DOCSLIB.ORG

Lagrange Multiplier Method: Max Or Min of a Function F X , Y Subject to the Constraint G X , Y  K

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Lagrange Multiplier Method: Max Or Min of a Function F X , Y Subject to the Constraint G X , Y  K Review: Lagrange multiplier method: Max or min of a function f x, y subject to the constraint g x, y k : At a critical point x00, y , the gradient of f is parallel to the gradient of g. f x0, y 0 g x 0 , y 0 for some 1) write the constraint as gk 0 (subtract the constant ) 2 set up the function F x, y , f x , y g x , y 3 solve FFFF 0 xy 0, 0, and 0 4 find all such points that satisfy these equations and evaluate f at these points 5) The largest/smallest value of f will be the max/min of fg subject to the constraint . Max or min of f( x , y , z ) subject to the constraint g( x , y , z ) k : At a critical poit, fg for some constant Use above method for Fxyz , , , fxyz , , gxyz , , Suppose that the temperature at a point x, y on a metal plate is T x, y 4 x22 4 xy y . Example: An ant, walking on the plate, traverses a circle of radius 5 centered at the origin. What are the highest and lowest temperatures encountered by the ant? f44 x22 xy y g x22 y 25 F x, y , 4 x2 4 xy y 2 x 2 y 2 25 Fx 8 x 4 y 2 x 0 8x 2 x 4 y F x22 y 25 Fy 4 x 2 y 2 y 0 2y 2 y 4 x solve: xy4 2 , yx12 and xy2225 2y 4y x yx12 1 (4 ) 4 (what if y 0?) 4 4 or 2 5 4 4 ( 5)=0 0 or 5 If 0 then yx 2 hence x2+4 x 2 5 x 2 25 or x2 5 and y2 20 critical points: 5,2 5 and 5, 2 5 If 5 then xy 2 hence 4y2 +y 2 5y 2 25 or y22 5 and x 20 TT 5,2 5 5, 2 5 0 2 5, 5 and 2 5, 5 TT2 5, 5 2 5, 5 125 highest temparature: 125, lowest temparature: 0 The same principle works for a function of 3 variables. Find min or max of f( x , y , z ) subject to the constraint g( x , y , z ) k : At a critical poit, fg for some constant Example: Find the point on the cone z2 x 2 y 2 which is closest to the point (3,4,0). Minimize f( x , y , z ) ( x 3)2 ( y 4) 2 z 2 subject to the constraint g(x , y , z ) x2 y 2 z 2 0 Find critical points of F( x , y , z ) ( x 3)2 ( y 4) 2 z 2 ( x 2 y 2 z 2 ) Fx 2( x 3) 2 x 0 or (xx 3) Fyy 2(y 4) 2 0 or (y 4) y Fz 2 z 2 z 0 or z z solve (x 3) x , (y 4) y and z z and z2 x 2 y 2 a) 1 3 (x 3) x , or 2 x 3, x and (y 4) y or 2 y 4, or y 2 2 2 9 25 5 3 5 3 5 zz 4 or critical points ,2, and ,2, 4 4 2 2 2 2 2 b) z 0 xy 0 from the constraint does not solve the other equations A picture: Find the point on the cone z2 x 2 y 2 which is closest to the point (3,4,0). Minimize f( x , y , z ) ( x 3)2 ( y 4) 2 z 2 3 5 3 5 ff,2, ,2, 2 2 2 2 2 35 2 3 (2 4) 22 9 25 36 4 4 13 4 4 4 3 5 3 5 distance = 13 critical points ,2, and ,2, 2 2 2 2 There are 2 points on the cone that are closest! 15.1 Double Integrals over rectangles Single variable integral : area Recall: under the graph of the function f and above the x-axis is found by using the area of “infinitely” many rectangles. 1) Divide the interval [a,b ] into n small intervals. * 2) Choose some point xk in the n-th interval. * 3) Add the area f( xkk ) x of the rectangles. 4) Take the limit as n . n n b * * A f xkk x Alim f xkk x A f() x dx n i1 i1 a Double variable integral : volume under the graph of the function (surface) and above the xy-plane found by using the volume of “infinitely” many rectangular prisms. Compute the volume under the graph of f and over the rectangle [a,b]x[c,d] 1) Divide the square [a,b ] [c,d] into n small squares. 