This is the slide masterClick to edit Master title style,*,Click to edit Master text styles,Second level,Third level,OptimizationLecture 2,Marco Haan,February 21,2023,Last week,Optimizing a function of more than 1 variable.,Determining local minima and local maxima.,First and second-order conditions.,Determining global extrema with direct restrictions on variables.,This week,Constrained problems.,The Lagrange Method.,Interpretation of the Lagrange multiplier.,Second-order conditions.,Existence,uniqueness,and characterization of solutions.,2,Suppose that we want to maximize some function f(x,1,x,2,)subject to some constraint g(x,1,x,2,)=0.,Example:A consumer wants to maximize utility,U(x,1,x,2,)=x,1,x,2,subject to budget constraint,2x,1,+3x,2,=10.,In this case:f(x,1,x,2,)=x,1,x,2,and g(x,1,x,2,)=,10,2x,1,3x,2,.,3,Suppose that,from g(x,1,x,2,)=0 we can write x,2,=,(,x,1,).,Take the total differential:dx,2,=,(,x,1,),dx,1,Also:g,1,(x,1,x,2,)dx,1,+g,2,(x,1,x,2,)dx,2,=0,We want to maximize f(x,1,x,2,)subject to g(x,1,x,2,)=0.,Hence:,We can now write the objective function as,:,Weve seen this in Micro 1!,4,Theorem 13.1,If(x,1,*,x,2,*)is a,tangency solution,to the constrained maximization problem,then we have that x,1,*and,x,2,*satisfy,5,Back to the example,f(x,1,x,2,)=x,1,x,2,and g(x,1,x,2,)=,10,2x,1,3x,2,.,We need,So,With,Hence,This yields,Note:this only says that this is a local optimum.,6,Lagrange Method,Again,we want to,Consider the function,The first two equalities imply,Lets maximize this:,Hence,we get exactly the conditions we need!,7,Definition 13.2,The,Lagrange method,of finding a solution(x,1,*,x,2,*)to the problem,consists of deriving the following first-order conditions to find the critical point(s)of the Lagrange function,which are,8,Back to the example,f(x,1,x,2,)=x,1,x,2,and g(x,1,x,2,)=,10,2x,1,3x,2,.,Again,this only says that this is a local optimum.,9,The method also works for finding minima.(Definition 13.2),The,Lagrange method,of finding a solution(x,1,*,x,2,*)to the problem,consists of deriving the following first-order conditions to find the critical point(s)of the Lagrange function,which are,10,The interpretation of*,*is the,shadow price,of the constraint.,It tells you by how much your objective function will increase at the margin as the the constraint is relaxed by 1 unit.,Later,we go into more details as to why this is the case.,In the consumption example,we had income 10 and*=0.204.,This tells us that as income increases by 1 unit,utility increases by 0.204 units.,In this example,this is not very informative,as the“amount of utility”is not a very informative number.,Yet,in the case of e.g.a firm maximizing its profits,this yields information that is much more useful.,11,The,Lagrange method,of finding a solution(x,1,*,.,x,m,*)to the problem,consists of deriving the following first-order conditions to find the critical point(s)of the Lagrange function,which are,It also works with more variables and more constraints.(Definition 13.3),12,Second-Order Conditions,With regular optimization in more dimensions,we needed some conditions on the Hessian.,We now need the same conditions but on the Hessian of the Lagrange function.,This is the,Bordered Hessian.,13,Theorem 13.3,A stationary value of the Lagrange function yields a,maximum if the determinant of the bordered Hessian is positive,minimum if the determinant of the bordered Hessian is negative.,14,Again back to the earlier example,f(x,1,x,2,)=x,1,x,2,and g(x,1,x,2,)=,10,2x,1,3x,2,.,Evaluate in,Thus,we now know that this is a local maximum.,15,With more than two dimensions.(Theorem 13.4),If a Lagrange function has a stationary value,then that stationary value is a maximum if the successive principal minor of|H*|alternate in sign in the following way:,It is a maximum if all the principal minors of|H*|are strictly negative.,Note:Both theorems only give,sufficient conditions.,16,Theorem 13.6,The Lagrange method works(in finding a local extremum)if and only if it is possible to solve the first-order conditions for the Lagrange multipliers.,17,Weierstrasss Theorem:,If f is a continuous function,and X is a nonempty,closed,and bounded set,then f has both a minimum and a maximum on X.,But when can we be sure that a minimum and a maximum really exist!?,18,Weierstrasss Theorem:,If f is a continuous function,and X is a nonempty,closed,and bounded set,then f has both a minimum and a maximum on X.,But when can we be sure that a minimum and a maximum really exist!?,f is continuous if it does not contain any holes,jumps,etc.,You cannot maximize the function f(x)=1/x on the interval-1,1.,But you,can,maximize the function f(x)=1/x on the interval 1,2.,19,Weierstrasss Theorem:,If f is a continuous function,and X is a nonempty,closed,and bounded set,then f has both a minimum and a maximum on X.,But when can we be sure that a minimum and a maximum really exist!?,X is nonempty if it contains at least one element.,Otherwise the problem does not make sense.,If there