Problem Set3 Solutions_Rev1

# Problem Set3 Solutions_Rev1 - BPUB250 Problem Set3 Due...

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Unformatted text preview: BPUB250 Problem Set3 Due: February 22-23, 2012 in Class Question 1 (a) MRS= -MUc/MUt= − =− To find demand, we use MRS=-1/p. This yields c=tp. Using the budget constraint, c+pt=20-8=12, we get 2pt=12. Therefore, t=6/p and c=6. (b) Before the subsidy, t=2 and c=6 so that Tom achieves a utility of 6√2 × 6 = 12√3 ≈ 20.78 Let I denote Tom’s money, then his budget constraint is c+pt=I-8.(remember he spends \$8 for the movie ticket). After subsidy, his demand for Twizzlers is (I-8)/2p=(I-8)/4, for popcorn is (I-8)/2. Therefore, his utility is 6 × = ( √ ) . Setting this to 12√3, we get I=17.80. In this case, he would end up consuming t=9.8/4=2.45 and c=9.8/2=4.9. (c) We would need to take away 20-17.80=2.20 after the subsidy, to make him equally well off as before the price change. Thus CV=2.20 (d) After the subsidy, t=3 and c=6 so that Tom achieves a utility of 6√3 × 6 = 18√2 ≈ 25.46. Let I denote Tom’s money. Before the subsidy, his demand for Twizzlers is (I-8)/2p=(I8)/6, for popcorn is (I-8)/2. Therefore, his utility is6 × = ( √ ) . Setting this to 18√2, we get I=22.70. EV=2.70. (e) Cost=(3-2)3=3 (f) The Movie Theater can instead give a cash grant of \$3 to Tom. With the cash grant, Tom’s budget constraint will be c+3t=20-8+3=15. MRS= -t/c= -1/3. This yields c=3t. Plugging into the budget constraint, we get t=5/2, c=15/2. Tom’s utility will be 6 × Question 2 (a) = = = √ ≈ 25.98 > 25.46 (utility under subsidy) = = (1 − ) (b) Yes. Check the derivatives of the marginal products with respect to the respective inputs. You will see that the marginal product of lemons is decreasing in lemons and the marginal product of lemon-squeezing hours is decreasing in lemon-squeezing hours. = ( − 1) < 0 because 0 < α < 1 = − (1 − ) =− (c) MRTS = = ( ) <0 =− ( ) =− ( ) > 0 so yes, the MRTS does diminish as H increases. Remember the lecture slide example? MRTS changes -4 => -1 => -0.25. Therefore the slope becomes flatter and this means >0 (d) We need to compare f(tL,tH) to tf(L,H) where t>1. = = ( , ) so this production function exhibiuts f(tL, tH) = ( ) ( ) constant returns to scale. (e) In order to find the short run cost function, first we have to write down H* in terms of y and plug the expression into the short run cost function. Given that L=3, we have y=3 from the production function. Rewriting it gives us this into the cost function, ( , , , ) = 3 + =3 + ∗ = . Plugging . (f) Since both variables can vary in the long run, Earl will choose the optimal production where MRTS = − Using the ( =y ( = − => H = production Rearranging gives us ∗ ) ) ∗ ( ) (i) function, y = =y ( ) ( = =y ( ) ) = ( ) . . Plug this into (i) gives us . (g) There are four changes we should consider: how demand for each of two inputs depends on each of two prices. ∗ = −y(1 − ) <0 ∗ ∗ ∗ = y(1 − α) = −yα ( = y(1 − α) ( >0 ) ) ( <0 ) >0 While these expressions look messy, we know that r, w, and y are all positive numbers, which simplifies our analysis. Not surprisingly quantity demanded of each input decreases with its own price, but imput demand incr reases as the price of the other input increases. (h) In order to find the long run cost function, first we have to write down H* and L* in terms of y and plug the expressions into the long run cost function. We already found H* and L* in terms of y in question (f). (1 − ) C( , , , ) = w ∗ + ∗ = y +y (1 − ) AC( , , , ) = MC( , , , ) = (, ,,) = (, ,,) = ( ) ( + ) + ( ) ( ) For