2) Choose some point (x ,y ) in the k-th square. kk Notation: V f( x , y ) dA 3) Add the volumes f( xk ,y k ) A k of the cylinders. R 4) Take the limit as n . n Vollim f xk , y k A k where A k 0 as n n i1 n Fubini' s Theorem for Rectangular Regions Let f be continuous over the rectangle R x, y a x b , c y d, then db f( x , y ) dA f ( x , y ) dx dy R c a Integrate w.r.t. x first Work inside out treating y as a constant b A y f( x , y ) dx cross sectional area a "Add" up the cross sectional areas: d Volume A( y ) dy c We can also do this the other way around: d b b d f(,) x y dA f (,) x y dx dy f(,) x y dy dx R c a a c Integrate w.r.t. y first treating x as a constant d A x f( x , y ) dy cross sectional area c "Add" up the cross sectional areas: b Volume A( x ) dx a We usually simply write: d b b d f(,) x y dA f (,) x y dx dy f(,) x y dy dx R c a a c Example : Find the volume V of the solid lying under the graph of the elliptic paraboloid z8 2 x22 y and above the rectangle R x, y 0 x 1,0 y 2 . V 82 x22 y dA R 21 8 2x22 y dxdy 00 2 3 x1 2x 2 8x xy dy 0 3 x0 2 2 3 2 2 2 22 2 22 y 44 8 8 y dy y dy y 3 0 0 3 330 33 Ay 12 13 1 Example : Find the value of the double interal dxdy 3 02 xy 4 13 13 1 x3 1 3 1 2 dxdy x4 y dxdy x4 y dy 3 02 xy 4 02 0 2 x2 1 1 22 3 4y 2 4 y dy 2 0 111 1 13 4yy 1 2 4 1 1 1 1 2 1 4 1 4 2 4 3 4yy 4 2 4 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 8 3 4yy 2 4 0 8 7 6 3 2 8 7 6 3 2 1 6 7 14 21 16 1 8 42 8 42 56 y Example : Find the value of the double integral dA 22 R xy 1 where R is the rectangle 0xy 2, 0 1 We have to decide if we integrate first with respect to xy or with respect to . 1 If we do x first, we have to use the indefinite integral dx arctan x 1 x2 21 y y dA dydx maybe y first: 22 22 R xy 1 00xy 1 substitute: u x2 y 2 1, du 2 x 2 ydy y0 u 1, y 1 u x2 1 2 2 2 1 x 1 x 1 x 1 2 y ydy du 1 du 1ux2 1 ln(x 1) 22 dy 2 ln(u ) xy 1 2xu 2 22u1 0 1 u 1 2xu1 22xx y 2 1 dAln( x2 1) dx use u' v uv uv ' 2 2 2 R x y120 x 1 2 2 u, v ln( x 1) 122 1 2x ln(5) 1 ln(x2 1) dx dx 2x 0022 2x0 2 x x 1 4 x 1 12x uv'22 , ' ln(5) 2 ln(5) 21xx arctan(x ) arctan(2) 4 0 4 f x, y is called separable if it can be writtten as f x, y g x h y . If fR is separable and integrable over a rectangle , then bd f x, y dA g x h y dA g x dx h y dy RR ac 5 7 e x Example : Evaluate dydx 1 e y 55 7 77ee 5 2 x 1 e x 5 49 1 dydx dy xdx ln y lnee ln e 22 11eeyy 2 1 48 51 96 2 15.2 Double Integrals over general regions For a double integral f x, y dA R over a more general region R in the x y plane we again divide R into many small rectangles We will miss smaller and smaller regions as the subdivision gets finer. We then add up the volume of the rectangular boxes n Vollim f xk , y k A k n i1 Compute the volume under fR and over a region bounded by the two curves y g12( x ) and y g( x ) by slicing: gx2 Cross sectional area A( x ) f x , y dy gx1 b Volume A( x ) dx a b gx2 f x,, y dA f x y dydx R a g1 x A picture in the x-y plane: “sweep” out the region b gx2 f x, y dA f x, y dydx R a g1 x Example : Compute the double intregral 4xy y3 dA R where R is the region bounded by y x and y x3 4xy y3 dydx R yx 1 x 1 4 1 2 12 3 2 y 27xx 4xy y dydx 2xy dx 22x x dx 4 3 44 0 x3 0 yx 0 2 1 7x 1 3 8 13 2 12 4 7xx7 7x x x 7 1 1 2x dx 12 4 52 0 440 12 4 52 7 3 1 11 52 3 55 12 52 3 52 156 156.
