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6. Lennys Car Wash uses two variable inputs: workers and materials (cloth, soap, etc). Lennys production function, in units
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6. (a) For the production level Q= 3, we have the input combinations as (W , M) = (1,3), (2,3), (3,3), (4,3), (5,3), (6,3), (7,3) . The most efficient combination would be what minimizes the W and M. As can be seen, combination (1,3) requires least amount of inputs to produce Q=3, since (1,3) \preceq (W,M) . Hence, the efficient way to produce Q=3 would be (W = 1 , M = 3) .

For the production level Q=5, we have the input combinations as (W , M) = (2,5), (3,5), (4,5), (5,5), (6,5), (7,5), (2,6), (3,6), (2,7), (3,7) . The efficience combination in this case would be (2,5) since (2,5) \preceq (W,M) . Hence, the efficient way to produce Q=5 would be (W = 2 , M = 5) .

(b) The table would be as below.

Q W M FC TC MC ATC AVC
0 0 0 50 50 - - -
1 1*20=20 1*10=10 50 80 30 80 30
2 1*20=20 2*10=20 50 90 10 45 20
3 1*20=20 3*10=30 50 100 10 33.33 16.67
4 1*20=20 4*10=40 50 110 10 27.5 15
5 2*20=40 5*10=50 50 140 30 28 18
6 4*20=80 6*10=60 50 190 50 31.66 23.33
7 7*20=140 7*10=70 50 260 70 37.14 30

We have MC = \frac{\Delta TC}{\Delta Q} , ATC = \frac{TC}{Q} and AVC = \frac{VC}{Q} = \frac{(20*W) + (10*M)}{Q} . The W and M here are the combination of minimum input required to produce the given output.

(c) The plot would be as below.

C18 Τ Α 1_(Cars) Ι MC - Β D | E | F | G | H Ι Ι Ι Κ Ι ι Μ Ν Ι ο Ι Ρ Ι ! 30 30 WNP = C ATCAVC 80 45 10 33.33 10 27.5 30 28 50

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