Question #319755

An American producer of steel produces and sells his products on two geographic regions: at home (H) and abroad (F). Demand on those markets is different. Also production costs in those two regions are different. Demand function on the home market is: PH = 260 – 0,1QH , and on the foreign market is PF = 240 (illimited amount can be sold for the price 240 dollars). Cost function on the home market is: CH = 1000 + 0,4Q2 H , and on the foreign market: CF = 5000 + 0,25Q2 F .

a. The products can not be exported due to high custom duties . Calculate amount of production and prices on home and foreign market to maximise profit. What will be profit? b. The company can export without limits and transport costs are low. Answer now the question from point a. c. Now suppose that transport costs are 16 dol. per ton. Answer the question from point a.


1
Expert's answer
2022-03-29T12:11:22-0400

PH=260-0.1QH

CH=1000+0.4QH2

PF=240

CF=5000+0.25QF2


A) TRH=260QH0.1QH2TR_H=260Q_H-0.1Q_H^2

MRH=2600.2QHMR_H=260-0.2Q_H

MCH=0.8QHMC_H=0.8Q_H

MRH=MCHMR_H=MC_H

2600.2QH=0.8QH260-0.2Q_H=0.8Q_H

QH=260{{Q_H}^*}=260

PH=2600.1(260)P_H=260-0.1(260)

PH=$234P_H=\$234

πH=TRHTCH\pi_H=TR_H-TC_H

πH=260(260)0.1(260)2(1000+0.4(260)2)\pi_H=260(260)-0.1(260)^2-(1000+0.4(260)^2)

πH=$32800\pi_H=\$32800


TRF=240QFTR_F=240Q_F

MRF=240MR_F=240

MCF=0.5QFMC_F=0.5Q_F

MRF=MCFMR_F=MC_F

240=0.5QF240=0.5Q_F

QF=480Q_F=480

PF=$240P_F=\$240

πF=240(480)(5000+0.4(480)2)\pi_F=240(480)-(5000+0.4(480)^2)

πF=$18040\pi_F=\$18040


B) P=PH+PFP=P_H+P_F

P=2600.1Q+240P=260-0.1Q+240

P=5000.1QP=500-0.1Q

MR=5000.2QMR=500-0.2Q

C=CH+CFC=C_H+C_F

C=1000+0.4Q2+5000+0.25Q2C=1000+0.4Q^2+5000+0.25Q^2

C=6000+0.65Q2C=6000+0.65Q^2

MC=1.3QMC=1.3Q

MR=MCMR=MC

5000.2Q=1.3Q500-0.2Q=1.3Q

Q=333Q=333

P=5000.1(333)P=500-0.1(333)

P=$466.7P=\$466.7

π=500(333)0.1(333)2(6000+0.65(333)2)\pi=500(333)-0.1(333)^2-(6000+0.65(333)^2)

π=$77333.33\pi=\$77333.33


C)if transportation costs are $16 per ton,

PH=2600.1QHP_H=260-0.1Q_H

QH=(260010PH)16=258410PHQ_H=(2600-10P_H)-16=2584-10P_H

PH=258.40.1QHP_H=258.4-0.1Q_H

TRH=258.4QH0.1QH2TR_H=258.4Q_H-0.1Q_H^2

MRH=258.40.2QHMR_H=258.4-0.2Q_H

MCH=0.8QHMC_H=0.8Q_H

258.40.1QH=0.8QH258.4-0.1Q_H=0.8Q_H

QH=258.4Q_H^*=258.4

PH=258.40.1(258.4)P^*_H=258.4-0.1(258.4)

PH=$232.56P_H^*=\$232.56

πH=258.4(258.4)0.1(258.4)2(1000+0.4(258.4)2)\pi_H=258.4(258.4)-0.1(258.4)^2-(1000+0.4(258.4)^2)

πH=$32385.28\pi_H=\$32385.28


PF=24016=224P_F=240-16=224

TRF=224QFTR_F=224Q_F

MRF=224MR_F=224

MCF=0.5QFMC_F=0.5Q_F

224=0.5QF224=0.5Q_F

QF=448Q_F^*=448


πF=224(448)(5000+0.4(448)2)\pi_F=224(448)-(5000+0.4(448)^2)

πF=$15070.40\pi_F=\$15070.40


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