Soln,
p=26−0.5Q
MC=6+6Q
TFC=4
a. MC=MR
P=AR
P=26−0.5Q
TR=P.Q(26−0.5Q)Q
TR=26Q−0.5Q2
MR=δQδTR=26−Q
MR=26−Q
Max at MC=MR
=6+6Q=26−Q=77Q=720
Q=2.86units.
b.TR greater than TC
TR=26−0.5Q2 BUT Q=2.86
26(2.86)−0.5(2.86)2
74.36−4.0898=78.4498
TR=78.4498
TC=∫(MC)dQ BUT MC=6+6Q
TC=4+6Q+6Q2=4+17.16+49.07=70.23
TC=70.23
TR−TC
=78.4498−70.23=8.2122
c. At break even point TC=TR
TC=4+6Q+6Q2
TR=26Q−0.5Q2
THUS, 26Q−0.5Q2=4+6Q+6Q2
=6.5Q2−20Q+4=0
=42.25-40Q+8=0
40Q=50.25
=1.26
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