(a)
MC = dC/dQ = 2Q
AC = C/Q = Q
P = 12 - Q
TR = PQ = 12Q - Q2
MR = dTR/dQ = 12 - 2Q
Graph:
(b)
Setting MR = MC,
12 - 2Q = 2Q
4Q = 12
Q = 3
P = 12 - 3 = 9
TR = 9 x 3 = 27
TC = 3 x 3 = 9
Profit = TR - TC = 27 - 9 = 36
In above graph, profit is maximized at point E with price P0 and output Q0.
(c)
Q = 12 - P
Elasticity (E) = (dQ/dP) x (P/Q) = - 1 x (9/3) = - 3
MR = 9 x [1 - (1/3)] = 9 x (2/3) = 6
From MR equation, MR = 12 - (2 x 3) = 12 - 6 = 6, hence proved.
(d)
When P = MC,
12 - Q = Q
2Q = 12
Q = 6
P = 6
Efficiency loss = (1/2) x Change in P x Change in Q = (1/2) x (9 - 6) x (6 - 3) = (1/2) x 3 x 3 = 4.5
In above graph, efficiency loss is area EFG.
NOTE: As per HOMEWORKLIB POLICY, only 1st 4 parts are answered.
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