Load more

Recommended publications
  • Introduction to the Modern Calculus of Variations
    MA4G6 Lecture Notes Introduction to the Modern Calculus of Variations Filip Rindler Spring Term 2015 Filip Rindler Mathematics Institute University of Warwick Coventry CV4 7AL United Kingdom [email protected] http://www.warwick.ac.uk/filiprindler Copyright ©2015 Filip Rindler. Version 1.1. Preface These lecture notes, written for the MA4G6 Calculus of Variations course at the University of Warwick, intend to give a modern introduction to the Calculus of Variations. I have tried to cover different aspects of the field and to explain how they fit into the “big picture”. This is not an encyclopedic work; many important results are omitted and sometimes I only present a special case of a more general theorem. I have, however, tried to strike a balance between a pure introduction and a text that can be used for later revision of forgotten material. The presentation is based around a few principles: • The presentation is quite “modern” in that I use several techniques which are perhaps not usually found in an introductory text or that have only recently been developed. • For most results, I try to use “reasonable” assumptions, not necessarily minimal ones. • When presented with a choice of how to prove a result, I have usually preferred the (in my opinion) most conceptually clear approach over more “elementary” ones. For example, I use Young measures in many instances, even though this comes at the expense of a higher initial burden of abstract theory. • Wherever possible, I first present an abstract result for general functionals defined on Banach spaces to illustrate the general structure of a certain result.
    [Show full text]
  • Finite Elements for the Treatment of the Inextensibility Constraint
    Comput Mech DOI 10.1007/s00466-017-1437-9 ORIGINAL PAPER Fiber-reinforced materials: finite elements for the treatment of the inextensibility constraint Ferdinando Auricchio1 · Giulia Scalet1 · Peter Wriggers2 Received: 5 April 2017 / Accepted: 29 May 2017 © Springer-Verlag GmbH Germany 2017 Abstract The present paper proposes a numerical frame- 1 Introduction work for the analysis of problems involving fiber-reinforced anisotropic materials. Specifically, isotropic linear elastic The use of fiber-reinforcements for the development of novel solids, reinforced by a single family of inextensible fibers, and efficient materials can be traced back to the 1960s are considered. The kinematic constraint equation of inex- and, to date, it is an active research topic. Such materi- tensibility in the fiber direction leads to the presence of als are composed of a matrix, reinforced by one or more an undetermined fiber stress in the constitutive equations. families of fibers which are systematically arranged in the To avoid locking-phenomena in the numerical solution due matrix itself. The interest towards fiber-reinforced mate- to the presence of the constraint, mixed finite elements rials stems from the mechanical properties of the fibers, based on the Lagrange multiplier, perturbed Lagrangian, and which offer a significant increase in structural efficiency. penalty method are proposed. Several boundary-value prob- Particularly, these materials possess high resistance and lems under plane strain conditions are solved and numerical stiffness, despite the low specific weight, together with an results are compared to analytical solutions, whenever the anisotropic mechanical behavior determined by the fiber derivation is possible. The performed simulations allow to distribution.
    [Show full text]
  • A Variational Approach to Lagrange Multipliers
    JOTA manuscript No. (will be inserted by the editor) A Variational Approach to Lagrange Multipliers Jonathan M. Borwein · Qiji J. Zhu Received: date / Accepted: date Abstract We discuss Lagrange multiplier rules from a variational perspective. This allows us to highlight many of the issues involved and also to illustrate how broadly an abstract version can be applied. Keywords Lagrange multiplier Variational method Convex duality Constrained optimiza- tion Nonsmooth analysis · · · · Mathematics Subject Classification (2000) 90C25 90C46 49N15 · · 1 Introduction The Lagrange multiplier method is fundamental in dealing with constrained optimization prob- lems and is also related to many other important results. There are many different routes to reaching the fundamental result. The variational approach used in [1] provides a deep under- standing of the nature of the Lagrange multiplier rule and is the focus of this survey. David Gale's seminal paper [2] provides a penetrating explanation of the economic meaning of the Lagrange multiplier in the convex case. Consider maximizing the output of an economy with resource constraints. Then the optimal output is a function of the level of resources. It turns out the derivative of this function, if exists, is exactly the Lagrange multiplier for the constrained optimization problem. A Lagrange multiplier, then, reflects the marginal gain of the output function with respect to the vector of resource constraints. Following this observation, if we penalize the resource utilization with a (vector) Lagrange multiplier then the constrained optimization problem can be converted to an unconstrained one. One cannot emphasize enough the importance of this insight. In general, however, an optimal value function for a constrained optimization problem is nei- ther convex nor smooth.
