5 Day 5: Maxima and minima for n variables.

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1 UNIVERSITAT POMPEU FABRA INTERNATIONAL BUSINESS ECONOMICS MATHEMATICS III. Pelegrí Viader Updated May 14, Day 5: Maxima and minima for n variables. The same kind of first-order and second-order conditions we used to find unconstrained max/min for two variables are good for three or more variables. FIRST-ORDER NECESSARY CONDITIONS. If f(x, y, z,...) has a max/min at a point (x, y, z,...) interior to f s domain, then (x, y, z,...) is a stationary point for f, that is f x (x, y, z,...) = 0 f y (x, y, z,...) = 0 f z (x, y, z,...) = CONCAVITY/CONVEXITY: GLOBAL SUFFICIENT CONDITIONS. If f is defined on an open convex set S and (x, y,...) is a stationary point of S, then 1. If f is concave in S then (x, y,...) is a max (global on S); 2. If f is convex in S then (x, y,...) is a min (global on S); This is equivalent to saying: SECOND-ORDER GLOBAL SUFFICIENT CONDITIONS. If f is defined on an open convex set S and (x, y,...) is a stationary point of S, and H(x, y,...) is the Hessian of f: 1. If H(x, y,...) is positive (definite or semidefinite) for all (x, y,...) in S then (x, y,...) is a global min of f on S. 2. If H(x, y,...) is negative (definite or semidefinite) for all (x, y,...) in S then (x, y,...) is a global max of f on S. If we do not have information about all of the domain S but just for the Hessian at (x, y,...), then the max/min are local: SECOND-ORDER LOCAL SUFFICIENT CONDITIONS. If f is defined on an open set S and (x, y,...) is a stationary point of S, and H(x, y,...) is the Hessian of f at (x, y,...): 1. If H(x, y,...) is positive definite then (x, y,...) is a local min of f. 2. If H(x, y,...) is negative definite then (x, y,...) is a local max of f.. If H(x, y,...) is indefinite then (x, y,...) is a saddle point of f. 1

2 Remember that we can check the sign (PD, ND, etc.) of the Hessian using the leading principal minors method (LPM). In the case D n (x, y,...) 0 but D 1 (x, y,...), D 2 (x, y,...),...,d n (x, y,...) not following the pattern for positivity or negativity, then (x, y,...) is a saddle point. Examples. Find max/min of f(x, y, z) = x + y + z 2xy 2xz 2yz. The domain of this function is R. The stationary points come from solving x 2 2y 2z = 0 y 2 2x 2z = 0 z 2 2x 2y = 0 This system is not easy to solve. A possible method would be to solve E1 for y and replace in E2 and E. This leads to a 4th degree equation a bit tiresome to solve. The similarity of the equations makes us think of subtracting: E1 E2 produces which can be written x 2 y 2 + 2x 2y = 0 (x 2 y 2 ) + 2(x y) = 0 (x y) [(x + y) + 2] = 0. Similarly E2 E and E1 E produce (x z) [(x + z) + 2] = 0 and (y z) [(y + z) + 2] = 0. Factoring each one of these equations we have x y = 0 E1 (x + y) + 2 = 0 ; E2 x z = 0 (x + z) + 2 = 0 ; E All in all we can combine these equations in 8 possible ways (2 2 2). x y = 0 1. Case x z = 0. Solution (z, z, z). y z = 0 x y = 0 2. Case x z = 0. Solution ( 1/, 1/, 1/). (y + z) + 2 = 0 x y = 0. Case y z = 0. Solution ( 1/, 1/, 1/). (x + z) + 2 = 0 x z = 0 4. Case y z = 0. Solution ( 1/, 1/, 1/). (x + y) + 2 = 0 x y = 0 5. Case (x + z) + 2 = 0. Solution ( 2/ z, 2/ z, z). (y + z) + 2 = 0 2 y z = 0 (y + z) + 2 = 0.

