6. (a) For the production level Q= 3, we have
the input combinations as
. The most efficient combination would be what minimizes the W and
M. As can be seen, combination
requires least amount of inputs to produce Q=3, since
. Hence, the efficient way to produce Q=3 would be
.
For the production level Q=5, we have the input combinations as
. The efficience combination in this case would be
since
. Hence, the efficient way to produce Q=5 would be
.
(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
,
and
. The W and M here are the combination of minimum input required to
produce the given output.
(c) The plot would be as below.
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need answer 1-3 answered please
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need problems 1-4 solved please.
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