    [Show full text]
  • Derivative, Gradient, and Lagrange Multipliers
    EE263 Prof. S. Boyd Derivative, Gradient, and Lagrange Multipliers Derivative Suppose f : Rn → Rm is differentiable. Its derivative or Jacobian at a point x ∈ Rn is × denoted Df(x) ∈ Rm n, defined as ∂fi (Df(x))ij = , i =1, . , m, j =1,...,n. ∂x j x The first order Taylor expansion of f at (or near) x is given by fˆ(y)= f(x)+ Df(x)(y − x). When y − x is small, f(y) − fˆ(y) is very small. This is called the linearization of f at (or near) x. As an example, consider n = 3, m = 2, with 2 x1 − x f(x)= 2 . x1x3 Its derivative at the point x is 1 −2x2 0 Df(x)= , x3 0 x1 and its first order Taylor expansion near x = (1, 0, −1) is given by 1 1 1 0 0 fˆ(y)= + y − 0 . −1 −1 0 1 −1 Gradient For f : Rn → R, the gradient at x ∈ Rn is denoted ∇f(x) ∈ Rn, and it is defined as ∇f(x)= Df(x)T , the transpose of the derivative. In terms of partial derivatives, we have ∂f ∇f(x)i = , i =1,...,n. ∂x i x The first order Taylor expansion of f at x is given by fˆ(x)= f(x)+ ∇f(x)T (y − x). 1 Gradient of affine and quadratic functions You can check the formulas below by working out the partial derivatives. For f affine, i.e., f(x)= aT x + b, we have ∇f(x)= a (independent of x). × For f a quadratic form, i.e., f(x)= xT Px with P ∈ Rn n, we have ∇f(x)=(P + P T )x.
    [Show full text]
  • Contactmethodsintegratingplasti
    Contact methodsintegrating plasticity B13/1 modelswith application to soil mechanics ChristianWeißenfels Leibniz Universität Hannover Contact methods integrating plasticity models with application to soil mechanics Von der Fakultät für Maschinenbau der Gottfried Wilhelm Leibniz Universität Hannover zur Erlangung des akademischen Grades Doktor-Ingenieur genehmigte Dissertation von Dipl.-Ing. Christian Weißenfels geboren am 30.01.1979 in Rosenheim 2013 Herausgeber: Prof. Dr.-Ing. Peter Wriggers Verwaltung: Institut für Kontinuumsmechanik Gottfried Wilhelm Leibniz Universität Hannover Appelstraße 11 30167 Hannover Tel: +49 511 762 3220 Fax: +49 511 762 5496 Web: www.ikm.uni-hannover.de © Dipl.-Ing. Christian Weißenfels Institut für Kontinuumsmechanik Gottfried Wilhelm Leibniz Universität Hannover Appelstraße 11 30167 Hannover Alle Rechte, insbesondere das der Übersetzung in fremde Sprachen, vorbehalten. Ohne Genehmigung des Autors ist es nicht gestattet, dieses Heft ganz oder teilweise auf photomechanischem, elektronischem oder sonstigem Wege zu vervielfältigen. ISBN 978-3-941302-06-8 1. Referent: Prof. Dr.-Ing. Peter Wriggers 2. Referent: Prof. Dr.-Ing. Karl Schweizerhof Tag der Promotion: 13.12.2012 i Zusammenfassung Der Schwerpunkt der vorliegenden Arbeit liegt in der Entwicklung von neuartigen Kon- zepten zur direkten Integration von drei-dimensionalen Plastizit¨atsmodellen in eine Kontaktformulierung. Die allgemeinen Konzepte wurden speziell auf Boden Bauwerk Interaktionen angewandt und im Rahmen der Mortar-Methode numerisch umgesetzt. Zwei unterschiedliche Strategien wurden dabei verfolgt. Die Erste integriert die Boden- modelle in eine Standard-Kontaktdiskretisierung, wobei die zweite Variante die Kon- taktformulierung in Richtung eines drei-dimensionalen Kontaktelements erweitert. Innerhalb der ersten Variante wurden zwei unterschiedliche Konzepte ausgearbeitet, wobei das erste Konzept das Bodenmodell mit dem Reibbeiwert koppelt. Dabei muss die H¨ohe der Lokalisierungszone entlang der Kontaktfl¨ache bestimmt werden.
