ECON 331 - Mathematical Economics - ANSWERS FINAL EXAM 1. a- The consumer's utility maximization problem is written as: max x,y [xy + y2 + 2x + 2y] s.t. 6x + 10y = m, x 0, y 0 - The associated Lagrangian function is dened as: Lx, y, λ = xy + y 2 + 2x + 2y λ [6x + 10y m] where λ is the Lagrange multiplier. - The Kuhn-Tucker conditions are written as: Lx, y, λ Lx, y, λ i x = 0, x 0, = y + 2 6λ 0 ii Lx, y, λ Lx, y, λ y = 0, y 0, = x + 2y + 2 10λ 0 iii Lx, y, λ = [6x + 10y m] = 0 λ - From iii: x = m/6 5/3y *. - Suppose y = 0. From * x = m/6 > 0 L = 0 λ = 1/3. Plug λ and x in ii: you nd L = m/6 4/3 0. However, remember that 0 < m < 40 0 < m/6 4/3 which is a contradiction. - Suppose y > 0 L = 0 x + 2y + 2 10λ = 0 **. Suppose also x = 0. Then from * y = m/10. And from ** λ = m/50 + 1/5. Plug y and λ in i. You nd L = m/50 + 4/5 0. However, remember that 8 < m < 40 0 < m/50 + 4/5 which is a contradiction. So suppose instead x > 0 L = 0 y = 6λ 2. And from * x = m/6 10λ+10/3. Plug x and y in ** and solve for λ. You nd: x = 40 m/24 > 0 b/c m < 40 y = m 8/8 > 0 b/c m > 8 λ = m + 8/48 NB: alternatively, you can directly plug the constraint into the objective function say by replacing y by a function of x. You end up with an optimization problem with one choice variable, x and one non-negativity constraint. b We have seen a result in class that states: Ux, y m = λ 1
We check by calculation that it works here: Ux, y = 40 m m 8 2 m 8 + + 40 m 24 8 8 12 Ux, y = 1 m 48 m + 1 = λ 6 Ux, y = 7 m m=20 12 + m 8 4 = 1 96 m2 + 1 6 m + 2 3 c - case where 0 m 8: previous solution does not work because it violates the positivity of x. So we need to study the behavior of U on the boundary.. x = 0 and y > 0: U0, y = y 2 + 2y with y > 0 and 10y = m. So y = m/10 and U 1 = m 2 /100 + m/5.. x > 0 and y = 0: Ux, 0 = 2x with x > 0 and 6x = m. So x = m/6 and U 2 = m/3 We can check that U 1 < U 2 when m 8, so the solution is: x = m/6 and y = 0. - case where m 40: similar study needs to be performed as the positivity of y found in b is violated:. same 2 cases and associated solutions! This time, U 1 > U 2 when m 40, hence the solution is x = 0 and y = m/10. 2. a I n + sm = I n + M 1 I n + smi n + M = I n I n + M + sm + sm 2 = I n 1 + sm + s3m = 0 b/c by assumption M 2 = 3M 1 + 4sM = 0 1 + 4s = 0 or M = 0 s = 1 4 or M = 0 To conclude, for any matrix M s.t. M 2 = 3M, I n M/4 is the inverse matrix of I n + M. b i AXB = XB + C AX XB = C A I n X = CB 1 b/c B is invertible by ass. X = A I 1 CB 1 b/c A I n is invertible by ass. ii A I n = 1 1 1 1 ; A I n 1 = 1/2 1/2 1/2 1/2 apply the inverse formula for 2,2-matrices 2
1 0 0 and B 1 = 0 2 0 0 0 4 apply the inverse formula for diagonal matrices X = = = 1 1 1 1 0 2 3/2 1 1 1/2 0 4 6 1 2 2 1 1 2 1 3 1 1 0 0 0 2 0 0 0 4 1 0 0 0 2 0 0 0 4 3. - Taylor's approximation: we dene a function and Taylor provides a polynomial approximation of this function around a point of interest. Here I dene fx = x 1/3 and I want to approximate it around x 0 = 8. The second-order approximation is: fx f8 + f 8x 8 + 1 2 f 8x 8 2 f8.1 2 + 1 3 8 2/3 8.1 8 1 2 2 8 5/3 3 3 = 2 + 1 12 0.1 1 144 2 0.12 = 2 + 0.008333 0.000035 = 2.0083 8.1 8 2 - Newton's algorithm: we dene an equation that has solution at x = 8.1 1/3 and Newton's provides an approximate solution. Here I dene gx = x 3 8.1 and the equation gx = 0 has solution at x = 8.1 1/3. The algorithm is given by: x i+1 = x i gx i g x i whenever g 0 x 1 = 2 8 8.1 3 2 2 = 241 120 2.00833 x 2 = 241 120 241/1203 8.1 3 241/120 2 2.0083 3