3 6. Case 7. Case 8. Case x z = 0 (x + y) + 2 = 0 (y + z) + 2 = 0 y z = 0 (x + y) + 2 = 0 (x + z) + 2 = 0 (x + y) + 2 = 0 (x + z) + 2 = 0 (y + z) + 2 = 0. Solution ( 2/ y, y, 2/ y).. Solution (x, 2/ x, 2/ x).. Solution ( 1/, 1/, 1/) Let us check on the original system the solutions we have obtained (remember this is not a linear system and spurious solutions may appear. Any possible solution has to be checked in the original system). Case 1: (z, z, z) leads to z 2 4z = 0 which gives two actual solutions: (0, 0, 0) and (4/, 4/, 4/). Cases 2,, 4, and 8: ( 1/, 1/, 1/) is not good. Replacing in E1, for instance, we have ( 1/) 2 2( 1/) 2( 1/) = 5/ 0. Case 5: ( 2/ z, 2/ z, z) is not good either. Replacing in E1 we end up with z 2 +4z+8/ = 0 that has no real solutions. Similarly for cases 6 and 7. This means only two stationary points (0, 0, 0) and (4/, 4/, 4/). Classifying the stationary points. The Hessian is 6x y 2. The LPM are 2 2 6z D 1 (x, y, z) = 6x, D 2 (x, y, z) = 6xy 4, D (x, y, z) = 216xyz 24y 24x 24z 16 At (0, 0, 0) we have D 1 (0, 0, 0) = 0, D 2 (0, 0, 0) = 4 < 0, D (0, 0, 0) = 16 < 0 Thus, (0, 0, 0) is a saddle point as D (0, 0, 0) < 0, and D 1 (0, 0, 0) and D 2 (0, 0, 0) do not follow the pattern < 0, > 0. At (4/, 4/, 4/) we have D 1 (4/, 4/, 4/) = 8 > 0, D 2 (4/, 4/, 4/) = 60 > 0, D (4/, 4/, 4/) = 400 > 0 At (4/, 4/, 4/) we have a min (local). The value of this min is: f (4/, 4/, /) = 2/9 =.555 This is NOT a global min as (for instance) f(x, 0, 0) = x as x!

4 5.1 Extreme-Value Theorem. It is valid for n variables: If f is a continuous function on a compact set (closed and bounded), S, f reaches a global maximum and a global minimum at points of S. If f is differentiable, the way to proceed is the same as for one and two variables: 1. Find the stationary points of f that are interior to S. 2. Find the max/min of f at the boundary of S.. Make a list of values of f at each of the points found in 1 and Choose the max and the min from the list in. This may be a little more difficult to do than in the one or two variable case. Think of the boundary of S: it may be a piece of surface in R (or a hypersurface in R n ) and we have to study f restricted to this piece of surface. Kind of complicated. We will see that Khun-Tucker s method will help us to solve this problem. 6 Day 6: Lagrange 6.1 Review: Lagrange for two variables Remember that the general problem is max/min f(x, y) s.t. g(x, y) = c. ( ) Functions f, g are twice differentiable functions and 0 g. 0 (Sydsaeter & Hammond, Example 18.7) Max/min f(x, y) = 2x + y s.t. x + y = 5. Remark. Notice that the constraint is only defined for x 0 and y 0 and is not differentiable at points (25, 0) and (0, 25). These two points will be corners of the constraint which will be needed to be taken into consideration alongside any candidate coming from Lagrange s system. We write down the Lagrangian auxiliary function: L(x, y) = 2x + y λ( x + y 5). We solve the system L x = 0 L y = 0 g(x, y) = c that is 2 λ 1 2 x = 0 (E1) λ 1 2 = 0 (E2) y x + y = 5 (E) 4

5 We solve E1 and E2 for λ and we equate both results (we eliminate λ). We get 4 x = 6 y. We use this new equation with E: 4 x = 6 y x = y x + y = 5 2 and y 2 These last two equations imply x = 9, y = 4 (recall that x, y 0). + y = 5 y = 2 and x = Notice that we treat the two equations above as if x, y were the unknowns. In this way we avoid a lot of unnecessary algebra and cancelations. Consequently, there is only one candidate: (9, 4). Are there any other possible points? The remark above has to be considered. Let us have a look at the constraint: A (0, 25) x + y = 5 B (25, 0) We see clearly that the constraint is a curve defined only for 0 x 25 and 0 y 25. It is a closed and bounded set of points in R 2 : a compact. So, the EVT (Extreme-Value Theorem) guarantees that our continuous function f has a global max and a global min on the constraint. We notice that the constraint contains two extreme points, (0, 25) and (25, 0) at which g(x, y) has one infinite derivative! These two points, in consequence, have to be added to any list of candidates for max/min. We have now to compare f(0, 25) = 75; f(25, 0) = 50; f(9, 4) = 0. The max is 75 at (0, 25) and the min is 0 at (9, 4). Notice that we have been able to find max and min because the EVT ensured we had a solution. In this way, we have had no need to check the character of the candidate provided by Lagrange s method: (9, 4). We could also use the graphical method: 5

