The Monthly Demand X Of A Compact Disc Is Related To The Price P Do
The monthly demand, x, of a compact disc is related to the price, p dollars per disc by the following equation: p(x) = -0.0004x + 6. The total monthly cost for pressing and packaging x copies is given by C(x) = 600 + 2x - 0.0002x^2. The assignment questions are:
A) What is the production level that minimizes the total cost?
B) What is the revenue function?
C) What is the profit function?
D) What is the production level that maximizes the profit? Show your work where possible.
Paper For Above instruction
Understanding the relationship between demand, cost, revenue, and profit is fundamental in optimizing production strategies. In this analysis, we explore the cost minimization and profit maximization of a compact disc production process based on the provided demand and cost functions.
Part A: Determining the Production Level That Minimizes Total Cost
The total cost function is given as C(x) = 600 + 2x - 0.0002x^2. To find the production level that minimizes cost, we need to identify the critical point of C(x), which involves taking its derivative with respect to x and setting it to zero:
C'(x) = d/dx [600 + 2x - 0.0002x^2] = 2 - 0.0004x
Setting C'(x) = 0 for the minimum:
2 - 0.0004x = 0
Solving for x gives:
x = \frac{2}{0.0004} = 5000
To verify that this critical point corresponds to a minimum, we check the second derivative:
C''(x) = -0.0004
Since C''(x) is negative, the function is concave down at that point, indicating a maximum. However, because of the nature of costs, this suggests the critical point is at the boundary of the feasible production
levels. Alternatively, considering the cost function's shape implies there's a minimum at some production level; the negative second derivative suggests the critical point is a maximum. Given the cost function is quadratic with a negative coefficient for x^2, it opens downward, so the parabola's vertex will give the maximum cost, not the minimum. But in this case, since the quadratic term is negative, the cost function opens downward, and the minimum cost occurs at the vertex's maximum point. Therefore, the minimum total cost occurs at the endpoints or by analyzing the domain. Typically, production cannot be negative, and the feasible domain is x ≥ 0. The vertex calculation suggests the maximum is at x=5000; hence, there might be a typo or a need to analyze endpoints. Assuming typical production constraints, the minimal cost will be at either x=0 or x=5000, with the specific choice depending on practical constraints, but based on this calculus, the cost is minimized at x=0 or x=5000. Usually, cost minimization occurs at the vertex point of the parabola if it opens upward. Since the quadratic coefficient is negative, the parabola opens downward, and the minimum occurs at the boundary, likely at x=0.
Part B: Revenue Function
The demand function relates the quantity demanded to the price: p(x) = -0.0004x + 6. The revenue function R(x) is calculated as:
R(x) = p(x) * x = (-0.0004x + 6) * x
Expanding:
R(x) = -0.0004x^2 + 6x
Part C: Profit Function
The profit function Π(x) is defined as revenue minus cost:
Π(x) = R(x) - C(x) = (-0.0004x^2 + 6x) - (600 + 2x - 0.0002x^2)
Simplify the expression:
Π(x) = -0.0004x^2 + 6x - 600 - 2x + 0.0002x^2
Combine like terms:
Π(x) = (-0.0004x^2 + 0.0002x^2) + (6x - 2x) - 600 = -0.0002x^2 + 4x - 600
Part D: Production Level That Maximizes Profit
To maximize profit, differentiate Π(x) and set the derivative to zero:
Π'(x) = d/dx [-0.0002x^2 + 4x - 600] = -0.0004x + 4
Set Π'(x) = 0:
-0.0004x + 4 = 0
Solving for x:
x = \frac{4}{0.0004} = 10,000
Since the second derivative is:
Π''(x) = -0.0004
which is negative, we confirm this critical point corresponds to a maximum profit. Therefore, the production level that maximizes profit is 10,000 units.
Conclusion
In this analysis, the minimal cost production level is at the boundary, specifically at the lowest feasible x (often x=0), owing to the quadratic nature of the cost function. The revenue function exhibits quadratic characteristics based on demand price, while the profit function's maximum occurs at 10,000 units, demonstrating the optimal production quantity for maximum profitability. Managers should focus on producing around 10,000 units to maximize profit, considering demand and cost constraints.
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