and, we have z=1.5x. Substituting in the constraint leads to, x=7.38 and z=11.07.

Transcription

1 EconS 526 Problem Set 2. Constrained Optimization Problem 1. Solve the optimal values for the following problems. For (1a) check that you derived a minimum. For (1b) and (1c), check that you derived a maximum. (a) min , (b) max , (c) max, , 4 20 (d) max.. (Hint: show all possible cases) Solution (a) min , min ,.. From and, we have z=1.5x. Substituting in the constraint leads to, x=7.38 and z= To verify that this is a minimum, check the bordered Hessian, The last n-k leading principal minors is 2-1=1. So we just need to know sign of bordered Hessian. For a minimum it has to have the sign (-1) k = (-1) 1 =-1. So it has to be negative. We have, 1

2 det The first two terms are negative given the sign of the inner terms. The last two terms are also negative. Therefore, the determinant is negative which verifies that this is a minimum. (b) max , max, From and, we have z=1+x. Substituting in the constraint leads to, x=10 and y=11. To verify that this is a maximum, check the bordered Hessian, The last n-k leading principal minors is 2-1=1. So we just need to know sign of bordered Hessian. For a maximum it has to have the sign (-1) n = (-1) 2 =1. So it has to be positive. We have, det 110 which is a maximum. (c) max, , 4 20 max,

3 Solve and from and, respectively. Substitute these solutions into to get, Factor out,..., to get, Simplifying yields, Or, From the constraints, and Substitute both in the above function yields, Simplifying yields, Or Using the quadratic formula, x= 6.12 or x= There are two sets of w and z as well. In this case, we have [40.82, 6.12, 3.47] or [13.64, 24.24, -1.06]. Note that the second set gives a negative value for the objective function. So it is most likely that [40.82, 6.12, 3.47] gives a maximum. Let us check by looking at the bordered Hessian, 3

4 The last n-k leading principal minors is 3-2=1. So we just need to know sign of bordered Hessian. For a maximum it has to have the sign (-1) n = (-1) 3 <0. So it has to be negative. We have, det which is not negative. Therefore our solution is not a maximum. (d) max.. (Hint: show all possible cases) max, Two cases: Then, z=0 from. Since z=0, there is a set of possible values for x that satisfies as long as b is not 1. Given the inequality constraint, x is any number from (-,m/c). Regardless of the value for x, the optimal value of the function would result in y= From and, we have z=(cb/d)(x+a). Substituting in the constraint leads to, x=(m- cba)/(c(1+b) and z=(cb/d)((m-cba)/(c(1+b)+a). As long as m>cba, then x and z are positive which leads to y>0. This is the likely optimal solution. 4

5 Problem 2. A farmer has a given amount of land, denoted by l max and can allocate it between two crops where l i i=1,2, is the amount of land allocated to crop i. The production function for crop i is l for all i=1,2 where 1 0. The net profit from one unit of land planted with crop 1 is r while the net profit from one unit of land planted with crop 2 is s. If the farmer wishes to maximize net profit from all his land, how much of each crop needs to be planted if the total available land is to be used? Prove that you derived optimal land allocations that maximize net profits. How does an increase in s affect land planted for crop 1 and 2? Prove your answer. Objective of the farmer: max l l,l l.. l l l, max l,l, l l l l l l 0 l l 0 l l l 0 l From l and l, we have l / l. Substituting in the constraint leads to, l l l and l /. To verify that this is a maximum, / / / check the bordered Hessian, l a 1l l The last n-k leading principal minors is 2-1=1. So we just need to know sign of bordered Hessian. For a maximum it has to have the sign (-1) n = (-1) 2 =1. So it has to be positive. We have, det 11a1l 0 since (a-1)<0 which is a maximum. An increase in s decreases land planted for crop 1 and increases land for crop 2. 5

6 Proof: l l 0 because a-1<0. l l 0 because a-1<0. Problem 3. A consumer has a utility function. She has a total income of 100. The price of x is 2 and the price of y is 3. She also has a maximum time 80 to consume both goods. It takes 1 unit of time to consume good x and 4 units of time to consume good y. She does not need to use all her income nor does she need to consume all her time. What is the level of x and y that maximizes her income given the constraints she faces? To fully solve the problem, show all possible combination of cases. Objective of consumer max , 4 80 max , Cases 1. 0, 0 x=y=0 leading to u= ; ;

7 2. 0, 0 Using the constraints, we find, x= 32, y=12 yielding u= , 0 From and, we have x=1.5y. Substituting in the budget constraint leads to, y= 16.67, x=25 yielding a u= Note however that these values for y and x are greater than the time constraint. Total time needed is so this cannot be a solution. 4. 0, 0 From and, we have x=4y. Substituting in the time constraint leads to, y= 10, x=40 yielding a u=400. Note that these values for x and y and greater than the budget. Total budget needed is110. Therefore, the optimal solution is x= 32, y=12 yielding u=384. Problem 4. A central planner uses labor L to produce two outputs, x and y, given the following production function, and where is the labor allocated in sector i=x,y. The price in sector x is 4 and the price in sector y is 5. The planner wishes to maximize aggregate revenues from both sectors in the economy. How much labor should be allocated in each sector of the economy to achieve the planner s goal if the maximum amount of labor is 20 and there is no need to produce output in each sector and not all labor needs to be employed? (Hint: use Kuhn Tucker conditions) Objective of planner max , 0, 0, max ,, ,

8 , ;. 20 Cases 1. 0, 0, 0. This yields R=0 2. 0, 0, 0. This yields 10. This yields R= , 0, 0. This yields This yields R= , 0, 0. This yields 10 and 5.33 and R= , 0, 0. No solution to this case. 6. 0, 0, 0. This yields 20. This yields R= , 0, 0. This yields 20. This yields R= , 0, 0. This yields and this yields R= The optimal solution is This yields 10 and this yields R= Deadline: Sept 21, pm. 8