(a) To determine the returns to scale, we compare f(λk, λl) to λf(k, L) with λ > 1.

Transcription

1 Problem Set : Solutions ECON 30: Intermediate Microeconomics Prof. Marek Weretka Problem (Cobb-Douglas) (a) To determine the returns to scale, we compare f(λk, λl) to λf(k, L) with λ >. For f(k, L) = K L : f(λk, λl) = (λk) (λl) = λ 4 K L = λ 4 f(k, L) > λf(k, L) = IRS For f(k, L) = K 3 L 3 : For f(k, L) = K 4 L 4 : f(λk, λl) = (λk) 3 (λl) 3 = λ K 3 L 3 = λf(k, L) = CRS f(λk, λl) = (λk) 4 (λl) 4 = λ K 4 L 4 = λ f(k, L) < λf(k, L) = DRS (b) For w L = w K =, we find the cost functions associated with the three Cobb-Douglas production functions using the first secret of happiness, cost minimization, which requires that T RS = w K w L (). For f(k, L) = K L, we have T RS = MP K MP L = L. So from equation () we have K L K = = K = L which gives us the cost minimizing proportion of K and L. We then plug K = L into the production (output) function to get both K and L in terms of output : = f(k, L) = K L = K K = K 4 = K = 4 and L = 4 (since K = L). Now we plug K = 4 and L = 4 into the cost function, getting c() = w K K + w L L = K + L = = 4. For f(k, L) = K 3 L 3, we have T RS = MP K MP L we have L K = = L = K = 3 L = L. So from equation () 3 K K

2 which gives us the cost minimizing proportion of K and L. We then plug L = K into the production (output) function to get both K and L in terms of output : = f(k, L) = K 3 L 3 = K 3 (K) 3 = 3 K = K = 3 and L = 3. Now we plug K = 3 and L = 3 into the cost function, getting c() = w K K + w L L = K + L = = ( ).9. For f(k, L) = K 4 L 4, we have T RS = MP K MP L we have = 4 L = L. So from equation () 4 K K L = = L = K K which gives us the cost minimizing proportion of K and L. We then plug L = K into the production (output) function to get both K and L in terms of output : = f(k, L) = K 4 L 4 = K 4 (K) 4 = K = K = and L =. Now we plug K = and L = into the cost function, getting c() = w K K + w L L = K + L = + =. (c) The cost functions we found in part (b) are shown below. Notice how the shape of the cost function is related to the returns to scale for each production function found above in part (a): c() = 4 c() =.9 c() = f(k, L) =K L f(k, L) =K 3 L 3 f(k, L) =K 4 L 4 IRS CRS DRS (d) The average and marginal cost functions are shown below. Remember that () = c() and () = c ().

3 () = 3 4 () =.9 () = () = 3 () =.9 () =4 4 = f(k, L) =K L f(k, L) =K 3 L 3 f(k, L) =K 4 L 4 IRS CRS DRS Problem (Perfect Complements) (a) Again, to determine the returns to scale, we compare f(λk, λl) to λf(k, L) with λ >. For f(k, L) = min{k, L}: f(λk, λl) = min{λk, λl} = λ min{k, L} = λf(k, L) = CRS For f(k, L) = (min{k, L}) : f(λk, λl) = (min{λk, λl}) = λ (min{k, L}) = λ f(k, L) > λf(k, L) = IRS For f(k, L) = min{k, L}: f(λk, λl) = (min{λk, λl}) = λ (min{k, L}) = λ f(k, L) < λf(k, L) = DRS (b) Our first step in finding the cost functions is to determine the cost-minimizing combination of K and L. For the production functions here, K and L are perfect complements and the cost-minimizing combination is such that K = L. (These production functions are associated with the L-shaped isoquants, just as when two goods were perfect complements in utilit theor, we saw L-shaped indifference curves. We had determined that optimal consumption was along the vertices of the indifference curves; the same thing is going on here, where optimal input combinations are along the vertices of the isoquants.) We can proceed now b substituting K = L into the three production functions to get K and L in terms of. For f(k, L) = min{k, L}: = f(k, L) = min{k, L} = min{k, K} = K = K = and L =. Plugging this into the cost function for K and L and w K = w L = we have c() = w K K + w L L = K + L = + =. 3

4 For f(k, L) = (min{k, L}) : = (K, L) = (min{k, L}) = (min{k, K}) = K = K = and L =. Plugging this into the cost function for K and L and w K = w L = we have For f(k, L) = min{k, L}: c() = w K K + w L L = K + L = + =. = f(k, L) = min{k, L} = min{k, K} = K = K = and L =. Plugging this into the cost function for K and L and w K = w L = we have c() = w K K + w L L = K + L = + =. (c) The cost functions we found in part (b) are shown below. Again, notice how the shape of the cost function is related to the returns to scale for each production function found above in part (a): c() = c() = c() = f(k, L) =min{k, L} f(k, L) =(min{k, L}) f(k, L) = p min{k, L} CRS IRS DRS (d) The average and marginal cost functions are shown below: () = () = () = () = () = () =4 = f(k, L) =min{k, L} f(k, L) =(min{k, L}) f(k, L) = p min{k, L} CRS IRS DRS 4

