Blok III Q0 Q1 Qm Q2 Second degree P cost minimal harga supaya tidak rugi Pa Pb

# Blok iii q0 q1 qm q2 second degree p cost minimal

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Blok III 0 Q0 Q1 Qm Q2 Second degree P= cost minimal harga supaya tidak rugi Pa Pb Pc Pc C MC A AC MC D=AR MR B First degree
Contoh soal : D P = 100 – Q S P = 20 – 3Q Ditanya : a. P dan Q Perfect Kompetisi b. P dan Q Monopsony Jawab : a. 100 – Q = P P = 100 – Q AR 1 = D 2 Q P P Monopsony ME ME = MV AE=S Q MC P MC MR AR=D Monopsony Power = p p v Monopoly Power = p MC p
20 + 3Q = P - = 100 - 20 80 – 4Q = 0 P = 80 80 = 4Q Q = 20 b. MV = ME S P = 20 – 3Q = AE Untuk monopsony pada TE yang dipakai P nya adalah Supplay TE = P.Q MV =ME PS = 20 + 3Q = (20 + 3Q) . Q 100 – Q = 20 + 6Q = 20 + 3(11,43) = 20Q + 3Q 2 80 = 7Q = 20 + 34,29 ME = 20 + 6Q Q = 11,43 P = 54,26 CONTOH SOAL TEORI EKONOMI MIKRO II 1. Diket : P = 300 – 3Q TC = Q 3 – 21 Q 2 + 333Q + 180 Tanya : a. Q, P, TR dan b. P 1 = 225 Q, P, TR, TC, 2 Jawab : A. MR = MC monopoli TR = P.Q = (300 – 3Q) Q = 300Q – 3Q 2 MR = Q dq dtr 6 300 TC = Q 180 333 21 2 1 3 Q Q MC = 3Q 2 – 42Q + 333 MR = MC 300 - 6Q = 3Q 2 – 42Q + 333
300 – 6Q = 3Q 2 + 42Q – 333 - 67 + 36Q – 3Q 2 / 33Q – 6Q – 3Q 2 + 42Q 3Q 2 – 36Q – 267 – 33 + 36Q – 3Q 2 / 3Q 2 – 36 + 33 = a c a b b . 2 . 4 2 = 3 . 2 36 / 167 3 . 4 1296 36 6 33 . 12 1296 = 6 124 1296 36 6 1172 36 = 6 23 , 34 36 6 900 36 = 6 30 63 = 11 Q 1 = 6 23 , 34 36 = 11,70 Q 2 = 6 23 , 34 36 = 0,295 P = 300 – 3 (Q 1 + Q 2 ) P = 300 – 3 (Q 1 + Q 2 ) = 300 – 3. 11,70 = 300 – 3 (11,995) = 300 – 35,1 = 300 – 35,985 = 264,9 = 264,015 TR = P.Q = P. (Q 1 + Q 2 ) = 264,015 . 11,995 = 3166,85 = TR – TC = 3166,85 – (11,995) 3 – 21 (11,995) 2 + 33 (11,995) 2 = 3166,85 – (1725,84) – 21 (143,88) + 395,83 = 3166,85 – 1725,84 -3021,48 + 575,83 = 3166,85 – 1725,84 – 2445,65 = 3166,85 – 4171,49 = -1004,64 B. P 225 Q,P,TR,TC dan
