Question

Please Answer Part 2.

The market for fabric has only one producer. Assume that daily market demand for fabric is y = 100,000 - 100p, where y denotes the quantity and p denotes the unit price. Also assume that producing y units of fabric costs 100y.

1. How many units of fabric should the producer produce and sell in order to maximize profits? Calculate the profit-maximizing price and the profit.

2. Now suppose that to produce one unit of fabric the producer needs to use one unit of an input, cotton. There is a single producer of cotton with a cost function c(x) = 100x where x is the quantity of cotton produced. (a) Suppose that the fabric producer purchases cotton at a price q. Calculate the equilibrium quantity of cloth as a function of q. (b) How many units of cotton should the input producer produce and sell in order to maximize its profits? How many units of fabric will be sold in the market? Calculate the equilibrium prices of cotton and fabric as well as the profits of both producers. How does the output of fabric compare to part (1)?

Answer 1

O Demandes: y = 100.000-100P >100P = 100,000-7 P= 1000-00ly e lonveres dernand » Total Revenue (TR) =py = 1000 y-0.014² Marginal Revenue (MR) - TR) TR) = 1000-0-02y Further, cost function () = 1004 > Marginal cost (*)(MC) y CCM) =100 At equilibrium, MC - MR -> 100 =1000 -0.02 -> 0.02y = 900 -> = 45,000 > P = 1000-0.01 (45,000) -1000-450 Profit - TR-TC = [1000 x 45000) - 0.01445.000 45000]-[100x45000] - 2,0 2.50,000 - Ai max profit maximizing level, 45,000 units of fabrics will be produced, prcice col be 550 and profit - 2,02,50,000 One unit of fabric needs 1 unit of input ay units of fabois need y units of input purchases (a) IP fabric producer Sonstuces cotton at price q >> Revenue fear cotton 2 xq.where x amant of cotton Also, cost por cotton is given as CCX) 2100X sofistit for other porducomes00100vossoux At maximum propria F43290 no 59too o For fabric producer, cost = q*y » preoft T = TR - TC = lood-old - 8 To maximize profit : on ay -> 10002-0.027-23 8 -> 0.02y = q +1000 = 509 +50000 So, the prooft maximizing y as a function of q is g = 50000tsoa (b) For cotton producen profit = X9-100*. At maximum proakt: (x9-100x) = 0 -) 9-100 -0 =) 9-100 Therefore, for fabrics: y - 50000+50 (100) = 55000 = P = 1000-0-01( 55000) = 450 and profit - (1000055060) -0.01253 006)]-[100x55000] = 1,9 2,50,000 Gred put scotton4-100, price of coita > output of cotton output of fabric = $8000, price of cotton = 9 = 100 > Revenue for cotton producen = 55000*100 = 55,00,000 Cosi Por caten producer = 100(55000) = 56,00,000 =) Profit fur cotto o =0