High School

A company manufactures and sells [tex]x[/tex] electric drills per month. The monthly cost and price-demand equations are given by:

\[ C(x) = 60000 + 70x \]
\[ p = 190 - \frac{x}{30} \]
\[ 0 \leq x \leq 5000 \]

(A) Find the production level that results in the maximum profit.

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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