A) P = 100 − 8 Q P=100-8Q P = 100 − 8 Q
At maximum, T R = P Q TR=PQ TR = PQ
T R = ( 100 − 8 Q ) Q = 100 Q − 8 Q 2 TR=(100-8Q)Q=100Q-8Q^2 TR = ( 100 − 8 Q ) Q = 100 Q − 8 Q 2
T C = 50 + 80 Q − 10 Q 2 + 0.6 Q 3 TC=50+80Q-10Q^2+0.6Q^3 TC = 50 + 80 Q − 10 Q 2 + 0.6 Q 3
Now M R = M C MR=MC MR = MC
Δ T R / Δ Q = Δ T C / Δ Q \Delta TR/\Delta Q=\Delta TC/\Delta Q Δ TR /Δ Q = Δ TC /Δ Q
100 − 16 Q = 80 − 20 Q + 1.8 Q 2 100-16Q=80-20Q+1.8Q^2 100 − 16 Q = 80 − 20 Q + 1.8 Q 2
80 − 20 Q + 1. 8 2 − 100 + 16 Q = 0 80-20Q+1.8^2-100+16Q=0 80 − 20 Q + 1. 8 2 − 100 + 16 Q = 0
1.8 Q 2 − 4 Q − 20 = 0 1.8Q^2-4Q-20=0 1.8 Q 2 − 4 Q − 20 = 0
Use quadratic formula to solve
Q = − b ± b 2 − 4 a c 2 a = − ( − 4 ) ± ( − 4 ) 2 − 4 ( 1.8 ) ( − 20 ) 2 ( 1.8 ) = 4 ± 12.64 3.6 Q=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=-\frac{(-4)\pm \sqrt{(-4)^2-4(1.8)(-20)}}{2(1.8)}=\frac{4\pm 12.64}{3.6} Q = 2 a − b ± b 2 − 4 a c = − 2 ( 1.8 ) ( − 4 ) ± ( − 4 ) 2 − 4 ( 1.8 ) ( − 20 ) = 3.6 4 ± 12.64
Q = 4.62 Q=4.62 Q = 4.62 or -2.4
Since the quantity demanded is not negative, the profit-maximizing level of output=4.62 units.
Now, P = 100 − 8 Q = 100 − 8 ( 4.62 ) = $ 63.04 P=100-8Q=100-8(4.62)=\$63.04 P = 100 − 8 Q = 100 − 8 ( 4.62 ) = $63.04
B) For revenue maximizing firm M R = 0 MR=0 MR = 0
100 − 16 Q = 0 100-16Q=0 100 − 16 Q = 0
16 Q = 100 16Q=100 16 Q = 100
Q = 6.25 Q=6.25 Q = 6.25
Now, P = 100 − 8 Q = 100 − 8 ( 6.25 ) = $ 50 P=100-8Q=100-8(6.25)=\$50 P = 100 − 8 Q = 100 − 8 ( 6.25 ) = $50
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