Math Problem Statement
Solution
Let's break down the problem based on the cost function , where represents the production level.
a) The cost at the production level 1950:
To find the cost at , substitute into the cost function : Let's calculate that:
= 62500 + 975000 + 3802500 = 4840000$$ So, the cost at production level 1950 is **4,840,000**. ### b) The average cost at the production level 1950: The average cost is given by $$AC(x) = \frac{C(x)}{x}$$. For $$x = 1950$$: $$AC(1950) = \frac{C(1950)}{1950} = \frac{4840000}{1950} \approx 2482.05$$ Thus, the average cost at production level 1950 is approximately **2482.05**. ### c) The marginal cost at the production level 1950: The marginal cost is the derivative of the cost function, $$C'(x)$$. Let's find $$C'(x)$$: $$C'(x) = \frac{d}{dx} \left( 62500 + 500x + x^2 \right) = 500 + 2x$$ Now, substitute $$x = 1950$$: $$C'(1950) = 500 + 2(1950) = 500 + 3900 = 4400$$ So, the marginal cost at production level 1950 is **4400**. ### d) The production level that will minimize the average cost: To minimize the average cost, we first find the average cost function: $$AC(x) = \frac{C(x)}{x} = \frac{62500 + 500x + x^2}{x} = \frac{62500}{x} + 500 + x$$ Now, differentiate $$AC(x)$$ with respect to $$x$$: $$AC'(x) = -\frac{62500}{x^2} + 1$$ Set $$AC'(x) = 0$$ to find the critical points: $$-\frac{62500}{x^2} + 1 = 0 \quad \Rightarrow \quad \frac{62500}{x^2} = 1 \quad \Rightarrow \quad x^2 = 62500 \quad \Rightarrow \quad x = 250$$ So, the production level that minimizes the average cost is **250**. ### e) The minimal average cost: Substitute $$x = 250$$ into the average cost function: $$AC(250) = \frac{62500}{250} + 500 + 250 = 250 + 500 + 250 = 1000$$ Thus, the minimal average cost is **1000**. --- Do you have any questions or want further details on any step? Here are five additional questions you might consider: 1. How does the marginal cost relate to the behavior of the cost function? 2. What does minimizing the average cost tell us about production efficiency? 3. Could the cost function have more than one minimum for the average cost? 4. What are the economic implications of the production level at $$x = 250$$? 5. How would changes in fixed costs affect the marginal cost? **Tip:** The marginal cost tells you how much additional cost is incurred by producing one more unit, which is useful for decisions regarding scaling production.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Cost Function
Derivatives
Marginal Cost
Optimization
Formulas
C(x) = 62500 + 500x + x^2
Average Cost: AC(x) = C(x)/x
Marginal Cost: MC(x) = dC(x)/dx = 500 + 2x
Theorems
Optimization using derivatives
Suitable Grade Level
Undergraduate (Economics, Business Calculus)
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