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Microeconomics Course E. John Hey. This week: The Firm. Tuesday Chapter 11: Cost minimisation and the demand for factors. Wednesday Chapter 12 : Cost curves. Thursday Exercise 4: A mathematical exercise on profit maximisation. Chapter 11.
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MicroeconomicsCourse E John Hey
This week: The Firm • Tuesday • Chapter 11: Cost minimisation and the demand for factors. • Wednesday • Chapter 12: Cost curves. • Thursday • Exercise 4: A mathematical exercise on profit maximisation.
Chapter 11 • In Chapter 10 we introduced the idea of an isoquant – the locus of the points (in the space (q1,q2) of the quantities of the inputs) for which the output is constant. • Also the production function: • y = f (q1,q2) where y denotes the output. • An isoquant is given by: • y = f (q1,q2) = constant.
Particular cases • Perfect substitutes 1 to a: isoquants are straight lines with slope a. • Perfect complements 1 with a: isoquants are L-shaped and the line joining the corners has slope a. • Cobb-Douglas with parameter a: isoquants are smoothly convex everywhere.
Two dimensions • The shape of the isoquants: depends on the substitution between the two inputs. (We call the slope of an isoquant the marginal rate of substitution between the inputs). • The way in which the output changes from one isoquant to another – depends on the returns to scale.
Returns to scale with Cobb-Douglas technology : examples • Case 1: f(q1,q2) = q10.4 q20.6 • Constant returns to scale. • Case 2: f(q1,q2) = q10.3 q20.45 • Decreasing returns to scale. • Case 3: f(q1,q2) = q10.6 q20.9 • Increasing returns to scale. • Note: the ratio of the exponents is the same: hence the shape of the isoquants is the same – but they have different returns to scale.
Chapters 11, 12 and 13 • We assume that a firm wants to maximise its profits. • We start with a small firm that has to take the price of its output and those of its inputs as given and fixed. • Given these prices, the firm must choose the optimal quantity of its output and the optimum quantities of its inputs.
Chapters 11, 12 and 13 • We will do the analysis in two stages… • …in Chapter 11 we find the optimal quantities of the inputs – given a level of output. • …in Chapters 12 and 13 we will find the optimal quantity of the output. • (Recall that we are assuming that all prices are given.)
Chapter 11 • So today we are finding the cheapest way of producing a given level of output at given factor (input) prices. • This implies demands for the two factors... • ... which are obviously dependent on the ‘givens’ – namely the level of output and the factor prices. • If we vary these ‘givens’ we are doing comparative static exercises. • The way that input demands vary depends upon the technology.
Chapter 11 • We use the following notation: • y for the level of the output. • p for the price of the output. • w1and w2for the prices of the inputs. • q1and q2for the quantities of the inputs. • We define an isocost by • w1q1+ w2q2 = constant • …a line with slope –w1/w2 • Let’s go to Maple…
Chapter 11 • The optimal combination of the inputs is given by the conditions: • The slope of the isoquant at the optimal point must be equal to to the relative prices of the two inputs. • (this assumes that the isoquants are strictly convex) • The output must be equal to the desired output.
Factor demands with CRS C-D • The production function: • y= q1a q2bwhere a + b =1 • The factor demands: • q1 = y (aw2/bw1)b • q2 = y (bw1/aw2 )a
Chapter 11 • What do we note? • The demand curve for an input is a function of the prices of the inputs and the desired output. • The shape of the function depends upon the technology. • From the demands we can infer the technology of the firm.
Compito a casa/Homework • CES technology with parameters c1=0.4, c2=0.5, ρ=0.9 and s=1.0. • The production function: • y =((0.4q1-0.9)+(0.5q2-0.9))-1/0.9 • I have inserted the isoquant for output = 40 (and also that for output=60). • I have inserted the lowest isocost at the prices w1 = 1 and w2 = 1 for the inputs. • The optimal combination: q1 = 33.38 q2 = 37.54 • and the cost = 33.58+37.54 = 70.92.
What you should do • Find the optimal combination (either graphically or otherwise) and the (minimum) cost to produce the output for the following: • w1 = 2 w2 = 1 y=40 • w1 = 3 w2 = 1 y=40 • w1 = 1 w2 = 1 y=60 • w1 = 2 w2 = 1 y=60 • w1 = 3 w2 = 1 y=60 • Put the results in a table.
Chapter 11 • Goodbye!