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Let's break down the problem based on the cost function \( C(x) = 62500 + 500x + x^2 \), where \( x \) represents the production level.

### a) The cost at the production level 1950:
To find the cost at \( x = 1950 \), substitute \( x = 1950 \) into the cost function \( C(x) \):
\[
C(1950) = 62500 + 500(1950) + (1950)^2
\]
Let's calculate that:
\[
C(1950) = 62500 + 500(1950) + 1950^2
= 62500 + 975000 + 3802500
= 4840000
\]
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.

For the given cost function C(x) = 62500 + 500x + x^2, find the following:
a) The cost at the production level 1950
b) The average cost at the production level 1950
c) The marginal cost at the production level 1950
d) The production level that will minimize the average cost
e) The minimal average cost

This problem focuses on solving a cost function C(x) = 62500 + 500x + x^2. Key tasks include finding the cost, average cost, marginal cost, and optimizing production levels to minimize the average cost. Ideal for undergraduate economics or business calculus students.

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