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ASSIGNMENT #4 PRODUCTION AND COSTS 1: _Let Q(L) = 6L² – (L³)/6 with costs = 300 + 100L. a) Complete the table on the nex
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- Assignment 4 Production And Costs 1 Let Q L 6l L 6 With Costs 300 100l A Complete The Table On The Nex 1 (107.76 KiB) Viewed 47 times

- Assignment 4 Production And Costs 1 Let Q L 6l L 6 With Costs 300 100l A Complete The Table On The Nex 2 (99.65 KiB) Viewed 47 times

- Assignment 4 Production And Costs 1 Let Q L 6l L 6 With Costs 300 100l A Complete The Table On The Nex 3 (72.55 KiB) Viewed 47 times
ASSIGNMENT #4 PRODUCTION AND COSTS 1: _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 AP₁. For labour below this we should see MPL > APL. For labour above we should see MPL < APL.) Confirm that AC is minimized where MC = AC d) e) Confirm that AVC is a minimum where AP is a maximum f) Confirm that MC is a minimum where MPL is a maximum
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 4 MPL Q(L) APL MP COST AC AVC MC 5 COST 300 |||||||||||||| = 6L² - (L³)/6 Q/L AQ/AL 300 + 100L COST/Q (COST-300)/Q ΔΕ/ΔΩ 6 7 AC AVC 8 MC
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