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Mathematical problems of calculus
Let L= profit, profit = total revenue-total cost, and total revenue = p * q.

From the demand function Q=2000-4P, we can solve P=500-Q/4.

l = P * Q-C(Q)=(500-Q/4)* Q-(50000+ 100 Q)

= -(Q^2)/4+400 Q-50000

The above formula takes the derivative of q and makes it equal to 0.

L'= -Q/2+400 = 0, and the solution is Q=800.

The maximum profit substituted into the profit formula is L= 1 10000 or the so-called best seller, that is, the supply is less than the demand, that is, there is no product backlog.

From the demand function Q=2000-4P.

P=500-Q/4

Let the profit be Pr (from English profit), then there is

Pr = pq-c = 500q-Q2/4-50000-100q, then

DPr/dQ=400-Q/2=0, and the solution is

Q=800, that is, when Q=800, the maximum profit is obtained. When Q=800 is substituted, the maximum profit is as follows.

Prmax= 1 10000