微觀經(jīng)濟(jì)理論英文版課件：ch02 THE MATHEMATICS OF OPTIMIZATION

1、Chapter 2THE MATHEMATICS OF OPTIMIZATIONCopyright 2005 by South-Western, a division of Thomson Learning. All rights reserved.1The Mathematics of OptimizationMany economic theories begin with the assumption that an economic agent is seeking to find the optimal value of some functionconsumers seek to

2、maximize utilityfirms seek to maximize profitThis chapter introduces the mathematics common to these problems2Maximization of a Function of One VariableSimple example: Manager of a firm wishes to maximize profits = f(q)Quantity*q*Maximum profits of* occur at q*3Maximization of a Function of One Vari

3、ableThe manager will likely try to vary q to see where the maximum profit occursan increase from q1 to q2 leads to a rise in = f(q)Quantity*q*1q12q24Maximization of a Function of One VariableIf output is increased beyond q*, profit will declinean increase from q* to q3 leads to a drop in = f(q)Quant

4、ity*q*3q35DerivativesThe derivative of = f(q) is the limit of /q for very small changes in qThe value of this ratio depends on the value of q16Value of a Derivative at a PointThe evaluation of the derivative at the point q = q1 can be denotedIn our previous example,7First Order Condition for a Maxim

5、umFor a function of one variable to attain its maximum value at some point, the derivative at that point must be zero8Second Order ConditionsThe first order condition (d/dq) is a necessary condition for a maximum, but it is not a sufficient conditionQuantity*q*If the profit function was u-shaped,the

6、 first order condition would resultin q* being chosen and wouldbe minimized9Second Order ConditionsThis must mean that, in order for q* to be the optimum, andTherefore, at q*, d/dq must be decreasing 10Second DerivativesThe derivative of a derivative is called a second derivativeThe second derivativ

7、e can be denoted by11Second Order ConditionThe second order condition to represent a (local) maximum is12Rules for Finding Derivatives13Rules for Finding Derivativesa special case of this rule is dex/dx = ex14Rules for Finding DerivativesSuppose that f(x) and g(x) are two functions of x and f(x) and

8、 g(x) existThen15Rules for Finding Derivatives16Rules for Finding DerivativesIf y = f(x) and x = g(z) and if both f(x) and g(x) exist, then:This is called the chain rule. The chain rule allows us to study how one variable (z) affects another variable (y) through its influence on some intermediate va

9、riable (x)17Rules for Finding DerivativesSome examples of the chain rule include18Example of Profit MaximizationSuppose that the relationship between profit and output is = 1,000q - 5q2The first order condition for a maximum isd/dq = 1,000 - 10q = 0q* = 100Since the second derivative is always -10,

10、q = 100 is a global maximum19Functions of Several VariablesMost goals of economic agents depend on several variablestrade-offs must be madeThe dependence of one variable (y) on a series of other variables (x1,x2,xn) is denoted by20The partial derivative of y with respect to x1 is denoted byPartial D

11、erivativesIt is understood that in calculating the partial derivative, all of the other xs are held constant21A more formal definition of the partial derivative isPartial Derivatives22Calculating Partial Derivatives23Calculating Partial Derivatives24Partial DerivativesPartial derivatives are the mat

12、hematical expression of the ceteris paribus assumptionshow how changes in one variable affect some outcome when other influences are held constant25Partial DerivativesWe must be concerned with how variables are measuredif q represents the quantity of gasoline demanded (measured in billions of gallon

13、s) and p represents the price in dollars per gallon, then q/p will measure the change in demand (in billiions of gallons per year) for a dollar per gallon change in price26ElasticityElasticities measure the proportional effect of a change in one variable on anotherunit freeThe elasticity of y with r

14、espect to x is27Elasticity and Functional FormSuppose thaty = a + bx + other termsIn this case,ey,x is not constantit is important to note the point at which the elasticity is to be computed28Elasticity and Functional FormSuppose thaty = axb In this case,29Elasticity and Functional FormSuppose thatl

15、n y = ln a + b ln xIn this case,Elasticities can be calculated through logarithmic differentiation30Second-Order Partial DerivativesThe partial derivative of a partial derivative is called a second-order partial derivative31Youngs TheoremUnder general conditions, the order in which partial different

16、iation is conducted to evaluate second-order partial derivatives does not matter32Use of Second-Order PartialsSecond-order partials play an important role in many economic theoriesOne of the most important is a variables own second-order partial, fiishows how the marginal influence of xi on y(y/xi)

