 My Math Forum Lagrange Multiplier with more constraint variables

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 September 11th, 2018, 08:11 AM #1 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 Lagrange Multiplier with more constraint variables Extrema of f(x,y) with constraint g(x,y)=0 $\displaystyle f_{x}dx+f_{y}dy=0$ at an extremum, but dx and dy not independent: $\displaystyle g_{x}dx+g_{y}dy=0$ $\displaystyle dy=-\frac{g_{x}}{g_{y}}dx$ $\displaystyle f_{x}dx+f_{y}(-\frac{g_{x}}{g_{y}}dx)=0$ $\displaystyle (f_{x}-\frac{f_{y}}{g_{y}}g_{x})dx=0$ for arbitrary dx $\displaystyle \rightarrow f_{x}-\frac{f_{y}}{g_{y}}g_{x}=0$ Let $\displaystyle \lambda =\frac{f_{y}}{g_{y}}$ Then $\displaystyle f_{y}-\lambda g_{y}=0$ $\displaystyle f_{x}-\lambda g_{x}=0$ $\displaystyle g(x,y)=0$ are 3 eqs in 3 unknowns x,y,$\displaystyle \lambda$ As a purely formal rule, the equations can be gotten by setting partial derivatives wrt x,y,$\displaystyle \lambda$ equal to zero of: L(x,y,$\displaystyle \lambda$)=f(x,y)-$\displaystyle \lambda$(g(x,y) ------------------------------------------------------------------------------------------------------ For extrema of f(x,y,z) with constraint g(x,y,z)=0: $\displaystyle f_{x}dx+f_{y}dy+f_{z}dz=0$ at an extremum, but dx,dy,dz not independent: $\displaystyle g_{x}dx+g_{y}dy+g_{z}dz=0$ $\displaystyle dy=-\frac{g_{x}}{g_{y}}dx-\frac{g_{z}}{g_{y}}dz$ $\displaystyle f_{x}dx+f_{y}(-\frac{g_{x}}{g_{y}}dx-\frac{g_{z}}{g_{y}}dz)+f_{z}dz=0$ $\displaystyle (f_{x}-\frac{f_{y}}{g_{y}}g_{x})dx+(f_{z}-\frac{f_{y}}{g_{y}}g_{z})dz=0$ arbitrary dx and dz $\displaystyle \rightarrow f_{x}-\frac{f_{y}}{g_{y}}g_{x}=f_{z}-\frac{f_{y}}{g_{y}}g_{z}=0$ Let $\displaystyle \lambda = \frac{f_{y}}{g_{y}}$ and you have Lagrange's multiplier. As a formal rule to get the equations, set to 0 partial derivatives wrt x,y,z,$\displaystyle \lambda$ of L(x,y,z,$\displaystyle \lambda$)=f(x,y,z)-$\displaystyle \lambda$ g(x,y,z): $\displaystyle \frac{\partial L }{\partial x}=f_{x}-\lambda g_{x}=0$ $\displaystyle \frac{\partial L }{\partial y}=f_{y}-\lambda g_{y}=0$ $\displaystyle \frac{\partial L }{\partial z}=f_{z}-\lambda g_{z}=0$ $\displaystyle \frac{\partial L }{\partial \lambda}=g(x,y,z)=0$ -------------------------------------------------------------------------------------- Now consider extrema of f(x,y) subject to g(x,y,z) = 0 $\displaystyle f_{x}dx+f_{y}dy=0$ $\displaystyle g_{x}dx+g_{y}dy+g_{z}dz=0$ To change independent variable to, say, dx and dz, solve latter equation for for dy and plug into former to get: $\displaystyle (f_{x}-\frac{f_{y}}{g_{y}}g_{x})dx-\frac{f_{y}}{g_{y}}g_{z}dz=0$ Now setting the coefficients of dx and dz (arbitray) to zero and letting $\displaystyle \lambda = \frac{f_{y}}{g_{y}}$, gives the modified form of Lagrange's equations for this case: $\displaystyle f_{x}-\lambda g_{x}=0$ $\displaystyle f_{y}-\lambda g_{y}=0$ $\displaystyle \lambda g_{z}=0$ $\displaystyle g(x,y,z)=0$ Formally, you get the equations, as before, from: L(x,y,z,$\displaystyle \lambda$) = f(x,y)-$\displaystyle \lambda$ g(x,y,z) by taking partial derivatives wrt x,y,z,$\displaystyle \lambda$ and setting them equal to 0. The latter procedure was used by JeffM1 in: Optimization with 2 variables in the objective function and 3 variables in constraint -------------------------------------------------------------------------------------- For the case of more than one constraint: Extrema of f(x,y,z,w) and g(x,y,z,w) and h(x,y,x,w) Solve for the differentials of dz and dw in terms of dx and dy from dg=0 and dh=0. Then substitute into df=0 and equate the coefficients of dx and dy to zero. It will be obvious what to label $\displaystyle \lambda$1 and $\displaystyle \lambda$2 as, and everything will fall into place. Forget the formulas, grasp the simple principle: At an extreme, df(x,y,z,w,..)=0 for arbitrary dx, dy, dz, dw... (independent). So their coefficients must be zero. I have taken pains to do this because I never came across Lagrange multipliers except as a rule which I didn't understand. Last edited by skipjack; September 11th, 2018 at 12:49 PM. Tags constraint, lagrange, multiplier, variables Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Caesar95 Calculus 2 May 2nd, 2015 08:04 AM Brazen Calculus 1 January 15th, 2013 10:29 AM trey01 Applied Math 2 March 25th, 2012 07:14 AM dk1702 Abstract Algebra 1 July 21st, 2010 05:25 AM roonaldo17 Calculus 0 November 16th, 2008 11:27 AM

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