225 = 300 – 3Q 3Q = 300 – 225 Q = 75/3 = 25 P = 300 – 3 (95) = 300 – 75 = 225 = TR – TC = P.Q – Q 3 – 21Q 2 + 333Q + 180 = 225.25 - 25 3 - 21 (25) 2 + 333 (25) + 180 = 5625 – 15625 – 13125 + 8325 + 180 = 5625 – 11005 = -5.380 Sebelumnya TR = P.Q = 225.25 = 5625 TC = 25 3 – 21 (25) 2 + 333 (25) + 180 = 15625 – 13125 + 8325 + 180 = 11005 2. Diket : Q = 144/P 2 AV C = Q 1/2 FC = 5 Tanya : a. Q – P Max b. P + 4 Q, P 2 Jawab : a. Q = 144/ P 2 TR = P.Q P 2 = Q 144 = Q Q . 12 2 / 1 P = Q 144 = 2 / 1 12 Q = 2 / 1 . 12 2 / 1 12 Q Q Q Q = 12Q 2 / 1 ......... 1 TC = VC + FC MR = MC
AC = AVC.Q + FC 6Q -1/2 = 3/2 Q 1/2 = Q 1/3 . Q + 5 2 / 1 6 Q = 3/2 Q 1/2 = Q 3/2 + 5 ........... 2 MC = TC 1 12 = 3Q = 3/2 .Q 1/2 ........... 3 Q = 4 ............. 5 MR = TR = 12 Q 1/2 = TR - TC = 6Q -1/2 ................. 4 = (P) . (Q) – (Q 3/2 + 5) P = Q 12 = 24 – (8 + 5) = 4 12 = 6 ............. 6 = 11 .............. 7 b. P = 4 P = 2 / 1 12 Q = TR - TC 4 = 2 / 1 12 Q = (4). (9) – (Q 3/2 + 5) Q 1/2 = 4 12 = 36 – 27 + 5 Q 1/2 = 3 = 36 -32 Q = 3 = 4 Q = 9 3. Diket : C = 100 – 5Q + Q 2 P = 55 – 2Q Tanya : a. Q 2P max , C5 2 DWL b. P 27 Q, P, CS 2 , DWL Jawab : A. TR = P.Q P = 55 – 2Q = (C5 – 2). Q = 55 – 2 (10)
= 55Q – 2Q 2 = 55 - 20 MR = 55.4Q = 35 MR = MC 55.4Q = -5 + 2Q 55+5 = 2Q + 4Q 60 = 6Q Q = 10 = TR – TC = P.Q – 100 - 5Q + Q 2 = 35.10 – 100 – 5 (10) + (10) 2 = 350 – 100 – 50 + 100 = 300 Coordinat P : Cordinat 5 : MC P = 55 – 2R P = -5 + 2Q P = 0 Q = 55/2 = 27,5 P = 0 Q = 2,5 Q = 0 P = 55 Q = 0 P = 5 C5 = ½ . a. t = ½ . 10. 55 – 35 = ½ . 10. 20 C5 = 5 . 20 = 100 DWL = ½ . a. t = ½ .5 . 20 = 2,5 . 20 = 50 ATC = Q TC = 100 150 = 15 MC = P MR = 55 – 4Q Q = 0 MR = 0 MR = 55 Q = 13,75
B. P = 27 P = 55 – 2Q 27 = 55 -2Q = 55 – 2 (14) 2Q = 55 – 27 = 55 – 28 2Q = 28 = 27 Q = 14 TC = 100 - 5Q + Q 2 = 100 – 5 (14) + (14) 2 = 100 – 70 + 196 = 226 = TR - TC = P.Q – 226 = 378 – 226 = 152 CS = ½ .a.t = ½ .14. 55 – 27 = 7. 28 = 196 PWL = ½ . a. t = ½ .20 .14 = 10. 14 = 140 4. Diket : MC 1 = 20 + 2Q 1 P = 20 – 3 (Q 1 + Q 2 ) MC 2 = 10 + 5Q 2 Tanya : a. Produksi b. P max Jawab :
TR = P.Q = (20 – 3Q).Q = 20Q – 3Q 2 MR = 30 – 3Q MR = MC 2 20 – 3Q = 10 + 5 Q 2 20 – 3 (Q 1 +Q 2 ) = 10 + 5Q 2 20 – 3Q 1 – 3Q 2 = 10 + 5Q 2 20 – 10 – 3Q 1 = 5Q 2 + 3Q 2 10 – 3Q 1 = 8Q 2 10 = 8Q 2 + 3Q 1 10 = 3Q 1 + 8 a2 MR = MC 1 20 – 3Q = 20 + 2Q 1 20 – 3 (Q 1 +Q 2 ) = 20 + 2Q 1 20 - 3Q 1 - 3Q

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