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Module 6 Session 12: Production with two inputs and returns to scale Production function with two variable inputs 1. You are given the following production functions. What types of production function they are? (i) Q=2L+K (ii) Q=min(2L,K) Solution: (i)
Production function Q=2L+K is a linear production function.This is proved as under. Let L and K are increased by a given number p Q’=2pL+Kp P can be factored out.Thus, Q’=p(2L+K)= pQ Thus, imcreasing each inputs by p,output also increases by p.This shows this production function is linear homogenous. Production function, Q=min(2L,K) is a Leontief production function. This is so called because a noted American economist Wassily Leontief used this production function to describe the American economy. This is a fixed proportion function in which labour and capital are combined in the ratio of 2L and 1K.
(ii)
Problem 2. The following production function is given: Q = L0.75 K0.25 I. II. III. IV. V.
Find the marginal product of labor and marginal product of capital. Show that law of diminishing return holds Show that if labor and capital are paid rewards equal to their marginal product total product would be exhausted. Calculate the marginal rate of technical substitution of capital for labor. Find out the elasticity of substitution.
Solution: I.
MPL = dQ/dL = 0.75L0.75-1 K0.25 =0.75L-0.25 K0.25 = 0.75(K/L)0.25 = 0.25L0.75 K-0.75 = 0.25(L/K)0.75
II.
III.
Law of diminishing returns holds, if marginal product of labor. Now given the constraint 0.75 and 0.25, if labor (L) increases, capital (K) held constant the term K/L will diminish and therefore the marginal product of labor (=0.75(K/L)0.25) too will decline. For the total product to be exhausted if the factor are paid reward equal to the marginal product, the following equation which Is called Euler’s theorem must hold” Q = MPL L + MPK K
Substituting the values of MPL and MPk obtained from the given production function the above section (i) In Euler’s theorem we have Q = 0.75(K/L)0.25 L +0.25(K/L)0.75 K Q = 0.75K0.25 L0.75 + 0.25L0.75K0.25 Q = L0.75 K0.25 (0.75 + 0.25) Q = L0.75 K0.25 Since L0.75 K0.25 = Q (given) Q=Q IV. V.
MRTS LK= MPL / MPK = {0.75(K/L)0.25} / 0.25 (L/K)0.75 3 (K/L) Elasticity of substitution = {Δ (K/L) / (K/L)} / {Δ MRS / MRS} Substituting the value of MRS = 3 (K/L) we have, Elasticity of Substitution = {Δ (K/L) / (K/L)} / {3 Δ (K/L) / 3 (K/L)} = 1
Optimal Use of two Inputs: 1. Suppose that a product requires two inputs for its production. Then is it correct to say that if the prices of the two inputs are equal, optimal behaviour on the part of the producer requires that these inputs be used in equal amounts? Solution: It is not correct to say that if the prices of the two inputs are equal, then these inputs should be used in equal amounts except in one case. We may explain as follows. Let us suppose that the production function of the firm is q = f(x, y) ……….(1)
where the symbols have their usual meanings. If the prices of the two inputs are r X and rY and if they are equal, then the slope of the firm’s iso-cost lines would be –r X/rY = -1. Therefore, at the optimal point of constrained output maximisation or cost minimisation, i.e., at the point of tangency between an isoquant and an iso-cost line, the slope of the isoquant should also be equal to —1. Now, if the firm is to use equal quantities of the two inputs, then at the point of tangency, we should have x = y. In other words, at the point of tangency, we should have
i.e., the isoquants of the firm should be rectangular hyperbolas. We may conclude, therefore, that the given statement is true, i.e., the firm should use the two inputs (with equal prices) in equal amounts if the production function of the firm gives rise to rectangular hyperbola isoquants, Otherwise the given statement is not correct. 2. A firm use two inputs X and Y for producing its output. The production function of the firm is q = f (x, y) = xy, and the market prices of the two inputs are given to be Rs 20 and Rs 10, respectively. Find the input combination the firm should use to obtain the maximum possible output with the help of an expenditure of Rs 2,000, and also find what would be the maximum possible output. Solution: The production function of the firm is q = f(x, y) = xy ……….(1) and the cost constraint is C° = rX x + rY y
or, 2,000 = 20x + 10y ………..(2) Here the relevant Lagrange function for obtaining the required input combination is V = xy+ λ(2,000-20x-10y) ………….(3) where λ is the Lagrange multiplier. Now, the first-order conditions for constrained output maximisation are
Putting this value of y in (6), we obtain 40 x = 2,000 => x = 50 units and putting the value of x in (7), we obtain y = 2 x 50= 100 units. Therefore, the FOCs give us that the firm would have to use the input combination (x = 50 units, y = 100 units) in order to produce the maximum possible output, and putting the values of x and y in the production function (1), we obtain the maximum possible amount of output to be q = 50 x 100 = 5,000 (units) we may also verify the second-order condition given by fxx f2y _ 2fxy fx fy + fyy f2x < 0 …………..(8) by putting the values fx = y, fy = x, fxx = 0, fyy = 0 and fxy = fyx = 1 in (8), we obtain: the left-hand side of (8) = – 2xy < 0 (v x = 50 > 0 and y =100 > 0). Therefore, the SOC is also verified.
