0$ $\displaystyle \lambda g_{x}dx+\lambda g_{y}dy+\lambda g_{z}dz=0$ $\displaystyle (f_{x}\lambda g_{x})dx+(f_{y}\lambda g_{y})dy\lambda g_{x}dz=0$ Let x and w be independent. Then eliminate dy by $\displaystyle (f_{y}\lambda g_{y})$ Then dx and dw independent require $\displaystyle f_{x}\lambda g_{x}=0$ and $\displaystyle g_{z}=0$ These three conditions with g(x,y,z)=0 give four equations in the four unknowns x,y,z,$\displaystyle \lambda$ It works with Example 2 Using f(x,y)$\displaystyle \lambda$g(x,y,z) and taking partial derivatives is a meaningless extension of Lagranges multiplier unless you can justify it. To do that you have to understand the rationale behind Lagranges multiplier. 
September 10th, 2018, 05:47 PM  #16 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550  Which only works if the maximizing values for x and y permit a solution for z in g(x,y,z) = 0. That is the problem with reasoning from examples. But the solution set of x,y pairs for solving g(x,y,z) = 0 may not intersect with the solution set of x,y pairs for optimizing f(x,y) at all. That is the rationale for using the Lagrangian process.

September 11th, 2018, 08:33 AM  #17 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
JeffM1. I acknowledge your method was correct, but i had to know why. So I figured out the rationale behind Lagranges multipliers here: Lagrange Multiplier with more constraint variables I posted it separately because I could never find a derivation. All that I could find was a rule and examples, to the point that I wonder if most authors understand it. After going through all that, i believe the better approach to OP type problem is to work with the differentials, as I outlined earlier. At least then you can see what's going on and think about it rather than using blind formalism for a case it wasn't derived for without a clue as to what's happening. Even better is simply solving constraint equation, plugging into object equation, and finding extrema. Outlined in latter part of my post#13. My thanks now are without reservations. 
September 11th, 2018, 11:44 AM  #18 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550 
@ Zylo I apologize most sincerely if I sounded a bit testy at times. Most of that was caused by the fact that I had already answered a very similar question by the same OP at another site. And that other question made clear that he was thinking in terms of theoretical economics so the OP did not give the context of the problem here. In theoretical economics, LaGrangian multipliers are standard technique for at least two important reasons. The first is that although some constraints, such as nonnegativity constraints, are fully specified, other constraints and the objective function itself are almost always only partially specified, and so the system cannot actually be solved numerically without additional information. What is being sought are certain general attributes of the maximum or minimum achievable under a general class of constraints. (As Samuelson said, theoretical economics is a qualitative calculus.) Furthermore, some of the constraints may not be relevant. Indeed you frequently have constraints such as x must be nonnegative but cannot exceed some positive function. Obviously, x cannot be both zero and positive:the constraints are literally contradictory. One may be binding and the other "slack," or both may be "slack." You can determine which constraints are "slack" from the sign of the Lagrangian multiplier. The second reason is that for binding constraints "The Lagrange Multiplier λ gives the ... [incremental] increase in the objective function from a unit relaxation of the constraint." You were correctly thinking about the numeric solution of fully specified problems (where LaGrangians may well be unnecessary) whereas I was thinking about the highly abstract problems of theoretical economics, where information about the multipliers may be the second most important result of the analysis. Had the OP shared the context of his concern, this might have been a less lengthy thread. Thank you for your kind words. 
September 15th, 2018, 10:36 AM  #19 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
An interestig example illustrating JeffM1's use of Lagrange multiplier for Subject is: Opt w=xy with $\displaystyle x^{2}+y^{2}+z^{2}k^{2}$=0 $\displaystyle L(x,y,z,\lambda)=xy\lambda (x^{2}+y^{2}+z^{2}k^{2})$ $\displaystyle \frac{\partial L }{\partial x}, \frac{\partial L }{\partial y},\frac{\partial L }{\partial z},\frac{\partial L }{\partial \lambda}$ all set equal to zero gives: $\displaystyle y2\lambda x=0$ $\displaystyle x2\lambda y=0$ $\displaystyle 2 \lambda z=0$ $\displaystyle x^{2}+y^{2}+z^{2}k^{2}=0 \rightarrow$ $\displaystyle z=0, x=\pm y, x^{2}=y^{2}$ $\displaystyle w=xy=x^{2}=k^{2}/2$ If you let x,y be independent. $\displaystyle dw=xdy+ydx =0 \rightarrow x=0, y=0, z^{2}=k^{2}$. This case is missed by Lagrange Multiplier. Same problem using differentials: Opt w=xy with $\displaystyle x^{2}+y^{2}+z^{2}k^{2}=0 $ dw=xdy+ydx 2xdx+2ydy+2zdz=0 Assume y and z are independent, solve for dx and substitute: $\displaystyle (x\frac{y^{2}}{x})dy\frac{yz}{x}dz=0$, and set coefficients of dy and dz equal to 0 since they are now independent. $\displaystyle x^{2}=y^{2}=k^{2}/2$ w=xy=k^{2}/2 For an overview of optimization see: https://en.wikipedia.org/wiki/Mathematical_optimization But this isn't necessary for OP, which is strictly a multivariable extremum problem with extra variables in the constraint. 

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constraint, function, objective, optimization, variables 
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