6 A (0, 25) max x + y = 5 2x + y = 0 f (9, 4) min (25, 0) B Or studied the convexity of the Lagrangian at (9, 4) with λ = 12 (The candidate. We cannot use this argument for corner points which are not candidates): L λ=12 (x, y) = 2x + y 12 x 12 y 60 a convex function as 2x + y 60 is linear (convex/concave) and 12 is a convex function of one variable. Lastly, the use of the bordered Hessian would have given information about (9, 4) but only LOCAL: 0 g x g 0 1/(2 x) 1/(2 y) y BH(x, y, λ) = g x L xx L xy = 1/(2 λ x) g y L xy L 4 x /2 0 yy 1/(2 λ y) 0 4 y /2 At (9, 4) and λ = /6 1/4 BH(9, 4, 12) = 1/6 1/27 0 < 0 min (local). 1/4 0 /8 6.2 Lagrange for n variables and m < n constraints. If we are interested in optimizing a three-variable function constrained by one or two constraints, we can use again Lagrange s method. Question: Find max/min of f(x, y, z) = x + 4y + z s.t. x 2 + y 2 + z 2 = 42 and x + 2y + z = 0. The two constraints are, respectively, the sphere of center (0, 0, 0) and radius 42 and a plane that goes through the sphere s center. Thus, the two constraints define a maximum circle on the sphere (think of the line of the equator on the Earth globe): 6

7 Sphere and plane. As such it is obviously a compact set. Our problem will have both a max and a min global solution. The Lagrangian is L(x, y, z) = x + 4y + z λ(x 2 + y 2 + z 2 42) µ(x + 2y + z). The candidates to max/min will come from solving the system of 5 equations and 5 unknowns 1 2λx µ = 0 4 2λy 2µ = 0 1 2λz µ = 0 x 2 + y 2 + z 2 = 42 x + 2y + z = 0 How do we solve this formidable system? It is not linear. We have to use substitution or some sort of reduction. The more sensible way to proceed is to solve E1 and E2 for λ and µ (eliminate λ and µ) and replace in E. In this way we will have a new equation in x, y, z to solve with the two constraints, E4 and E5. A little thinking, though, make things somewhat easier. If we want to solve E1, E2 and E for λ, µ we can write these three equations as if the unknowns were precisely λ, µ: 2x λ + µ = 1 2y λ + 2 µ = 4 2z λ + µ = 1 This is a 2-unknowns, λ and µ, -equations system. It will have a solution if and only if the rank of the system matrix equals that of the enlarged matrix. This implies that the determinant of the enlarged matrix must be 0: 2x 1 1 2y 2 4 = 20x + 4z + 4y = 0 or 5x + z + y = 0. 2z 1 This gives us the equation in x, y, z we sought in order to solve with E4 and E5. We now must solve 5x + y + z = 0 x + 2y + z = 0 x 2 + y 2 + z 2 = 42 We solve the first two equations for x, y in terms of z and we have x = z/11; y = 16z/11 Replacing in the third equation, z z z2 = 42 z = ± 11 7

8 We have two candidates: ( 1, 16 ), 11 λ = 14, µ = 6 7 and ( 1, 16, 11 ) We have found the corresponding values of λ and µ by solving λ 2 + µ = 1 λ 2 + µ = 1 2 λ + 2 µ = 4 and 2 λ + 2 µ = 4 22 λ + µ = 1 22 λ + µ = 1 The canceled equation is a linear combination of the other two. λ = 14, µ = 6 7. Remark. Check the note at the end of these notes for a general procedure to solve the Lagrangian systems. Using the EVT, we only need to find the value of f at each one of these points ( f 1, 16, 11 ) ( 1 = 18; f, 16 ), 11 = 18. ( 1 Global max is 18 at, 16 ) (, 11 and global min is 18 at 1, 16, 11 ). Alternatively, let us appeal to the concavity/convexity of the Lagrangian in its domain, R. For the candidate ( 1, 16, 11 ), the Lagrangian is L λ= /14,µ=6/7 (x, y, z) = = x + 4y + z + 14 (x2 + y 2 + z 2 42) 6 (x + 2y + z) = 7 = 1 14 (x2 + y 2 + z 2 + 2x + 2y 22z 126) a convex function in R (the sum of convex x 2 + y 2 + z 2 and linear 2x + 2y 22z 126). Our candidate then is a min (global). For the candidate ( 1, 16 ), 11, the Lagrangian is L λ=/14,µ=6/7 (x, y, z) = = x + 4y + z 14 (x2 + y 2 + z 2 42) 6 (x + 2y + z) = 7 = 1 14 ( x2 y 2 z 2 + 2x + 2y 22z 126) a concave function in R (the sum of concave x 2 y 2 z 2 and linear 2x + 2y 22z 126). Our candidate then is a max (global). 6. Interpretation of the Lagrange multipliers. As in the case of two variables, Lagrange multipliers can be interpreted as shadow prices of each constraint. Let us consider the problem of maximizing f(x, y,...) s.t. different constraints 8

