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MARKET EQUILIBRIUM FOR TWO COMMODITIES. Asumption : demand to a certain good just affected by its value, and the factor is fixed (usually more than 1 factor affects the good values)
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MARKET EQUILIBRIUM FOR TWO COMMODITIES Asumption : demand to a certain good just affected by its value, and the factor is fixed (usually more than 1 factor affects the good values) When there is a relation between two goods, the demand is affected by its value (price) and the other value which is the relation could be substitutive (e.g rice and corn / coffee and tea) or complemented (e.g. a car and gazoline / coffee and sugar) Demand function of each : Qdx = f (Px, Py) Qdx : demand quantity of x Qdy = f ( Py , Px ) Qdy : demand quantity of y Px : price of x/unit Py : price of y/unit
E.g 1. Demand of X Qdx = 5 – 2 Px + Py Demand of Y Qdy = 6 + Px – Py Supply of X Qsx = -5 + 4 Px – Py Supply of Y Qsy = -4 – Px + 3 Py What is the market equilibrium Solution : There are 2 equations with 2 unknown values those are for Px and Py 1: Qdx = Qsx -6 Px = -10 – 2 Py 2: Qdy = Qsy 2 Px = -10 + 4 Py Check whether Px = 3 Py = 4 Qx= 5 – 2 Px + Py = 3 Qy= 6 + Px – Py = 5
(other type of writing) • Commodity A : D XA = 4 – 2 PA + PB S XA = 4 PA Commodity B : D XB = 20 + PA – 5 PB S XB = - 1 + 6 PB Market eq. DA = SA DB = SB Solve the equation Market equilibrium A : (XA , PA) B : (XB , PB)
TAX & SUBSIDY TO TWO COMMODITIES Commodity A : D X = 5 – PA + PB Commodity B : D X = 10 – PA - PB S X = -5 + PA + PB S X = -2 – PA + 2 PB Tax = 0.5 Subsidy = 0.5 ME pre tax & sub : A : DA = SAB : DB = SB solve the equation(s) PA = 5 PB = 4 XA = 4 XB = 1 (4 ; 5) (1 ; 4) ME post tax & sub : can you solve the problem ? Remember that the demand function is not changed
Post A : S X = -5 + (PA-0.5) + (PB + 0.5) B : S X = -2 – (PA-0.5) + 2 (PB + 0.5) ....................................... .................................. Eq D = S for both of A and B You should find PA = 5 and PB = 3.5 check it out ! XA = 3.5 XB = 1.5 Market Equilibrium : A ( 3.5 ; 5 ) B ( 1.5 , 3.5 )
Problems using differentiation and integration • The unit price p of a product is related to the number of unit sold, x, by the demand equation p = 400 – x/1000. The cost of producing x units is given by C(x) = 50x + 16000. The number of units produced and sold, x, is increasing at a rate of 200 units per week. When the number of units produced and sold is 10000, determine the rate of change with respect to time, t (in weeks) of : a. Revenue b. Cost c. Profit Solution : a. R = px = 400x – x2/1000 dR/dt = (400 – x/500) dx/dt when x=10000 and dx/dt = 200 dR/dt = 76000 Means : the revenue is increasing at a rate of 7600 per week
C(x) = 50x +16000 d(C)/dt = 50 dx/dt + 0 when dx/dt = 200, dC/dt = 50 * 200 = 10000 So the cost is increasing at a rate of 10000 per week • Profit : P = R – C dP/dt = dR/dt – dC/dt = 76000 – 10000 = 66000 Means that the profit is increasing at a rate of 66000/week
The demand y for a commodity is y = 12/x, where x is tha price. Find the rate at which the demand changes when the price is 4 Solution : the rate of change of the demand y with respect to the price is dy/dx dy/dx = 12 / x2 So the rate of change of demand with respect to price x is -12/x2 When the price is 4, an increase in price by 1% will result in the fall of demand by 0.75%
A firm produces x tonnes of output at a total cost C = (1/10 x3 – 5x2 + 10x + 5) At what level of output will the Marginal Cost attain the minimum Solution : MC = d(C)/dx= 3/10 x2 – 10x + 10 (=y) minimum first derivative =0 and second derivative > 0 dy/dx = 0 3/5 x – 10 = 0 x = 50/3 When x=50/3, d2y/dx2 = 3/5 >0 MC is minimum So MC attain its minimum at x=50/3 units
The cost function of a firm is C=1/3 x3 -5x2 + 28x + 10 where x is the output. A tax at 2 per unit of output is imposed and the producer adds it to his cost. If the market demand function is given by p=2530 – 5x where p is the price per unit of output, find the profit maximising output and price Solution total revenue R = px = 2530x – 5x2 Total cost after tax : C = 1/3 x3 -5x2 + 28x + 10 + 2x = .......... Profit : P = R – C = -1/3 x3 + 2500x – 10 dP/dx = -x2 + 2500 Max dP/dx = 0 x = ±500 d2P/dx2 = -2x When x=50 d2P/dx2 = -100 < 0 P is maximum So profit maximising output is 50 units and when x=50, price p = 2530 –(5*50) = 2280
Partial derivative The revenue derived from selling x calculators and y adding machines is given by R(x,y) = -x2 + 8x -2y2 + 6y + 2xy + 50 If 4 calculators and 3 adding machines are sold, find the Marginal Revenue of selling : a. One more calculator b. One more adding machine Solution : • Rx=σ(R)/σx= -2x +8 – 0 + 0 + 2(1)(y) Rx(4,3) = 6 So, at (4,3), revenue is increasing at the rate of 6 calculators sold MR=6 • Ry= σ(R)/σy = 0+0 – 4y + 6 + 2x(1) Ry (4,3) = 2 Means : at (4,3), revenue is increasing at the rate of 2 per adding machine MR = 2
Integration The marginal cost function of manufacturing x units of a commodity is 6 + 10x – 6x2. Find the total cost and average cost, given that the total cost of producing 1 unit is 15 Solution : MC = 6 + 10x – 6x2 C = ʃ(MC) dx + k = 6x + 10 x2/2 – 6x3/3 + k = ............ Given x=1, C=15 15 = 6 + 5 -2 k k=6 Total cost function : C= 6x + 5x2 – 2x3 + 6 Average Cost Function = C/x, x≠0 = ...................
The marginal Cost Function of manufacturing x units of a commodity is 3x2 – 2x + 8. Ifre is no fixed cost, find the total cost & average cost function Solution : MC = 3x2 – 2x + 8 C = ʃ (MC) dx + k = x3 – x2 + 8x + k No fixed cost k=0 Hence, Total Cost C = x3 – x2 + 8x Average Cost = C/x = x2 – x + 8
Monopoly is a situation in which there is a single seller of a product for which there are no good substitutes. What is bad about monopoly? • - Consumer options are limited. • Profits do not signal firms to enter the industry. (They can’t get in because of the barriers to entry.) • The price depends on the producer • when S decrease P will be arised • S increase P will be decreased • Read about monopoly, how to get the profit and how much it is • ᴨ(read as phi) = R - C