    [Show full text]
  • Dual-Dual Formulations for Frictional Contact Problems in Mechanics
    View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Institutionelles Repositorium der Leibniz Universität... Dual-dual formulations for frictional contact problems in mechanics Von der Fakultät für Mathematik und Physik der Gottfried Wilhelm Leibniz Universität Hannover zur Erlangung des Grades Doktor der Naturwissenschaften Dr. rer. nat. genehmigte Dissertation von Dipl. Math. Michael Andres geboren am 28. 12. 1980 in Siemianowice (Polen) 2011 Referent: Prof. Dr. Ernst P. Stephan, Leibniz Universität Hannover Korreferent: Prof. Dr. Gabriel N. Gatica, Universidad de Concepción, Chile Korreferent: PD Dr. Matthias Maischak, Brunel University, Uxbridge, UK Tag der Promotion: 17. 12. 2010 ii To my Mum. Abstract This thesis deals with unilateral contact problems with Coulomb friction. The main focus of this work lies on the derivation of the dual-dual formulation for a frictional contact problem. First, we regard the complementary energy minimization problem and apply Fenchel’s duality theory. The result is a saddle point formulation of dual type involving Lagrange multipliers for the governing equation, the symmetry of the stress tensor as well as the boundary conditions on the Neumann boundary and the contact boundary, respectively. For the saddle point problem an equivalent variational inequality problem is presented. Both formulations include a nondiffer- entiable functional arising from the frictional boundary condition. Therefore, we introduce an additional dual Lagrange multiplier denoting the friction force. This procedure yields a dual-dual formulation of a two-fold saddle point structure. For the corresponding variational inequality problem we show the unique solvability. Two different inf-sup conditions are introduced that allow an a priori error analysis of the dual-dual variational inequality problem.
    [Show full text]
  • 18.02SC Notes: Lagrange Multipliers
    Lagrange Multipliers We will give the argument for why Lagrange multipliers work later. Here, we’ll look at where and how to use them. Lagrange multipliers are used to solve constrained optimization problems. That is, suppose you have a function, say f(x; y), for which you want to find the maximum or minimum value. But, you are not allowed to consider all (x; y) while you look for this value. Instead, the (x; y) you can consider are constrained to lie on some curve or surface. There are lots of examples of this in science, engineering and economics, for example, optimizing some utility function under budget constraints. Lagrange multipliers problem: Minimize (or maximize) w = f(x; y; z) constrained by g(x; y; z) = c. Lagrange multipliers solution: Local minima (or maxima) must occur at a critical point. This is a point where rf = λrg, and g(x; y; z) = c. Example: Making a box using a minimum amount of material. A box is made of cardboard with double thick sides, a triple thick bottom, single thick front and back and no top. It’s volume is fixed at 3. What dimensions use the least amount of cardboard? Answer: We did this problem once before by solving for z in terms of x and y and substi­ tuting for it. That led to an unconstrained optimization problem in x and y. Here we will do it as a constrained problem. It is important to be able to do this because eliminating one variable is not always easy. The box shown has dimensions x, y, and z.
