7. Mercantile Metalworks, Inc. manufactures wire carts for grocery stores. The p

Question

7. Mercantile Metalworks, Inc. manufactures wire carts for grocery stores. The production manager at Mercantile wishes to estimate an empirical production function for the assembly of carts using the following time series data for the last 22 days of assembly operations. Lis the daily number of assembly workers employed, and is the number of carts assembled (completely) for that day Mercantile pays its assembly workers $140 per day in wages and benefits. Using E-views (a computer regression package) to estimate short-run cubic production function, we obtained the results in the table below: QAL3 BL2 Dependent Variable: Q Method: Least Squares Date: 12/02/16 Time: 16:05 Sample: 1 22 Included observations: 22 Coefficient Std. Error t-Statistic Prob. 0.036843 0.013958 2.639548 0.0157 2.532560 0.532178 4758856 0,0001 R-squared Adjusted R-squared S.E. of regression Sum squared resid Log likelihood Durbin Watson stat 0.612286 Mean dependent var 1090.227 0.592900 S.D. dependent var 696.8685 444.6326 Akaike info criterion 15.11888 15.21807 164.3077 HannanQuinn crite 15.14225 3953962. Schwarz criterion 2.368164 What are the estimated total, average, and marginal product functions from your regression results? a. i. Total Product ii. Average Product iii. Marginal Product b. At what level of labor usage does average product reach its maximum value? In a day, how many carts per worker are assembled when c. What is short-run marginal cost when average product is maximized? d. Beyond what level of labor employment does the law of diminishing average product is maximized? returns set in? Beyond what level of output?

Explanation / Answer

a) We are given Q = AL^3 + BL^2 from the regression result A = -0.0346843 and B = 2.532560

Therefore, total product = -0.0346843*L^3 + 2.532560*L^2

Average product will be = Q/L = AL^3/L + BL^2/L = AL^2 + BL thus,

-0.0346843*L^2 + 2.532560*L

Marginal production will be dQ/dL = 3AL^2 + 2BL = -0.0346843*3*L^2 + 2.532560*2*L thus,
-0.1040529L^2 + 5.06512*L

b) Average product will be = Q/L = AL^3/L + BL^2/L = AL^2 + BL

Maximizing AP = 2AL + B = 0
L = -B/2A = -2.532560/(-2*0.0346843) = 36 units of labor

c) As given in question MC is $140 per worker we found in part B that AP is maximum at 36 units of labor thus the MC will be 36&140 = $5040 at this point.

d) Marginal production will be dQ/dL = 3AL^2 + 2BL at maximum MPL
6AL + 2B = 0
L = -2B/6A = -1/3A = -1/(3*-0.0346843) = 10 units of labor therefore, diminishing returns set in after 10 units of labor

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