3. If the marginal product of L is MPL = 100 K – L and the marginal product of K is MPK = 100 L – K, then find the maximum possible output when the total amount that can be spent on L and K is $ 1,000, and the price of L (P L) is $2 and that of K (PK) is $5 with comments, if any. Solution: Let us suppose, the production function is
Therefore, the SOC is satisfied, and so, the output-maximising quantities of the inputs are L = 247.5 (units) and K = 101.0 (units).
Now, the maximum output may be obtained here as total product of labour (TP L) at L = 247.5, K remaining fixed at K = 101.0, or, it may be obtained as total product of capital (TP K) at K = 101.0, L remaining fixed at L = 247.5. In the first case, the MPL (= fL) function would have to be integrated over the interval 0 and 247.5 units of L, and, in the second case, the MP K (= fK) function would have to be integrated over the interval 0 and 101.0 units of K. Let us proceed along the first course and obtain the amount of maximum output in the following way.
Comments: In order to obtain the maximum output we might as well integrate the MP K function over the interval 0 and 101. Although in both cases (of integration of the MP L function and that of the MP« function), the (maximum) output quantity would be obtained at L = 247.5 and K = 101, the results would be different (may be by a relatively small extent) because in the first case labour is treated as a variable input and capital as a fixed input and in the second case capital is treated as a variable input and labour as a fixed input. Returns to Scale: 1. Given the production function Q=f(x1,x2)=x1^2.x2^3 where x1,x2 are two inputs and Q is production.Does this show constant, decreasing or increasing returns to scale? Solution: Q= x1^2.x2^3 Increasing both inputs x1 and x2 by m, we get Q’= (mx1)^2.(mx2)^3= m^5.x1^2.x2^3= m^5Q Thus increase in both inputs by m causes output to increase by m^5.Hence, increasing returns to scale prevail in this case. In Cobb-Douglas equation, if sum of exponents is greater than one, increasing returns to scale. 2. Consider production function Q=5L^0.5K^0.3.Does it represent increasing,decreasing or constant returns to scale? Solution: Increasing both labour and capital inputs by m,we have:
Q’=5(mL)^0.5.(mK)^0.3 Q’=m^0.8Q That is, by increasing labour and capital by m, output increases by m^0.8, that is less than m. Thus, in this case decreasing returns to scale occurs. 3. Suppose a commodity is produced with two inputs, labour and capital, and the production function is given by Q=10*sqrt(L*K) Where Q is the output and L and K are amounts of labour and capital. What type of returns to scale does it exhibit? Solution: The above production function can be rewritten as Q= 10L^0.5*K^0.5 Hence, this a Cobb-Douglas production function. To show the nature of return to scale, let us multiply L and K by p Q’= 10(pL)^0.5(pK)^0.5=pQ Increasing L and K by p results in increase in output(Q) by p. Hence, this shows constant returns to scale.