A. 
Profit=Revenue−Cost 
P(Q)=R(Q)−TC(Q) 
R(Q)=p(Q)Q=(10−0.001Q)Q 
=10Q−0.001Q2  
P(Q)=10Q−0.001Q2−(5000+2Q) 
=8Q−0.001Q2−5000 
P(Q)=8Q−0.001Q2−5000 B.
Q≥0,p(Q)≥0 
10Q−0.001Q2≥0 
0≤Q≤10000 
P′(Q)=(8Q−0.001Q2−5000)′ 
=8−0.002Q Find critical number(s)
P′(Q)=0=>8−0.002Q=0=>Q=4000If 0≤Q≤4000,P′(Q)>0,P(Q) increases.
If 4000≤Q≤10000,P′(Q)<0,P(Q) decreases.
The function P(Q) has a local maximum at Q=4000. 
Since the function P(Q) has the only extremum on [0,10000], then the function P(Q) has the absolute maximum at Q=4000. 
P(4000)=8(4000)−0.001(4000)2−5000 
P(4000)=$11000 
p(4000)=10−0.001(4000) 
p(4000)=$6
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