this production function, which generates a total cost function that is constant multiple of quantity, average cost and marginal cost will be the same. Question 3 (a) Pizza Rustica Alegro Ed’s (b) Pizza Rustica q = 8 0.4 0.6 = 8 0.6 q/ ∗ = 8 Plugging L* into the cost function give us SRTC(q, w) = w ∗ + 96 = 32 q / + 96 Alegro q = min(100 , 2 ) = min(100,2 ) We cannot produce q>100. If we want to produce q ≤ 100, from the production q=2L to minimize the cost. ∗ = Plugging L* into the cost function give us SRTC(q, w) = w ∗ + 96 = + 96 for q ≤ 100 Ed’s q = 30 + = 30 + ∗ = − 30 Plugging L* into the cost function give us SRTC(q, w) = w ∗ + 96 = ( − 30) + 96 for q ≥ 30 (Ed’s is guaranteed to produce 30 units because it has one unit of automated pizza making machine. (c) and (d) Pizza Rustica MRTS = − =− . 0.4 −0.4 . −0.6 0.6 =− =− K= (1) We also have to satisfy the production function. Plug (1) into the production function, q=8 ∗ = 0.4 0.6 0.4 =8 0.6 = 1.7 . Plugging L* into (1) we get ∗ = = . . Plugging L* and K* into the cost function give us LRTC(q) = 3 ∗ + 96 ∗ =3 +96 . . =2.94q Alegro Using the logic in PS1 and PS2, we know that the optimal choice has to lie at the kinked point in the isoquant graph, satisfying 100K = 2L = (2) We also have to satisfy the production function. Plug (2) into the production function, q = min(100 , 2 ) = min 100 × ∗ = min(2 , 2 ) = 2 ,2 = Plugging L* into (2) we get ∗ = = Plugging L* and K* into the cost function give us LRTC(q) = 3 ∗ + 96 ∗ =3 +96 =2.46q Ed’s MRTS = − =− =− and Using the logic in PS1 and PS2, | lie in the x-axis(K*=0). =− = |>| | implies that the optimal choice has to We also have to satisfy the production function. q = 30 ∗ =q Plugging L* and K* into the cost function give us LRTC(q) = 3 ∗ + 96 ∗ = 3 + 96× 0 =3q ∗ + = Question 4 (a) q = 3 ∗ 1/3 2/3 =3 1/3 = Plugging S* into the cost function give us SRTC(q, w, r, F) = w ∗ + + = ++ (b) (q) = pq − SRTC(q, w, r, F) = 8q − 2 (q) q =8−2 3 ∂q ∗ = 6, ∗ = + 16 + 30 =0 =8 (q) = 8 × 6 − 2 × 8 + 16 + 30 = −14 Although Michelle makes negative profit, this is still better than shutting down, q=0, because then the profits will be -30 (or -16-30=-46 depending on how you assume about K usage) (c) MRTS = − =− −2/3 1/3 1/3 −2/3 =− =− B= (3) We also have to satisfy the production function. Plug (3) into the production function, q=3 (d) ∗ = 1/3 2/3 =3 2/3 1/3 =3 2/3 42/3 Plugging S* into (3) we get ∗ = 42/3 Plugging S* and B* into the cost function give us LRTC(q) = 2 ∗ + 16 ∗ =2 42/3 + 16 42/3 = 5.04 (e) In the long run, profit will be zero, because Michelle operates in the long-run competitive equilibrium. (q) = pq − LRTC(q) = pq − 5.04 = 0 ∗ = 5.04 Question 5 = (tx)2/3[(ty)+2(tz)]1/3 = t2/3x2/3[t (y+2z)] 1/3 = t2/3x2/3t1/3 (y+2z) 1/3 = t[x2/3(y+2z) 1/3] < tf(x,y,z) = t[x2/3(y+2z) 1/3] Therefore it exhibits constant returns to scale. (a) f(tx,ty,tz) (b) x=1, thus q= (y+2z)1/3 Since all input prices are the same and equal to \$w and y+2z is a linear function (y,z are perfect substitutes), the firm will only use z and not y. Thus y*=0 and q=(2z)1/3 z* = q3/2. S(w,q) = wx+wy+wz = w+0+w[q3/2] =w[1+ q3/2] (c) Π = pq - w[1+ q3/2] ∂Π/∂q = p – (3/2)wq2 = 0 q* = (2p/3w)1/2 (d) With the same logic, y*=0 in the long run. Thus min wx + wz, subject to q= x2/3(2z)1/3 MRTS = -MPx/MPz = - [(2/3)x-1/3(2z)1/3 ] / [(2/3)x2/3(2z)-2/3] = - 2z/x = -w/w= -1 x=2z Plugging this into the production function, q=(2z)2/3(2z)1/3=2z z*= q/2 x*= q The long-run total cost function is C(w,q) = wx* + wz* = 2wz* + wz* = 3wz* = (3/2)wq ...
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