17、changes as the value of xi increasesa value of fii 0 indicates diminishing marginal effectiveness33Total DifferentialSuppose that y = f(x1,x2,xn)If all xs are varied by a small amount, the total effect on y will be34First-Order Condition for a Maximum (or Minimum)A necessary condition for a maximum

18、(or minimum) of the function f(x1,x2,xn) is that dy = 0 for any combination of small changes in the xs The only way for this to be true is if A point where this condition holds is called a critical point35Finding a MaximumSuppose that y is a function of x1 and x2y = - (x1 - 1)2 - (x2 - 2)2 + 10y = -

19、 x12 + 2x1 - x22 + 4x2 + 5First-order conditions imply thatOR36Production Possibility FrontierEarlier example: 2x2 + y2 = 225Can be rewritten: f(x,y) = 2x2 + y2 - 225 = 0Because fx = 4x and fy = 2y, the opportunity cost trade-off between x and y is37Implicit Function TheoremIt may not always be poss

20、ible to solve implicit functions of the form g(x,y)=0 for unique explicit functions of the form y = f(x)mathematicians have derived the necessary conditionsin many economic applications, these conditions are the same as the second-order conditions for a maximum (or minimum)38The Envelope TheoremThe

21、envelope theorem concerns how the optimal value for a particular function changes when a parameter of the function changesThis is easiest to see by using an example39The Envelope TheoremSuppose that y is a function of xy = -x2 + axFor different values of a, this function represents a family of inver

22、ted parabolasIf a is assigned a specific value, then y becomes a function of x only and the value of x that maximizes y can be calculated40The Envelope TheoremOptimal Values of x and y for alternative values of a41The Envelope TheoremAs a increases,the maximal valuefor y (y*) increasesThe relationsh

23、ipbetween a and yis quadratic42The Envelope TheoremSuppose we are interested in how y* changes as a changesThere are two ways we can do thiscalculate the slope of y directlyhold x constant at its optimal value and calculate y/a directly43The Envelope TheoremTo calculate the slope of the function, we

24、 must solve for the optimal value of x for any value of ady/dx = -2x + a = 0 x* = a/2Substituting, we gety* = -(x*)2 + a(x*) = -(a/2)2 + a(a/2)y* = -a2/4 + a2/2 = a2/444The Envelope TheoremTherefore,dy*/da = 2a/4 = a/2 = x*But, we can save time by using the envelope theoremfor small changes in a, dy

25、*/da can be computed by holding x at x* and calculating y/ a directly from y45The Envelope Theoremy/ a = xHolding x = x*y/ a = x* = a/2This is the same result found earlier46The Envelope TheoremThe envelope theorem states that the change in the optimal value of a function with respect to a parameter

26、 of that function can be found by partially differentiating the objective function while holding x (or several xs) at its optimal value47The Envelope TheoremThe envelope theorem can be extended to the case where y is a function of several variablesy = f(x1,xn,a)Finding an optimal value for y would c

27、onsist of solving n first-order equations y/xi = 0 (i = 1,n)48The Envelope TheoremOptimal values for theses xs would be determined that are a function of ax1* = x1*(a)x2* = x2*(a)xn*= xn*(a).49The Envelope TheoremSubstituting into the original objective function yields an expression for the optimal

28、value of y (y*)y* = f x1*(a), x2*(a),xn*(a),aDifferentiating yields50The Envelope TheoremBecause of first-order conditions, all terms except f/a are equal to zero if the xs are at their optimal valuesTherefore,51Constrained MaximizationWhat if all values for the xs are not feasible?the values of x m

29、ay all have to be positivea consumers choices are limited by the amount of purchasing power availableOne method used to solve constrained maximization problems is the Lagrangian multiplier method52Lagrangian Multiplier MethodSuppose that we wish to find the values of x1, x2, xn that maximizey = f(x1

30、, x2, xn) subject to a constraint that permits only certain values of the xs to be usedg(x1, x2, xn) = 053Lagrangian Multiplier MethodThe Lagrangian multiplier method starts with setting up the expressionL = f(x1, x2, xn ) + g(x1, x2, xn) where is an additional variable called a Lagrangian multiplie

31、rWhen the constraint holds, L = f because g(x1, x2, xn) = 054Lagrangian Multiplier MethodFirst-Order ConditionsL/x1 = f1 + g1 = 0L/x2 = f2 + g2 = 0.L/xn = fn + gn = 0.L/ = g(x1, x2, xn) = 055Lagrangian Multiplier MethodThe first-order conditions can generally be solved for x1, x2, xn and The solutio

