# Eco 507 Midterm

April 28, 2018 General Studies

P=-1. 2 %? P=4. 375% (iii). a. Maximize…Z = M + . 5S + . 5MS – S? Subject to 30000S + 60000M = 1200000 Lagrangean…L=M+. 5S+. MS-S2+? 1,200,000-30,000S-60,000M ?L? S=0. 5+0. 5M-2S-30,000? ?L? M=1+0. 5S-60,000? ?L?? =30,000S+60,000M Equating ? , I get 1 + 0. 5S/60000 = 0. 5 + 0. 5M – 2S M = 4. 5S By substituting into budget constraint, I get 30000S + 60000 * 4. 5S = 1200000 S = 4 M = 18 b. Cost function = 30000S + 60000M Marginal cost of S = 30000 Marginal cost of M = 60000 Total marginal cost = 90000 c. (iv. ) a. Demand…Q = a – bP

E = (P/Q)*(? Q/? P) E = -b (P/Q) -0. 4 = -b(4/2) b = 0. 2 a = Q + bP = 2 + 0. 2 * 4 a = 2. 08 Demand Equation…Q = 2. 08 – 0. 2P 2. (i) Q = LK ?Q? L = K ?2Q? L2 = 0 The second order derivative did not give a negative value, so it ignores the condition of diminishing marginal productivity of labor. b. Q (L, K) = LK Q (mL, mK) = m? LK The output increases more than proportionally, there are increasing returns to scale. c. Q = LK TC = wL + rK L = wL + rK + ? (Q-LK) ?L? L = w + ? (K) =0 ?L? K = r + ? (L) =0 w /r = K/L =RTS

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In this equation, the firm should use K and L as given that ratio to minimize cost of production. The Lagrangean Multiplier is marginal cost of any input to marginal benefit of any input should be same for any input. It explains if marginal cost –benefit ratio is greater for K than L, we have to substitute L for K to minimize cost. d. 225 = LK 225 = 16L+144K L = 16L+144K + ? (225-LK) ?L? L = 16 + ? (K) =0 ?L? K = 144 + ? (L) =0 K/L =0. 11 K = 0. 11 L L (0. 11L) = 225 0. 11 L^2 = 225 L^2= 2045. 46 L = 45. 23 45. 23K = 225 K = 4. 97 TC = 16*45. 23+144*4. 7 TC = \$1439. 36 e. (ii) X dollars increase in the daily rate above \$60, there are x units vacant. So 60+X= 80-X 2X=20 X=10 If they charge 60+10=\$70, 10 rooms will be vacant and 70- rooms will be occupied. The profit for 80 rooms occupation at \$60 per room, TR= 80*60= \$4800 TC= 4*80= \$320 Profit = \$4480 The profit for 70 rooms at the price of \$70 TR= 70*70= \$4900 TC= 4*70= \$280 Profit= \$4900 -\$280= \$4620 In this case the profit will also be maximized. 3. i) a) Maximize Y = 2Ty – . 001Ty^2 S. t. 100Ty + 25Tz = 1300 Also Maximize Z= 20 Tz – . 1 Tz^2 S. t 100Ty + 25Tz = 1300 b) I used the Lagrangean to get: L = 2Ty – . 001Ty^2 + 20 Tz – . 01 Tz^2 +? (1300 – 100 Ty- 25Tz) dL/dTy = 2 – 0. 002Ty – ? (100) = 0 dL/dTz = 20 – 0. 02Tz -? (25) = 0 Also 100Ty + 25Tz = 1300 Divide the first two equation to get : 2 – 0. 002Ty = ? (100) 20 – 0. 02Tz =? (25) 2- 0. 002Ty = 100 20- 0. 02Tz = 25 2-0. 002Ty /20- 0. 002Tz = 4 2- 0. 002Ty = 80 – 0. 008Tz 0. 008 Tz – 0. 002Ty = 78 100Ty + 25Tz = 1300 So T*y = 2. 28 and Tz = 42. 88 ii) a) Q= 10 L – 0. 1L ^2 Wage rate = 12

Now Q = 250 Then L required Then L* = 50 And Labor price is 12 so total cost = 12×50 = 600 &lt; 500. You should not accept the offer b) Optimal amount of labor will be the one that equates MPL with wage ratio MPL = 10 – 0. 2L = 2 8 = 0. 2 L L* = 40 And wage paid = 80 This is the optimal point and I should accept the offer as 80 &lt; 500 Profit = 500 – 80 = 420 iii) To calculate the optimal price I used the markup formula that says that P – MC/ P = – 1/ed Put the values to get P- 10/P = -1/1. 5 1. 5 P – 15 = -P 2. 5 P = 15 P* = 6

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