    [Show full text]
  • ECON 301 Two Variable Optimization (With- and Without- Constraints)
    ECON 301 Two Variable Optimization (with- and without- constraints) Review of Some Derivative Rules 1. Partial Derivative Rules: U = xy ∂U/∂x = Ux = y ∂U/∂y = Uy = x a b a 1 b a b 1 U = x y ∂U/∂x = Ux = ax − y ∂U/∂y = Uy = bx y − a b xa a 1 b a b 1 U = x y− = b ∂U/∂x = Ux = ax − y− ∂U/∂y = Uy = bx y− − y − U = ax + by ∂U/∂x = Ux = a ∂U/∂y = Uy = b 1/2 1/2 1 1/2 1 1/2 U = ax + by ∂U/∂x = Ux = a 2 x− ∂U/∂y = Uy = b 2 y− 2. Logarithm (Natural log) ln x ¡ ¢ ¡ ¢ (a) Rules of natural log If Then y = AB ln y =ln(AB)=lnA +lnB y = A/B ln y =lnA ln B y = Ab ln y =ln(Ab−)=b ln A NOTE: ln(A + B) =lnA +lnB 6 (b) derivatives IF THEN dy 1 y =lnx dx = x dy 1 y =ln(f(x)) = f 0(x) dx f(x) · (c) Examples If Then 2 1 y =ln(x 2x) dy/dx = (x2 2x) (2x 2) 1/−2 1 1 − 1 −1 y =ln(x )= 2 ln xdy/dx= 2 x = 2x ¡ ¢¡ ¢ 3. The Number e dy if y = ex then = ex dx f(x) dy f(x) if y = e then = e f 0(x) dx · (a) Examples 3x dy 3x y = e dx = e (3) 7x3 dy 7x3 2 y = e dx = e (21x ) rt dy rt y = e dt = re 1 Using Calculus For Maximization Problems OneVariableCase If we have the following function y =10x x2 − we have an example of a dome shaped function.
    [Show full text]
  • Math 241: Multivariable Calculus, Lecture 15 Lagrange Multipliers
    Math 241: Multivariable calculus, Lecture 15 Lagrange Multipliers. go.illinois.edu/math241fa17 Wednesday, October 4th, 2017 go.illinois.edu/math241fa17. Math 241: Problem of the day Problem: Find the absolute maximum and minimum value of the function f (x; y) = x3 − x + y 2x + y + 2 on the unit disk D = f(x; y) j x2 + y 2 ≤ 1g. What guarantees that the max and min exist? go.illinois.edu/math241fa17. rf (x; y) = h3x2 − 1 + y 2; 2yx + 1i −1 If this is the zero vector we get 2xy + 1 = 0, or y = 2x . (Note that x nor y can be zero, so dividing by x or y is no problem.) Substitute in the first component to get 1 1 3x2 − 1 + = 0 =) 3x4 − x2 + = 0: 4x2 4 2 2 1 Put z = x , get 3z − z + 4 = 0. The discriminant b2 − 4ac = 1 − 3 = −2, so there are NO critical points. So we need to look at the boundary! Solution Part 1: Critical points If f (x; y) = x3 − x + y 2x + y + 2 the gradient vector is go.illinois.edu/math241fa17. −1 If this is the zero vector we get 2xy + 1 = 0, or y = 2x . (Note that x nor y can be zero, so dividing by x or y is no problem.) Substitute in the first component to get 1 1 3x2 − 1 + = 0 =) 3x4 − x2 + = 0: 4x2 4 2 2 1 Put z = x , get 3z − z + 4 = 0. The discriminant b2 − 4ac = 1 − 3 = −2, so there are NO critical points. So we need to look at the boundary! Solution Part 1: Critical points If f (x; y) = x3 − x + y 2x + y + 2 the gradient vector is rf (x; y) = h3x2 − 1 + y 2; 2yx + 1i go.illinois.edu/math241fa17.
    [Show full text]
  • Notes on Calculus and Optimization
    Economics 101A Section Notes GSI: David Albouy Notes on Calculus and Optimization 1 Basic Calculus 1.1 Definition of a Derivative Let f (x) be some function of x, then the derivative of f, if it exists, is given by the following limit df (x) f (x + h) f (x) = lim − (Definition of Derivative) dx h 0 h → df although often this definition is hard to apply directly. It is common to write f 0 (x),or dx to be shorter, or dy if y = f (x) then dx for the derivative of y with respect to x. 1.2 Calculus Rules Here are some handy formulas which can be derived using the definitions, where f (x) and g (x) are functions of x and k is a constant which does not depend on x. d df (x) [kf (x)] = k (Constant Rule) dx dx d df (x) df (x) [f (x)+g (x)] = + (Sum Rule) dx dx dy d df (x) dg (x) [f (x) g (x)] = g (x)+f (x) (Product Rule) dx dx dx d f (x) df (x) g (x) dg(x) f (x) = dx − dx (Quotient Rule) dx g (x) [g (x)]2 · ¸ d f [g (x)] dg (x) f [g (x)] = (Chain Rule) dx dg dx For specific forms of f the following rules are useful in economics d xk = kxk 1 (Power Rule) dx − d ex = ex (Exponent Rule) dx d 1 ln x = (Logarithm Rule) dx x 1 Finally assuming that we can invert y = f (x) by solving for x in terms of y so that x = f − (y) then the following rule applies 1 df − (y) 1 = 1 (Inverse Rule) dy df (f − (y)) dx Example 1 Let y = f (x)=ex/2, then using the exponent rule and the chain rule, where g (x)=x/2,we df (x) d x/2 d x/2 d x x/2 1 1 x/2 get dx = dx e = d(x/2) e dx 2 = e 2 = 2 e .
