Answer:
The answer is: 1,800 electric drills per month
Explanation:
C(x) = 60,000 + 70x
p = 190 - (x / 30)
0 ≤ x ≤ 5000
To determine the production level that maximized profit, we need to solve the following equation:
profit = px - c
profit = [190 - (x / 30)]x - (60,000 + 70x)
profit = 190x - x²/30 - 60,000 - 70x
profit = 120x - x²/30 - 60,000
profit' = 120 - x/15
0 = 120 - x/15
x/15 = 120
x = 1,800
