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 unction for the assembly of carts using the following time-series data for the days of assembly operations. Lis the daily number of assembly workers last 22 t ci pays its assembly workers $140 per day in wages and benefits. Using E-views (a employed, and is the number of carts assembled (completely) for that day Me computer regression package) to estimate short-run cubic production function, we obtained the reaults in the table below -AL +BL Dependent Variable: Q Method: Least Squares Date: 12/02/16 Time: 16:05 Sample 1 22 Included obaervations: 22 Coefficient Std. Error t-Statistic Prob. 0.036843 0.013958 2.639548 0.0157 2.532560 0.532178 4.758856 0.0001 R-squared 0.612286 Mean dependent var 1090.227 Adjusted R squared 0.592900 S.D. dependent var696.8685 S. E. of rogression Sum squared resid Log likelihood Durbin-Watson stat 444.6326 Akaike info criterion 15.11888 16.21807 164.3077 Hannan-Quinn criter. 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 ii. 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 average product is maximized? c. What is short run marginal cost when average product is maximized? d. Beyond what level of labor employment does the law of diminishing 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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