32、n will have two properties:the xs will obey the constraintthese xs will make the value of L (and therefore f) as large as possible56Lagrangian Multiplier MethodThe Lagrangian multiplier () has an important economic interpretationThe first-order conditions imply thatf1/-g1 = f2/-g2 = fn/-gn = the num

33、erators above measure the marginal benefit that one more unit of xi will have for the function fthe denominators reflect the added burden on the constraint of using more xi57Lagrangian Multiplier MethodAt the optimal choices for the xs, the ratio of the marginal benefit of increasing xi to the margi

34、nal cost of increasing xi should be the same for every x is the common cost-benefit ratio for all of the xs58Lagrangian Multiplier MethodIf the constraint was relaxed slightly, it would not matter which x is changedThe Lagrangian multiplier provides a measure of how the relaxation in the constraint

35、will affect the value of y provides a “shadow price” to the constraint59Lagrangian Multiplier MethodA high value of indicates that y could be increased substantially by relaxing the constrainteach x has a high cost-benefit ratioA low value of indicates that there is not much to be gained by relaxing

36、 the constraint=0 implies that the constraint is not binding60DualityAny constrained maximization problem has associated with it a dual problem in constrained minimization that focuses attention on the constraints in the original problem61DualityIndividuals maximize utility subject to a budget const

37、raintdual problem: individuals minimize the expenditure needed to achieve a given level of utilityFirms minimize the cost of inputs to produce a given level of outputdual problem: firms maximize output for a given cost of inputs purchased62Constrained MaximizationSuppose a farmer had a certain lengt

38、h of fence (P) and wished to enclose the largest possible rectangular shapeLet x be the length of one sideLet y be the length of the other sideProblem: choose x and y so as to maximize the area (A = xy) subject to the constraint that the perimeter is fixed at P = 2x + 2y63Constrained MaximizationSet

39、ting up the Lagrangian multiplierL = xy + (P - 2x - 2y)The first-order conditions for a maximum areL/x = y - 2 = 0L/y = x - 2 = 0L/ = P - 2x - 2y = 064Constrained MaximizationSince y/2 = x/2 = , x must be equal to ythe field should be squarex and y should be chosen so that the ratio of marginal bene

40、fits to marginal costs should be the sameSince x = y and y = 2, we can use the constraint to show thatx = y = P/4 = P/865Constrained MaximizationInterpretation of the Lagrangian multiplierif the farmer was interested in knowing how much more field could be fenced by adding an extra yard of fence, su

41、ggests that he could find out by dividing the present perimeter (P) by 8thus, the Lagrangian multiplier provides information about the implicit value of the constraint66Constrained MaximizationDual problem: choose x and y to minimize the amount of fence required to surround the fieldminimize P = 2x

42、+ 2y subject to A = xySetting up the Lagrangian:LD = 2x + 2y + D(A - xy)67Constrained MaximizationFirst-order conditions:LD/x = 2 - Dy = 0LD/y = 2 - Dx = 0LD/D = A - xy = 0Solving, we getx = y = A1/2The Lagrangian multiplier (D) = 2A-1/268Envelope Theorem & Constrained MaximizationSuppose that we wa

43、nt to maximizey = f(x1,xn;a) subject to the constraintg(x1,xn;a) = 0 One way to solve would be to set up the Lagrangian expression and solve the first-order conditions69Envelope Theorem & Constrained MaximizationAlternatively, it can be shown thatdy*/da = L/a(x1*,xn*;a) The change in the maximal val

44、ue of y that results when a changes can be found by partially differentiating L and evaluating the partial derivative at the optimal point70Inequality ConstraintsIn some economic problems the constraints need not hold exactlyFor example, suppose we seek to maximize y = f(x1,x2) subject tog(x1,x2) 0,

45、x1 0, andx2 071Inequality ConstraintsOne way to solve this problem is to introduce three new variables (a, b, and c) that convert the inequalities into equalitiesTo ensure that the inequalities continue to hold, we will square these new variables to ensure that their values are positive72Inequality

46、Constraintsg(x1,x2) - a2 = 0;x1 - b2 = 0; andx2 - c2 = 0Any solution that obeys these three equality constraints will also obey the inequality constraints73Inequality ConstraintsWe can set up the LagrangianL = f(x1,x2) + 1g(x1,x2) - a2 + 2x1 - b2 + 3x2 - c2This will lead to eight first-order conditi

