Question

Exercise 4. Cobb-Douglas and Increasing Returns to Scale (Mining)

Exercise 4. Cobb-Douglas and increasing Returns to Scale (Mining) Consider mining output, y (measured in million tons of ore), as ultimately the function of two essential inputs, extraction, Xı, and exploration, X2. Both inputs are indispensable, as there would be nothing to excavate without a geologic survey to know where to dig, and conversely finding ore without then quarrying it also produces no mining output. Both inputs also present diminishing marginal returns, therefore assume the production function is y = x;"x2. a) What is the marginal product for extraction, dy/dxı? Holding the number of discovered quarries fixed (X2), can you give a realistic justification of why increased extraction (xi) presents diminishing marginal returns? b) What is the marginal product for excavation, dy/dx2? Holding the quarrying capability fixed (X,), can you give a realistic explanation of why increased exploration (xa) presents diminishing marginal returns? c) Assume that geologic exploration has found eight quarry sites (x2=8), and the mining company has equipment and workforce consisting of X1-4. How much ore, y, can this mining company produce? Suppose the mining company expands, doubling quarry sites (Xa'=16) and extraction equipment and workforce (Xi'=8). Does production double as well, more than double or less than double? d) Suppose market ore price is p=3, côst per unit of extraction input is w=2, and the amortization cost of exploration for each mine is r=1. How much profit is this mining company doing, given its current input levels of Xi=4 and X2=8? e) Is this mining company maximizing profits? Suppose in the short run this firm can only change xı, and x2=8 cannot be changed. What is the optimal xı? What is the maximum short-run profits that can be made? 1) How about the long run? What is the maximum profit when the firm can scale up/down in both xa and xa? Careful not to find a inimum profit instead of a maximum.

Answer 1

y = x₁^1/2 x₂^2/3

1. Marginal return to extraction = dy/dx₁ = (1/2)x₁^-1/2 x₂^{2/3 ­}

Now Increasing extraction with fixed quarrying, x₂ would give diminishing marginal returns.

Because d²y/dx₁² = (-1/4)x₁^-3/2 x₂^2/3 < which is less than 0

Realistically the marginal return gets diminished because the as one extracts up a given mine the ore amout left in it is lesser and lesser and hence the effort as well as cost of mining any additional ore increses hence the utility or marginal return generated keeps on depleting .

2. Marginal product for exploration =dy/dx₂ = (2/3)x₁^1/2 x₂^-1/3 ;

d²y/dx₂² = -(2/9) x₁^1/2 x₂^-4/3 which is always < 0 given any fixd value of x₁

Thus the realistic explanation that any increased exploration exhausts more and more options of where can one find the presence of ore. So even though exploration is successful or not in finding new places for quarrying, any increased level of exploration than earlier shall reveal more and more about where the ore can be found and not and hence , it becomes rarer to carry out successful explorations.

3. Given the first set of alues :- y* = 4 ^½ * 8^2/3

                                                                 = 2* 4 = 8

The second set of values yield :- y ‘’ = 8^1/2 * 16^2/3

                                                                       = 2*4*2^1/2*2^2/3 =8*2^7/6 = 16 *2^1/6 > 16 Hence output increases more than double since production function has is increasing return to scale typr.

4. P= 3 ; w= 2, r=1

At input levels x₁ =4; x₂ = 8, output : y= 8

Therefore Revenue = 3* 8 = 24

Cost = w*x₁ + r*x₂ = 2*4+ 1*8 = 16

Hence Profit = revenue – cost = 24 – 16 = 8