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Solutions for Homework 1

Chapter 2: Problem 4

a. Good Z is a substitute for X, while good Y is a complement for X.

b. X is a normal good.

c. QXd = 6,000 - ½(5,230) - 6,500 + 9(100) + 1/10(70,000) = 4,785

d. QXd = 6,000 - ½Px - 6,500 + 9(100) + 1/10(70,000) => QXd = 7,400 – ½PX and PX = 14,800 – 2QXd

Chapter 2: Problem 5

a. Solve the demand function for PX to get inverse demand function: PX = 150 – ½QXd.

b. QXd|Px=45 = 300 – 2PX = 210, y intercept of PX is 150 => CS = 210(150 – 45)/2 = $11,025.

c. QXd|Px=30 = 300 – 2PX = 240 => CS = 240(150 – 30)/2 = $14,400.

d. As long as low of demand holds a decrease in price leads to an increase in consumer surplus, and vice
versa. There is an inverse relationship between the price of a product and consumer surplus.

Chapter 2: Problem 6

a. Qd = 60 – P = P – 20 = Qs => Pe = 40 => Qe = 60 – 40 = 20

b. Effective Pf = 50 > Pe = 40: Qd = 60 – 50 = 10 < 30 = 50 – 20 = Qs => Q = 10 and surplus = 20.

c. Effective Pc = 32 < Pe = 40: Qd = 60 – 32 = 28 > 12 = 32 – 20 = Qs => Q = 12 and shortage = 16.
Full economic price Pfe is inverse demand at exchanged quantity Pd = 60 – Q|Pc = 60 – 12 = 48 = Pfe.

Surplus is Qs|Pf=12 – Qd|Pf=12 = 2. Shortage is Qd|Pc=6 – Qs|Pc=6 = 4 – 1 = 3 => Pfe = Pd|Q=1 = 12. total tax revenue is only $24. One unit is sold. Consumers pay P = $12. CS = 2(14 – 10)/2 = 4 PS = 2(10 – 2)/2 = 8.Chapter 2: Problem 7 a. Since only two units are sold after the tax and the tax rate is $12 per unit. b. d. b. c.5 = 18. Qd =14 – ½P = 1/4P – 4 = Qs => P = 24 => d Q = 14 – 24/2 = 2. e. No. Chapter 2: Problem 8 a.5 – 1 = 1. At Pc = $2 no output is produced. c. Producers receive P .T = 6. Tax shifts S0 by 6 to S1. Qd =14 – ½P = 1/4P – 1 = Qs => Pe = $20 => Qd = 14 – ½(20) = 4.5 => Government cost is 12*1. . Tax of 12 increases Intercept of the inverse supply to P = (12+4) + 4Qs => Qs = ¼P – 4.

20 CSc = Qs|Pc(966.104 = 126.817 c.530. ESf = Qs|Pf – Qd|Pf = 4*600 – 120 – (5.67 – 600)/2 = 403. Pd = 966. Ps = 30 + 1/4Q Pd = 966.34 – 500)/2 = 28.904.34 DWLc = (Qe – Qs|Pc)(Pfe – Pc)/2 = (2.688 CSe + PSe = 1.Problem Set: Problem 1 a.337 PSf = Qd|Pf(Ps|Qdf – 30)/2 + Qd|Pf(Pf – Ps|Qdf) = 2.96 e.104(556 – 30)/2 = 553.200(966.248(966.240 DWLt = t(Qe – Qt)/2 = 60(2.823.34 – 500) = 288.880 = 920 Pfe = Pd|Qs = 966.052.880(653.248 CSe = 2. TRt + DWLt + CSt + PSt (+ rounding error) = 1.800 – 6*600) = 2.67 – 616)/2 = 368.352 .000 DWLf = (Qe – Qd|Pf)(Pd|Qd – Ps|Qd)/2 = (2.880(966.248 – 2.880/6 = 653.880(500 – 30)/2 = 441.84 .67 – 592)/2 = 421.104 and Pb = 616 Pp = Pb – t = 616 – 60 = 556 TRt = t*Qt = 60*2.67 – 1.67 – Pf)/2 = 2.67–90)/5 = 2.052.880)(653.052.800 – 6P = 4P – 120 = Qs => Pe = 5.34)/2 = 294.20 PSc = Qs|Pc(Pc – 30)/2 = 1.816.76 – 653.214.7 – 1/6Q b.104)/2 = 4.320 CSt = Qt(966. ED = Qd|Pc – Qs|Pc = 5.200(580 – 30)/2 + 2.67–1/6Q = (30+60)+¼Q = Ps’ => Qt = 12(966.052.200)(600 – 580)/2 = 480 CSf = Qd|Pf(966.920/10 = 592 and Qe = 4*592 – 120 = 2.56 BMGc = Qs|Pc(Pfe – Pc) = 1.248 – 2.129 PSe = 2. Qd = 5.800 – 6*500 – (4*500 – 120) = 2.200(600 – 580) = 649.200 = 80 GSf = Pf(QS|Pf – QD|Pf) = 600*80 = 48.84 PSt = Qt(Pp – 30)/2 = 2.000 DWLf + CSf + PSf = 1.817 d.104(966.248(592 – 30)/2 = 631.00 DWLc + BMGc + CSc + PSc (+ rounding error) = 1.248 – 1.279.7 – Pfe)/2 = 1.800.67 – Pb)/2 = 2.800 – 1.280 – 2.