5 Problem 3 (Perfect Substitutes) (a) To determine the returns to scale, we compare f(λk, λl) to λf(k, L) with λ >. For f(k, L) = K + 0.5L: f(λk, λl) = λk + 0.5(λL) = λ(k + 0.5L) = λf(k, L) = CRS For f(k, L) = (K + 0.5L) : f(λk, λl) = (λk + 0.5(λL)) = (λ(k + 0.5L)) = λ f(k, L) > λf(k, L) = IRS For f(k, L) = K + 0.5L: f(λk, λl) = (λk+0.5(λl)) = (λ(k+0.5l)) = λ f(k, L) < λf(k, L) = DRS (b) For all of these production functions, MP K > MP L (capital is relativel more productive at all levels), and since the costs are equal (w K = w L = ), onl capital should be used as an input. Hence the cost minimizing choice of inputs requires L = 0. (As with Problem part (b), this is analogous to the argument in utilit theor when two goods were perfect substitutes.) For f(k, L) = K + 0.5L: = f(k, L) = K + 0.5L = K + 0.5(0) = K = K = Plugging this into the cost function for K and L and w K = w L = we have For f(k, L) = (K + 0.5L) : c() = w K K + w L L = K + 0 =. = f(k, L) = (K + 0.5L) = (K + 0.5(0)) = K = K = Plugging this into the cost function for K and L and w K = w L = we have For f(k, L) = K + 0.5L: c() = w K K + w L L = K + 0 =. = f(k, L) = (K + 0.5L) = (K + 0.5(0)) = K = K = Plugging this into the cost function for K and L and w K = w L = we have c() = w K K + w L L = K + 0 =. 5

6 (c) The cost functions we found in part (b) are shown below. Again, notice how the shape of the cost function is related to the returns to scale for each production function found above in part (a): c() = c() = c() = f(k, L) =K +0.5L f(k, L) =(K +0.5L) f(k, L) = p K +0.5L CRS IRS DRS (d) The average and marginal cost functions are shown below: () = () = () = () = () = () = = f(k, L) =K +0.5L f(k, L) =(K +0.5L) f(k, L) = p K +0.5L CRS IRS DRS Problem 4 (Cost Curves) (a) The production function associated with cost curve c() = 4 must exhibit decreasing returns to scale. With this cost curve, doubling output more than doubles the cost; this is a result of the fact that, with decreasing returns to scale, doubling output would require more than doubling inputs, and since input costs are linear, the cost will more than double. (b) The total cost function is T C() = and so T C() =, T C() = 0, and T C(4) = 6 as shown below along with the fixed and variable cost curves: 6

7 6 TC() = (c) The average fixed cost is AF C() = F C = 4, so AF C() = 4, AF C() =, and AF C(4) =. As becomes infinitel large, AF C goes to zero (the fixed cost of 4 is being spread over a ver large number of units); when is ver close to zero, AF C is infinitel large (lim = ). Average fixed cost is shown below: AF C (d) The average variable cost is AV C() = V C() and AV C(4) = 6: = 4 = 4, so AV C() = 4, AV C() =, AV C

8 (e) Average total cost is AT C() = AV C()+AF C() = 4+ 4, so AT C() =, AT C() = 0, and AT C(4) = 7, as shown below: AT C AV C AF C For smaller levels of output, the average variable cost is negligible and hence AT C is dominated b the high AF C. When production is large, AT C becomes large due to the high AV C for large. (f) The minimum efficient scale occurs at the output for which AT C obtains its minimum (i.e., the such that AT C () = 0). Here AT C () = 0 = = 0 = = ( = also solves this equation, but we cannot have negative production!). So MES = and AT C MES = AT C( MES ) =. (g) We know that () = T C () so () =. This is shown below along with AT C. Notice that the intersect at MES =. AT C AT C MES MES MES (h) Wh, intuitivel, the curve intersects AT C at its minimum, MES : For output < MES, () is below AT C(), so producing an additional unit is cheaper than the average cost per unit, which brings that average down. This means AT C() is decreasing to the left of MES. For output > MES, () is above AT C(), so producing an additional

9 unit is more costl than the average cost per unit, which brings that average up. This means AT C() is increasing to the right of MES. Since AT C() is decreasing to the left of MES and increasing to the right of MES, the minimum of AT C(), which is AT C MES, must be at MES. (i) If the fixed cost is F, AT C() = V C() + F C = 4 + F. Then MES solves AT C () = 0 so MES = F and AT C( MES ) = 4 F. For higher fixed costs, the minimum efficient scale increases. Problem 5 (Suppl Curve of G) (a) With T C() = 4 + 4, condition p = () (the first-order condition for profit maximization) becomes p = = = p. For p = 4, = 4 =, giving profit π = p T C() = 4 (4 resulting in profit π = 0. For p =, = =, resulting in profit so = is produced with π = 0. For p = 6, = 6 =, giving profit ( ) + 4) < 0 = should produce = 0 instead, π = p T C() = ( 4 () + 4 ) = 0, π = p T C() = 6 ( 4 () + 4 ) =. (b) The suppl function (p) will correspond with the marginal cost curve for all prices that give positive profits, and will be (p) = 0 for all prices ielding strictl negative profits. Positive profits are associated with all p above AT C( MES ) = AT C() =. Or, alternativel prices such that profits are positive: p (4 + 4) 0 = p ( p) (4 ( p) + 4) 0 (using p = condition) = p. Negative profits are associated with p < and hence the optimal choice is = 0 for those prices. Our suppl function is then: { 0 for p < (p) = p for p 9

10 (c) The suppl curve and AT C are shown below: (p) AT C (p) (d) For a new fixed cost F =, the slope of the suppl curve will not change, however the price at which the firm will choose not to produce ( = 0) does change. In part (b), we found that the firm will onl produce when p > AT C MES =. Now, we need to find the MES with the new F = (at which point = AT C): () = AT C() = = + 4 = =, then MES = and so AT CMES = AT C( MES ) = ( ) / + 4 = 4. This gives us the following piecewise defined suppl function: { 0 for p < 4 (p) = p for p 4 0