4. a - We can build up the following table of revenues for successive years: Y ear 0 A Revenue 1 A pa = 1 pa 2 1 pa p[1 pa] = 1 pa[1 p] = 1 p 2 A. n. 1 p n A So the revenue in year n can be expressed as R n = 1 p n A. formula can be proved by induction - not requested here. - The company operates as long as the yearly revenue exceeds the yearly cost, ie as long as 1 p n A > K. So we need to nd the largest integer n representing the largest number of years that satises this equation. 1 p n A > K ln [1 p n A] > lnk K > 0!! n ln1 p + lna > lnk n ln1 p > lnk lna K n ln1 p > ln A n < ln K A 0 < 1 p < 1, so ln1 p < 0, so inequality changes!! ln1 p So n is the largest integer satisfying the above equation. b The total prot from year 0 to n is simply the dierence between the total revenue and the total cost from year 0 to n. The total revenue is the sum of the yearly prots from year 0 to n, ie: R = n i=0 n R i = [1 p i A] = i=0 [ n ] 1 p i A = 1 1 +1 pn A use the hint with x = 1 p 1 1 p = 1 1 +1 pn A p The cost is the same each year, so the total cost is simply: C = n + 1 K Finally, the total prot is: i=0 π = 1 1 +1 pn A n + 1 K p c i Plug the given values in the equation found in a to get: ln K A ln1 p = ln 5000000 7000000 = 16.65 ln0.98 4
n is the largest integer that is smaller than 16.65, so n = 16. ii The associated prot is: π = 1 0.9817 0.02 with n = 14, we get: π = 6.5 10 6. 7000000 17 5000000 = 16.74 10 6 5. a The coecient matrix of the system is: 1 3 4 A = 2 2 1 3 3 9 and its determinant is calculated through an expansion in co-factors according to the rst column other expansions according to any line or column is acceptable - same nal answer: 2 1 deta = 3 9 2 3 4 3 9 + 3 3 4 2 1 = 18 + 3 2 27 + 12 + 33 8 = 0 bi From a, the coecients matrix associated with the system has a zero-determinant. Hence, the system does not have a unique solution: it may have no solution or an innite number of solutions, but we cannot tell yet. It cannot be solved by matrix-inversion and the Cramer's rule cannot be used. ii We solve the system using the general Gaussian elimination method: x + 3y + 4z = b 1 2x + 2y + z = b 2 3x 3y 9z = b 3 x + 3y + 4z = b 1 4y 7z = b 2 2b 1 L 2 2L 1 12y 21z = b 3 3b 1 L 3 3L 1 x + 3y + 4z = b 1 y + 7/4z = b 1 /2 b 2 /4 L 2 /4 0 = b 3 3b 1 3b 2 + 6b 1 L 3 3L 2 The system has solutions if and only if the coecients b 1, b 2, and b 3 satisfy the last equation: 3b 1 3b 2 + b 3 = 0. 5
6. a The domain is dened by couples x, y that satisfy 2 equations: i x 2 +y 2 1; ii x 0. To sketch the domain, we use the hint! We know that couples x, y satisfying x 2 + y 2 = 1 lie on a circle with radius 1 and centre at 0,0. Naturally, the couples x, y satisfying i lie inside the circle with radius 1 and centre 0,0 you can also take a test point, say 1/2,1/2: 1/2 2 + 1/2 2 = 1/2 1. We also need to consider ii. Finally, the domain is the semi-circle as displayed in the gure below. b To nd the stationary points, we need to solve the rst-order conditions: gx,y = 0 gx,y = 0 3x 2 2x = 0 2y = 0 x3x 2 = 0 y = 0 x = 0 x = 2/3 or y = 0 y = 0 We have found 2 stationary points: 0,0 and 2/3,0. To classify them, we need to consider the second-order conditions. We calculate the second-order derivatives: g xxx, y = 6x 2; g yyx, y = 2; g xyx, y = g yxx, y = 0; hx, y = g xxx, yg yyx, y [g xyx, y] 2 = Evaluate the SOC at the 2 stationary points: - at 0,0: g xx0, 0 = 2 < 0; g yy0, 0 = 2 < 0; and h0, 0 = 4 > 0. So 0,0 is a local maximum point. - at 2/3,0: g xx2/3, 0 = 0; g yy2/3, 0 = 2 < 0; and h2/3, 0 = 0. So 2/3,0 is a saddle point. c We need to consider the behavior of g on the boundary. We decompose the boundary into 2 zones: - zone 1: x = 0, 1 < y < 1. The function g is rewritten as: g0, y = 3 y 2 ; this is function a quadratic function for 1 y 1. Its minimum is reached when y = ±1: the associated points are 0,-1 and 0,1 and the function equals 2. Its maximum is reached when y = 0: the associated point is 0,0 and we already know it is a local 6
maximum function equals 3. - zone 2: x 2 + y 2 = 1 and 1 x 0. The function g can be rewritten as: gx, y = 3 + x 3 x 2 1 x 2 = 2 + x 3 ; this function is strictly increasing for 0 x 1. Its minimum is reached when x = 0: so the point is 0,1 and the function equals 2; and the maximum is reached when x = 1: so the point is 1,0 and the function equals to 3. Conclusion: global minimum at 0,1 and 0,-1; global maximum at 0,0 and 1,0. 7