Price/profit math problem

A manufacturer of tennis rackets finds that the total daily cost in dollars, C(q), of producing
q rackets is given by C(q) = 18,000 - 10q + .001q 2. Each racket can be sold at a price p in
dollars, given by p = 10 - .004q. If all of the rackets that are manufactured can be sold, find
the daily level of production (q) that will maximize profit.

Profit = Revenue - Cost
Since you know both the revenue function and the cost function you can find the profit function.
P(q) = R(q) - C(q)
P(q) = [10 - 0.004q] - [18,000 - 10q + 0.001q²]
P(q) = -17,990 + 9.996q - 0.001q²

To find the maximum of this function, find the derivative, set it equal to zero and solve for q.
P(q) = -17,990 + 9.996q - 0.001q²
dP/dq = 9.996 - 0.001*2q
0 = 9.996 - 0.002q
.002q = 9.996
q = 4998

So in order to maximize the profit daily level production of rackets should be q = 4,998 units

Hope this helped

Vick