9 g 1 (x, y,...) = c 1, g 2 (x, y,...) = c 2,... and let λ 1, λ 2,... be the corresponding multipliers. If the solution to our problem is f (maximum/minimum), we can imagine f as a function of c 1, c 2,.... Then λ i = f c i. So, if c i undergoes a change i (small compared to c i ) in its value, the max/min will change as f (c 1 + 1, c 2 + 2,...) f (c 1, c 2,...) + λ λ Question: In the problem above, max f(x, y, z) = x + 4y + z s.t. x 2 + y 2 + z 2 = 42 and x + 2y + z = 0 how will the max change if the constraints change to x 2 + y 2 + z 2 = 44 and x + 2y + z = 1? We had found that the max was f = 18 at and 2 = 1. ( 1, 16 ), 11 with λ = /14 and µ = 6/7. Now, 1 = 2 The new max will be ( 1) = Solving again the problem for the new RHS of the constraints we obtain a max (true value) of In the case of the min, its value was f = 18 and λ = /14 and µ = 6/7. Again, 1 = 2 and 2 = 1. The new min will be ( 1) = Solving again the problem for the new RHS of the constraints we obtain a min (true value) of Problems. x + 2y + z = 1 1. (a) Solve max/min f(x, y, z) = x 2 + y 2 + z 2 s.t. 2x y z = 4. x + 2y + z = 1.2 (b) How do the optimal values change if the constraints change to 2x y z =.9? 2. Solve max / min f(x, y, z) = x + y + z s.t. x 2 + y 2 + z 2 = 1 x y z = 1. A student used Lagrange with the problem max/min f(x, y, z) = x + y + z s.t. g(x, y, z) = c and she found the two points ( ) ( ) c c c c c c,, and,,. Which is the max and which the min? For each point, find the corresponding value of λ if c = 1. 9

10 7 Day 7: Kuhn-Tucker s method (KT). Kuhn-Tucker s method is a systematic method to solve general optimization problems of the form (standard): g 1 (x 1, x 2,...,x n ) c 1 g max/min f(x 1, x 2,...,x n ) s.t. 2 (x 1, x 2,...,x n ) c g m (x 1, x 2,...,x n ) c m Any point in R n that satisfies all the constraints is called a feasible (or admissible) point and the set of all feasible points, S, is the feasible set. If all the functions g i are differentiable, the feasible set is always a closed set (because of the = in which includes all the frontier curves). It may be bounded or not. Notice that all constraints are written using. If you are given a constraint like g(x, y,...) c, change it into g(x, y,...) c. As usual, we say that a constraint is active at a point (x, y,...) if g i (x, y,...) = c i. If g i (x, y,...) < c i, we say that the constraint is inactive. We assume all the functions involved are differentiable. KT establishes a protocol that helps us to obtain feasible candidate points to max/min. These candidates may be interior to S (all constraints inactive) or belong to the boundary of S (at least one constraint active). (See the figure below: P is an interior point to S and Q is a boundary point that makes two constraints active and the third inactive.) Q g 1 (x,y,...) c 1 S g 2 (x,y,...) c 2 P g (x,y,...) c KT uses Lagrange s method to study feasible points on the boundary of S. If we have m constraints, we use m Lagrange multipliers. We define L(x, y,...) = f(x, y,...) λ 1 [g 1 (x, y,...) c 1 ] λ m [g m (x, y,...) c m ] Let (x, y,...) be a candidate (solution to the Lagrangian system). Two possibilities: 1. Point (x, y,...) is interior to S. Then all constraints will be inactive and (x, y,...) will be a stationary point of L with λ 1 = λ 2 = = λ m = 0 (actually, it will be a stationary point of f as all the λ s are 0). 10

