Answer :
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
Final answer:
The maximum profit is achieved at a certain level of production, obtained by deriving the profit function, equating it to zero and solving for the quantity produced.
Explanation:
The subject matter here revolves around maximizing profit, a basic principle in business mathematics. The profit P(x) can be obtained from the difference of revenue R(x) and cost C(x). Here, Revenue R(x), can be calculated with price p times the quantity x, or R(x) = p*x. Since we have equations for C(x) and p, we can substitute them into the profit equation P(x)=R(x)-C(x). This gives us the profit function P(x)=(190 - x/30)x - (60000 + 70x). Getting the derivative P'(x), we can equate it to zero and solve for x to find the profit maximizing level. This level would be the production level when their profit is at its maximum.
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