1. Solve the intertemporal consumption choice problem: Max U = In(C, ) + B In (C2) subject to (1 + r) C1+ C2 = (1+ r)/1 + 12. a) Set up Lagrangian function and use the first order condition to find the stationary pointb) Use the second order condition to conrm if the stationary point is a maximum . Step 1: Set Up Lagrangian function o L = In(C1) + BIn(C2)+2 ((1+r)1+12 - ((1+r)(1+Cz)) o L = In(C1) + BIn(C2) + a((1+r)1+ 12 -(1+r)(1 -C2) . Step 2: Solve the first order conditions O - = 0 SC1 1 - 1(1 +r) = 0 1 = 1(1 +r) 1 C1 (1+r) O = 0 SC2 B - 2=0 C2 B C2 o Set 1 equal to each other and solve for CZ1 B C1(1+r) C} = C1(1-r) B o Plug C2 back in and solve for C1 " (1+r)1+12 - ((1+r)c1 + Ci(1-r) = 0 - (1+r)li+ 12 = (1+r)C, + Ci(1-r) B " C* = Bl2+8(1+r)/1 Br+1 o Plug C1 back in and solve for C2 Bl2+B (1+r)11)(1-r) C2 = Br+1 B C* = 2+1-r12-1211 Br+1 o Plug C2 back in and solve for a B = 1 12+1-r12-r21 Br+1 2* = B'r+ B 12+11-r12-r2/1 O = 0 sa " (1+r)/1+ 12 - ((1+r)(1+ Cz) =0. Step 3: Find the second order conditions -(1+r) -1 Laa Lac1 LAC2 1 -(1+ r) 0 C2 o H = LC12 LC , C1 Lc C2 = B 0 LCza LC2C1 LC2 C2. -1 Co |H| = (-1)1+2 . (-(1 +r)) . (1+r) 0 -1 B + (-1)1+3 . (-1). c3 -(1 +r) -1 o . |HI = -(-(1+7)) . ((-(1+r)) (-")-(0)(-1) )+(-1). ((-(1+r))(0) - (-2) (-1) ) - |HI = -(-(1+r)) (8) (B(1+r) ) + (-1) . (-27) " [H| = 1 + B(1+r)2 CZ o Plug in C1 and C2 from part 1 E B(1+r)2 2 + (Bl2+B (1+r)/1) Br+1 (12+1-r12-1211 Br+1 (Br+1)2 + B(1+r)2(Br+1)2 (B12 +B(1+r) 1 1)2 (12+1-rl2-r21 1)2 . Since (Br+1)2 (B12 +B(1+r)/1)2 + B(1+r)2 (Br+1)2 (12+/1-r12-r211) > 0, this means that it is a maximum at the stationary points (C1, C2) = (2+but)l1 12+1-r12-r211) | Br+1 Br+1