Question

1. Suppose that the production function for lava lamps is given by Q = KL -ľ, where is the number of lamps produced per year, K is the machine-hours of capital, and L is the man-hours of labor. Suppose K = 600. a. Draw a graph of the production function over the range L = 0 to L = 500, putting L on the horizontal axis and on the vertical axis. Over what range of L does the production function exhibit increasing marginal returns? Diminishing marginal returns? Diminishing total returns? b. Derive the equation for average product of labor and graph the average product of labor curve. At what level of labor does the average product curve reach its maximum? c. Derive the equation for marginal product of labor. On the same graph you drew for part b, sketch the graph of the marginal product of labor curve. At what level of labor does the marginal product curve appear to reach its maximum? At what level does the marginal product equal zero? d. Relate your answer to part c to your answer to part a.

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Answer 1

1. (a) The production function would be $\dpi{80} Q = 600 L^2 - L^3$ . The graph is as below.

3x107 2x107 11x10 200 400 6po

The marginal return would be $\dpi{80} \frac{\mathrm{d} Q}{\mathrm{d} L} = \frac{\mathrm{d} }{\mathrm{d} L}(600 L^2 - L^3)$ or $\dpi{80} \frac{\mathrm{d} Q}{\mathrm{d} L} = 1200 L - 3L^2$ . The marginal return itself is increasing where $\dpi{80} \frac{\mathrm{d} }{\mathrm{d} L}\left (\frac{\mathrm{d} Q}{\mathrm{d} L} \right ) > 0$ or $\dpi{80} \frac{\mathrm{d} }{\mathrm{d} L} (1200 L - 3L^2)> 0$ or $\dpi{80} 1200 - 6L > 0$ or $\dpi{80} 6L < 1200$ or $\dpi{80} L < 200$ . This is before point A. The marginal return is decreasing where $\dpi{80} \frac{\mathrm{d} }{\mathrm{d} L}\left (\frac{\mathrm{d} Q}{\mathrm{d} L} \right ) < 0$ or $\dpi{80} \frac{\mathrm{d} }{\mathrm{d} L} (1200 L - 3L^2)< 0$ or $\dpi{80} 1200 - 6L < 0$ or $\dpi{80} 6L > 1200$ or $\dpi{80} L > 200$ . This is between point A and B.

The total return is diminishing where $\dpi{80} \frac{\mathrm{d} Q}{\mathrm{d} L} = < 0$ or $\dpi{80} 1200 L - 3L^2 < 0$ or $\dpi{80} 3L^2 > 1200 L$ or $\dpi{80} L > 400$ . This is after point B.

(b) The APL would be $\dpi{80} AP_L = \frac{Q}{L}$ or $\dpi{80} AP_L = \frac{600 L^2 - L^3}{L}$ or $\dpi{80} AP_L = 600 L - L^2$ . The graph is as below.

100000 APL 180000- 60000- 140000 2000o- 200 400 6 po

The APL reaches maximum where $\dpi{80} \frac{\mathrm{d} }{\mathrm{d} L}(AP_L) = 0$ or $\dpi{80} \frac{\mathrm{d} }{\mathrm{d} L}(600L - L^2) = 0$ or $\dpi{80} 600 - 2L = 0$ or $\dpi{80} 2L = 600$ or $\dpi{80} L = 300$ .

(c) The The marginal product of labor is $\dpi{80} MP_L = \frac{\mathrm{d} }{\mathrm{d} L}(600 L^2 - L^3)$ or $\dpi{80} MP_L = 1200 L - 3L^2$ . The graph is as below.

150000 MPL,APE 100000 MPL APL 500bo- 200 40B 600

The MPL reaches maximum where $\dpi{80} \frac{\mathrm{d} }{\mathrm{d} L}(MP_L) = 0$ or $\dpi{80} \frac{\mathrm{d} }{\mathrm{d} L}(1200 L - 3L^2) = 0$ or $\dpi{80} 1200 - 6L = 0$ or $\dpi{80} 6L = 1200$ or $\dpi{80} L = 200$ . The MPL reaches zero where $\dpi{80} MP_L = 0$ or $\dpi{80} 1200 L - 3L^2 = 0$ or $\dpi{80} 3L^2 = 1200 L$ or $\dpi{80} L = 400$ .

(d) Note that the labels in the graph in part c are related to that of labels in the graph in part a. The marginal return is increasing where MPL has positive slope, while the marginal return is diminishing where MPL has negative slope. As can be suspected, the MPL reaches maximum between these points, labelled as A in part a graph and as A' in part c graph. Also, the total return is diminishing where MPL is negative, which is labelled as B in part a graph and B' in part b graph. After B', the MPL is negatitve, and the total return (Q) decreases.