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40 .Pd = 966.50) = 118.25 = 139 and Qsq = 124 + 4*2.34 – 558.25 = 133 Windfall = W = q(Pq – Pw) = 6(2.5 s s PIt’ = t(Q t’ – Q w’)/2 = 0.045.248 – 2.25*133 – 0.625 d.25(160 – 155)/2 = 0.25(144 – 139)/2 = 0.25*6 = 1.25(133 – 132)/2 = 0.125 Tt’ = t*Qst’ – PIt’ = 0.67–33)/5.25*133 – 0.114(558.67(2.5 and Qe = 134 b.412.25: Mt’ = Qdt’ – Qst’ = 200 – 20(2+0.25) – [124 + 4(2 + 0.114(966.25) – [124 + 4(2 + 0.9167 = 135.67)(2. Qd = 184 – 20P = 124 + 4P = Qs => Pe = 2.67 – Pb)/2 = 2.9167 – 2)/2 = 1.5 – 30)/2 = 558.9167 – 2)/2 = 8.1(30+¼Q) = (1+v)Ps => Qv = 12(966.25 – 2) – 0.25(133 – 132)/2 = 0.34 Pp = Ps|Qv = 30 + 2.3 = 2.67 – 132)(2.625 e.68 Tq’ = Qsq’(Pq’ – Pw) – PIq = 135.5 Production inefficiencies = PIt = t(Qst – Qsw)/2 = 0.114/4 = 558. Increased demand: Qd’ = 200 – 20P With Pw = 2: Mw’ = Qdw’ – Qsw’ = 200 – 20*2 – (124 + 4*2) = 160 – 132 = 28 With tariff = 0.9167 = 141.125 Tq = Qsq(Pq – Pw) – PIq = 133(2.125 CIt’ = t(Qdw’ – Qdt’)/2 = 0.50 TRv = Qv(Pb – Pp) = 2.35 Problem Set: Problem 2 a.125 Transfer to PS = Tt = t*Qst – PIt = 0.5 PIq’ = (Qsq’ – Qsw’)(Pq’ – Pw)/2 = (135.67–1/6Q = 1.25)] = 139 – 133 = 6 Tax revenue = G = t*Mt = 0.25 – 2) = 1. With Pw = 2: Mw = Qdw – Qsw = 184 – 20*2 – (124 + 4*2) = 144 – 132 = 12 c.67 – 2.25*22 = 5.114/6 = 614.50)/2 = 3.9167 – 2) – 1.052.34)/2 = 372.625 With quota = 6: Qdq’ = 200 – 20P = 6 + 124 + 4P = q + Qsq => Pq’ = 2.68 = 122.624.67 and Qsq’ = 124 + 4*2.76 DWLv = (Qe – Qv)(Pb – Pp)/2 = (2.824.81 PSt = Qv(Pp – 30)/2 = 2.25)] = 155 – 133 = 22 G’ = t*Mt’ = 0.114(614.741.28 CSv = Qv(966.114 Pb = Pd|Qv = 966.50 TRv + DWLv + CSv + PSv (+ rounding error) = 1.25 – 2)/2 = 0.125 = 33.125 CIq = (Qdw – Qdq)(Pq – Pw)/2 = (144 – 139)(2.125 Consumption inefficiencies = CIt = t(Qdw – Qdt)/2 = 0.25: Mt = Qdt – Qst = 184 – 20(2 + 0.5 PIq = (Qsq – Qsw)(Pq – Pw)/2 = (133 – 132)(2.34 – 558.9167 Qdq’ = 200 – 20*2.125 = 33.69 CIq’ = (Qdw’ – Qdq’)(Pq’ – Pw)/2 = (160 – 141.25 Qdq = 184 – 20*2.67 – 614.25 – 2)/2 = 0.67 W’ = q(Pq’ – Pw) = 6(2. With tariff = 0.114)(614.9167 – 2) = 5. With quota = 6: Qdq = 184 – 20P = 6 + 124 + 4P = q + Qsq => Pq = 2.125 = 33.