11 2. Point (x, y,...) belongs to the frontier of S, and as such, satisfies r constraints g i1 (x, y,...) = 0;...; g ir (x, y,...) = 0 leaving the other m r inactive (<). Then we have to solve the actual Lagrangian problem for λ i1,...,λ ir as multipliers and the other m r λ s = 0. In the first case, KT leads to a stationary point of f, interior to its domain S. All λ s are 0 and the candidate may be a max, a min or neither (a saddle point). In the second case, KT leads to a candidate for which the λ s outside λ i1,...,λ ir have to be 0. At the candidate, λ i1,...,λ ir may be positive, negative or 0. The candidate may be a candidate to max or a candidate to min or not a valid candidate to either thing. How do we tell? The shadow price interpretation of the λ s will help us to see how that works. We know that if f = f(x, y,...) is the optimal value (max or min) of our objective function, then λ i = f c i. If constraint j is active, increasing the value of c j, enlarges set S. That means that if our problem is a max one, f can only increase with c j : consequently λ j 0. If our problem is a min one, the enlargement of S can only produce a decrease in f. Consequently λ j 0. In any case, if g j is inactive at a solution (x, y,...), clearly λ j = 0 and if λ j 0, g j must be active. So, g j (x, y,...) c j and λ j cannot be 0 at the same time. KT imposes this condition and calls it the complementary slackness condition for each constraint. For a max candidate: For a min candidate: λ j 0 and λ j [g j (x, y,...) c j ] = 0. λ j 0 and λ j [g j (x, y,...) c j ] = 0. If some λ j are > 0 and some are < 0, the candidate is not valid. 7.1 Kuhn-Tucker: General procedure. 1. We define the Lagrangian auxiliary function: L(x 1,..., x n ) = f(x 1,...,x n ) λ 1 [g(x 1,...,x n ) c 1 ] λ m [g(x 1,...,x n ) c m ] 2. The candidates to max [min] will come from solving the system given by the KT conditions: (Block 1) L xi (x 1,...,x n ) = 0 (the n partial derivatives of L = 0); (Block 2) λ j [ ]0 and λ j [g j (x 1,...,x n ) c j ] = 0 (Complementary slackness); (Block ) g j (x 1,...,x n ) c j (Constraints). Block 1 are the usual first-order conditions that make our candidates stationary points for L. Block are the constraints that have to be satisfied: our candidates must be feasible points. And lastly block 2 are the special KT conditions that state that the multipliers have to be non-negative for a max candidate [respectively, non-positive for a min candidate] and must also satisfy the complementary slackness conditions. Notice that the complementary slackness max condition for constraint j says that either λ j > 0 or g j (x 1,...,x n ) < c j but not at the same time; that means that both cannot be inactive at the same time. 11

12 . We have to add to the list of candidates any extreme points not found in the previous process. Basically points where at least one of the g i (x, y,...) is not differentiable or points where the gradients of g i that are not independent vectors. We will not enter too deep into this area. 4. We establish the character of each candidate using either the EVT (if the domain is a compact) or considering the concavity/convexity of L in all its domain for the λ j specific for each candidate. If the problem is a two-variable problem, we can also use the graphical method. Remark. A good way of not forgetting any case in the KT protocol is to use the following simple schemes. Let A stand for ACTIVE and I stand for INACTIVE. Number of constraints: 2. Cases to study 2 2 = 4: AA; AI; IA; II. Number of constraints:. Cases to study = 8: the four cases above adding first an A and then an I: AAA; AIA; IAA; IIA and AAI; AII; IAI; III Each inactive case implies the corresponding λ = 0. Each active case, implies the corresponding constraint to be active (the = stands). Thus a case like AAI means that g 1 = c 1 ; g 2 = c 2 ; λ = 0. The best way to understand how KT works is with an example. We start with a simple two-variable one. Example 1. max/minf = x 2 + y 2 x s.t. x 2 + y 2 1. Notice that the constraint is the whole circle of center (0, 0) and radius 1. We can apply the EVT to this problem and solve it using the techniques we already know (stationary interior points, study of the function restricted to the boundary; list of values). Nevertheless, we are going to use KT. The Lagrangian is L(x, y) = x 2 + y 2 x λ (x 2 + y 2 1). The candidates will come from solving the system given by the KT conditions: 2x 1 2λx = 0 (Block 1) the two partial derivatives of L = 0; 2y 2λy = 0 [ (Block 2) λ 0 [λ 0] and λ x 2 + y 2 1 ] = 0 complementary slackness; (Block ) x 2 + y 2 1 constraints. Block 1 are the usual first-order conditions that make any candidates a stationary point for L. Block is the constraint that has to be satisfied: our candidates must be feasible points. And lastly, block 2 is the special Kuhn-Tucker condition that states that λ has to be non-negative for a max candidate [respectively, non-positive for a min candidate] and satisfy the complementary slackness condition which says that λ and x 2 + y 2 1 cannot be both different from zero. We proceed systematically in order to find candidates: 12