    [Show full text]
  • Unit #23 : Lagrange Multipliers Goals: • to Study Constrained Optimization
    Unit #23 : Lagrange Multipliers Goals: • To study constrained optimization; that is, the maximizing or minimizing of a function subject to a constraint (or side condition). Constrained Optimization - Examples - 1 In the previous section, we saw some of the difficulties of working with optimization when there are multiple variables. Many of those problems can be cast into an important class of problems called constrained optimization problems, which can be solved in an alternative way. Examples of Problems with Constraints 1. Show that the rectangle of maximum area that has a given perimeter p is a square. The function to be maximized: A(x; y) = xy The constraint: 2x + 2y = p Constrained Optimization - Examples - 2 2. Find the point on the sphere x2 + y2 + z2 = 4 that is closest to the point (3; 3; 5). Here we want to minimize the distance D(x; y; z) = p(x − 3)2 + (y − 3)2 + (z − 5)2 subject to the constraint x2 + y2 + z2 = 4 Lagrange Multipliers - 1 Lagrange Multipliers To solve optimization problems when we have constraints on our choice of x and y, we can use the method of Lagrange multipliers. Suppose we want to maximize the function f(x; y) subject to the constraint g(x; y) = k. We consider the relative positions of some sample level curves on the next page. Lagrange Multipliers - 2 Black lines: contours of f(x; y). Blue lines: the constraint g(x; y) = k. f = 10 f = 20 f = 30 g(x,y)= k f = 40 f = 50 f = 60 On the diagram, locate the maximum and minimum of f(x; y) (label with M and m respectively) on the level curve g(x; y) = k.
    [Show full text]
  • A Global Divergence Conforming DG Method for Hyperbolic Conservation Laws with Divergence Constraint
    A global divergence conforming DG method for hyperbolic conservation laws with divergence constraint Praveen Chandrashekar∗ Abstract We propose a globally divergence conforming discontinuous Galerkin (DG) method on Cartesian meshes for curl-type hyperbolic conservation laws based on directly evolving the face and cell moments of the Raviart-Thomas approximation polyno- mials. The face moments are evolved using a 1-D discontinuous Gakerkin method that uses 1-D and multi-dimensional Riemann solvers while the cell moments are evolved using a standard 2-D DG scheme that uses 1-D Riemann solvers. The scheme can be implemented in a local manner without the need to solve a global mass matrix which makes it a truly DG method and hence useful for explicit time stepping schemes for hyperbolic problems. The scheme is also shown to exactly pre- serve the divergence of the vector field at the discrete level. Numerical results using second and third order schemes for induction equation are presented to demonstrate the stability, accuracy and divergence preservation property of the scheme. Keywords: Hyperbolic conservation laws; curl-type equations; discontinuous Galerkin; constraint-preserving; divergence-free; induction equation. 1 Introduction Constraint-preserving approximations are important in the numerical simulation of prob- lems in computational electrodynamics (CED) and magnetohydrodynamics (MHD). The time domain Maxwell equations used in CED for the electric and magnetic fields may be written in non-dimensional units as @E @B B = J; + E = 0 @t − r × − @t r × with the constraint that @ρ B = 0; E = ρ, + J = 0 r · r · @t r · where ρ is the electric charge density and J is the current which can be related to E, arXiv:1809.03294v1 [math.NA] 10 Sep 2018 B through Ohm's Law.
    [Show full text]