47、ons74Inequality ConstraintsL/x1 = f1 + 1g1 + 2 = 0L/x2 = f1 + 1g2 + 3 = 0L/a = -2a1 = 0L/b = -2b2 = 0L/c = -2c3 = 0L/1 = g(x1,x2) - a2 = 0L/2 = x1 - b2 = 0L/3 = x2 - c2 = 075Inequality ConstraintsAccording to the third condition, either a or 1 = 0if a = 0, the constraint g(x1,x2) holds exactlyif 1 =

48、 0, the availability of some slackness of the constraint implies that its value to the objective function is 0Similar complemetary slackness relationships also hold for x1 and x276Inequality ConstraintsThese results are sometimes called Kuhn-Tucker conditionsthey show that solutions to optimization

49、problems involving inequality constraints will differ from similar problems involving equality constraints in rather simple wayswe cannot go wrong by working primarily with constraints involving equalities77Second Order Conditions - Functions of One VariableLet y = f(x)A necessary condition for a ma

50、ximum is thatdy/dx = f (x) = 0To ensure that the point is a maximum, y must be decreasing for movements away from it78Second Order Conditions - Functions of One VariableThe total differential measures the change in ydy = f (x) dxTo be at a maximum, dy must be decreasing for small increases in xTo se

51、e the changes in dy, we must use the second derivative of y79Second Order Conditions - Functions of One VariableNote that d 2y 0 implies that f (x)dx2 0Since dx2 must be positive, f (x) 0This means that the function f must have a concave shape at the critical point80Second Order Conditions - Functio

52、ns of Two VariablesSuppose that y = f(x1, x2)First order conditions for a maximum arey/x1 = f1 = 0y/x2 = f2 = 0To ensure that the point is a maximum, y must diminish for movements in any direction away from the critical point81Second Order Conditions - Functions of Two VariablesThe slope in the x1 d

53、irection (f1) must be diminishing at the critical pointThe slope in the x2 direction (f2) must be diminishing at the critical pointBut, conditions must also be placed on the cross-partial derivative (f12 = f21) to ensure that dy is decreasing for all movements through the critical point82Second Orde

54、r Conditions - Functions of Two VariablesThe total differential of y is given bydy = f1 dx1 + f2 dx2The differential of that function is d 2y = (f11dx1 + f12dx2)dx1 + (f21dx1 + f22dx2)dx2d 2y = f11dx12 + f12dx2dx1 + f21dx1 dx2 + f22dx22By Youngs theorem, f12 = f21 and d 2y = f11dx12 + 2f12dx1dx2 + f

55、22dx2283Second Order Conditions - Functions of Two Variablesd 2y = f11dx12 + 2f12dx1dx2 + f22dx22For this equation to be unambiguously negative for any change in the xs, f11 and f22 must be negativeIf dx2 = 0, then d 2y = f11 dx12for d 2y 0, f11 0If dx1 = 0, then d 2y = f22 dx22for d 2y 0, f22 0the

56、second partial derivatives (f11 and f22) must be sufficiently negative so that they outweigh any possible perverse effects from the cross-partial derivatives (f12 = f21)85Constrained MaximizationSuppose we want to choose x1 and x2 to maximizey = f(x1, x2)subject to the linear constraintc - b1x1 - b2

57、x2 = 0We can set up the LagrangianL = f(x1, x2) + (c - b1x1 - b2x2)86Constrained MaximizationThe first-order conditions aref1 - b1 = 0f2 - b2 = 0c - b1x1 - b2x2 = 0To ensure we have a maximum, we must use the “second” total differentiald 2y = f11dx12 + 2f12dx1dx2 + f22dx2287Constrained MaximizationO

58、nly the values of x1 and x2 that satisfy the constraint can be considered valid alternatives to the critical pointThus, we must calculate the total differential of the constraint-b1 dx1 - b2 dx2 = 0dx2 = -(b1/b2)dx1These are the allowable relative changes in x1 and x288Constrained MaximizationBecaus

59、e the first-order conditions imply that f1/f2 = b1/b2, we can substitute and getdx2 = -(f1/f2) dx1Sinced 2y = f11dx12 + 2f12dx1dx2 + f22dx22 we can substitute for dx2 and getd 2y = f11dx12 - 2f12(f1/f2)dx12 + f22(f12/f22)dx1289Constrained MaximizationCombining terms and rearrangingd 2y = f11 f22 - 2

60、f12f1f2 + f22f12 dx12/ f22Therefore, for d 2y 0, it must be true thatf11 f22 - 2f12f1f2 + f22f12 0This equation characterizes a set of functions termed quasi-concave functionsany two points within the set can be joined by a line contained completely in the set90Concave and Quasi-Concave FunctionsThe