Let Q(L) = 6L2 – (L3)/6 with costs = 300 + 100L. a) Complete the table on the next page. You can do this in excel or by

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Let Q(L) = 6L2 – (L3)/6 with 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 APL is maximized where MPL = APL. (You can plot APL
and MPL versus L. Or
you can use the table directly. Find the maximum APL. For labour
below this we should
see MPL > APL. For labour above we should see MPL <
APL.)
d) Confirm that AC is minimized where MC = AC
e) Confirm that AVC is a minimum where APL is a maximum
f) Confirm that MC is a minimum where MPL is a maximum

Let Q L 6l2 L3 6 With Costs 300 100l A Complete The Table On The Next Page You Can Do This In Excel Or By 1 (76.16 KiB) Viewed 32 times

TABLE: Column 1: Labour input from 1-24 Column 2: Output resulting from L: Column 3: Average Product of Labour Column 4: Marginal Product of Labour: Column 5: Cost: Column 6: Average Cost: Column 7: Average Cost: Column 8: Marginal Cost: 2 3 Q AP₁ 0 1 L 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 4 MP Q(L) = AP₁ = MP = COST = = AC AVC = MC = 5 COST 300 6L²-(L³)/6 Q/L AQ/AL 300 + 100L COST/Q (COST-300)/Q AC/AQ 6 7 AC AVC 8 MC