13 1. The constraint is active (A): x 2 + y 2 = 1. We solve B1 and B: 2x 1 2λx = 0 (E1) 2y 2λy = 0 (E2) x 2 + y 2 = 1 (E) From E2, y = 0 or λ = 1: Case y = 0. Replacing in E, x = ±1. For x = 1, from E1, λ = 1/2 0. Candidate to max: (1, 0), with λ = 1/2. For x = 1, from E1, λ = /2 0. Candidate to max: ( 1, 0), with λ = /2. Case λ = 1. Replacing in E1, 1 = 0. Impossible. We have no candidates from here. 2. The constraint is inactive (I): x 2 + y 2 < 1, and so λ = 0. We solve block one (B1) to find stationary points of L (or f) 2x 1 = 0 x = 1/2; y = 0. 2y = 0 The only stationary point is (1/2, 0) and λ = 0. It satisfies the constraint: (1/2) < 1. We have a candidate to max/min. Two ways of finishing our study. As S = (x, y) : x 2 + y 2 1} is a compact set, we apply the EVT to our candidates. The list of values is f(1/2, 0) = 1/4; f(1, 0) = 0; f( 1, 0) = 2. The global max is 2 at the (boundary) point ( 1, 0). The global min is 1/4 at the (interior) point (1/2, 0). Candidate (1/2, 0) is interior to the feasible region. For it, λ = 0. So, L λ=0 = f. And f is convex on its domain (it is the sum of convex x 2 +y 2 and linear x). So any stationary interior point is a global min on the whole domain: min = 1/4 at (1/2, 0). The global max has to be on the boundary: it must be either f(1, 0) = 0 or f( 1, 0) = 2. It is clearly 2 at ( 1, 0). 8 Day 8: Problems on KT max/min f(x, y, z) = 1 2 x y s.t. x + e x y + z 2 0 x 0 Notice that f is a three-variable function and that constraint 2 must be written x 0. Now, the Lagrangian is L(x, y, z) = 1 2 x y λ(x + e x y + z 2 ) µ( x) = 1 2 x y λ(x + e x y + z 2 ) + µx 1

14 and KT necessary conditions will be 1 2 λ(1 e x ) + µ = λ = 0 2λz = 0 λ [ ]0 and λ (x + e x y + z 2 ) = 0 Constraint 1 µ [ ]0 and µ ( x) = 0 Constraint 2 There are 4 different possibilities: 1. Constraint 1 is active and 2 is active: AA. The system is 1 2 λ(1 e x ) + µ = λ = 0 2λz = 0 x + e x y + z 2 = 0 x = 0 The only solution is x = 0; y = 1; z = 0; λ = 1; µ = 1/2. It is not a valid candidate as λ and µ have different signs. 2. Constraint 1 is active and 2 is inactive: AI. That means µ = λ(1 e x ) = λ = 0 2λz = 0 x + e x y + z 2 = 0 x > 0 We have immediately λ = 1; z = 0. From E1 we have e x = 1/2 and taking logarithms x = ln(1/2) which is x = ln 2. This satisfies constraint 2 and replacing in constraint 1 we get y = 1/2 + ln 2. We have a candidate to max: (ln 2, 12 + ln 2, 0 ) with λ = 1; µ = 0.. Constraint 1 is inactive and 2 is active: IA. That means λ = 0. The second equation in the system above becomes inconsistent: 1 = 0. No candidates. 4. Constraints 1 and 2 are inactive: II. That means λ = µ = 0. The second equation in the system above becomes inconsistent: 1 = 0. No candidates. We are left with a single candidate to max: (ln 2, 12 + ln 2, 0 ) with λ = 1; µ = 0. For λ = 1; µ = 0 the Lagrangian is L λ=1,µ=0 (x, y, z) = 1 2 x y (x + e x y + z 2 ) = 1 2 x e x z 2. This is concave as (1/2)x is linear and e linear and z 2 are concave (e linear and z 2 are convex). Our candidate is a global max. The value of the max is ln

