Let Q(L) = 6L² - (L³)/6with costs = 300 + 100L.

(a)Complete the table on the next page. You can do this in excel or by hand.

(b)Why is L = 25 or larger not something we should consider?

(c)Confirm that AP_L is maximized where MP_L = AP_L. (You can plot AP_L and MP_L versus L. Or you can use the table directly. Find the maximum AP_L. For labour below this we should see MP_L > AP_L. For labour above we should see MP_L L.)

(d)Confirm that AC is minimized where MC = AC

(e)Confirm that AVC is a minimum where AP_L is a maximum

(f)Confirm that MC is a minimum where MP_L is a maximum

TABLE: Column 1: Labour input from 1-24 Column 2: Output resulting from L: Q(L) 612 - (L3)/6 Column 3: Average Product of Labour APL Q/L Column 4: Marginal Product of Labour: MP 40/AL Column 5: Cost: COST 300 + 100L Column 6: Average Cost: AC COST/Q Column 7: Average Cost: AVC = (COST-300)/Q Column 8: Marginal Cost: MC AC/AQ 1 2 3 4 5 6 7 8 a AP MP COST AC AVC MC 0 0 ----- 300 ----- ---- 4 CO 9 10 11