15 8.1 Problems. Remember that you can use in order to find max/min. The syntax for our last problem would be: extrema x/2-y on x+exp(-x)-y+z^2<=0,x>=0 1. max/min f(x, y) = 1 x 2 y 2 s.t. 2. max/min f(x, y) = x 2 + y s.t. x 0 y 0 x x + y 2. max/min f(x, y) = x/2 y s.t. 4. max/min f(x, y) = x 2 + 2y s.t. 5. max/min f(x, y) = y x 2 s.t. x + e x y x 0 x 2 + y 2 5 y 0 y x 2 y 2 x y 0 6. max/min f(x, y, z) = x 2 + y 2 + e z s.t. x + y + z 1 e x+y+z 7. max/min f(x, y, z) = e x y + z 2 s.t. x + y + z 1 x 2 + y 2 8. Difficult problem An aircraft manufacturing firm can operate plants in either of two countries. In country A, its cost function as a function of output x 0 is C A (x) = ln(1 + x/100). In country B, its cost as a function of output y 0 is C B (y) = 2 ln(1+y/100). The firm allocates production between the two plants in order to minimize the total cost of producing at least q units of output worldwide where q > 0. (a) Show that the firm s cost-minimizing choices of x and y must solve a particular constrained optimization problem with non-negativity constraints. (b) Use the Lagrange multiplier method to show that there are one, two, or three solution candidates satisfying Kuhn-Tucker conditions, depending on the value of q. (c) Show that the firm uses only the plant in country A for levels of output below some critical level q, and only the plant in country B when q > q. (d) Find the firm s minimum cost as a function of q, and show that it is not differentiable at q. 8.2 Note on the solution of Lagrangian systems Let us consider the problem maxf(x, y, z) s.t. g(x, y, z) = c 1, h(x, y, z) = c 2. 15

16 We assume that f, g, and h are continuously differentiable on an open convex set S. We also assume that the gradients g, h are linearly independent in S. We will see later the necessity of this condition. The Lagrange function is L(x, y, z) = f(x, y, z) λ(g(x, y, z) c 1 ) µ(h(x, y, z) c 2 ). The candidates to max/min will come from solving the system of 5 equations and 5 unknowns f x (x, y, z) λ g x (x, y, z) µ h x (x, y, z) = 0 f y (x, y, z) λ g y (x, y, z) µ h y (x, y, z) = 0 f z (x, y, z) λ g z (x, y, z) µ h z (x, y, z) = 0 g(x, y, z) = c 1 h(x, y, z) = c 2. In order to eliminate λ and µ from the first equations, we consider them as a linear system in the unknowns λ and µ: λ g x + µ h x = f x λ g y + µ h y = f y λ g z + µ h z = f z The matrices of this system are A = g x g y h x h y and A = g x h x f x g y h y f y. g z h z g z h z f z The rank of these two matrices must coincide in order to have solutions for λ and µ. As for A, rank(a) = 2 (remember the assumption about the independence of the gradients of g and h). Thus rank(a ) = 2 which implies g x h x f x g y h y f y = 0. g z h z f z Let us call the above determinant, det(g, h, f ). This is the equation that must be solved together with the two constraints in order to find the Lagrange candidates: det(g, h, f ) = 0 g = 0 h = 0 Let (x, y, z ) be one of the solutions of this system. The corresponding values of λ and µ come from solving λ g x(x, y, z ) µ h x(x, y, z ) = f x(x, y, z ) λ g y (x, y, z ) µ h y (x, y, z ) = f y (x, y, z ) λ g z (x, y, z ) µ h z (x, y, z ) = f z (x, y, z ) One of the equations above is redundant. Assume it is the third. The values of λ and µ may be obtained by Cramer s rule: f x(x, y, z ) h x(x, y, z ) g x(x, y, z ) f x(x, y, z ) f y λ = (x, y, z ) h y (x, y, z ) g y g x (x, y, z ) h x (x, y, z ) ; µ = (x, y, z ) f y (x, y, z ) g x (x, y, z ) h x (x, y, z ). g y (x, y, z ) h y (x, y, z ) g y (x, y, z ) h y (x, y, z ) 16