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Let's solve this step by step using the given cost function:

\[
C(x) = 62500 + 500x + x^2
\]

### Part a) The cost at the production level 1950
We calculate \( C(1950) \) by substituting \( x = 1950 \) into the cost function:

\[
C(1950) = 62500 + 500(1950) + (1950)^2
\]

\[
C(1950) = 62500 + 975000 + 3802500 = 4840000
\]

Thus, the cost at the production level 1950 is **4,840,000**.

---

### Part b) The average cost at the production level 1950
The average cost \( AC(x) \) is given by:

\[
AC(x) = \frac{C(x)}{x}
\]

Substitute \( x = 1950 \) and \( C(1950) = 4840000 \):

\[
AC(1950) = \frac{4840000}{1950} \approx 2482.05
\]

So, the average cost at the production level 1950 is approximately **2482.05**.

---

### Part c) The marginal cost at the production level 1950
The marginal cost \( MC(x) \) is the derivative of the cost function \( C(x) \):

\[
C'(x) = 500 + 2x
\]

Substitute \( x = 1950 \):

\[
C'(1950) = 500 + 2(1950) = 500 + 3900 = 4400
\]

So, the marginal cost at the production level 1950 is **4400**.

---

### Part d) The production level that will minimize the average cost
To minimize the average cost \( AC(x) = \frac{C(x)}{x} \), we first find its derivative, set it to zero, and solve for \( x \).

1. \( AC(x) = \frac{62500}{x} + 500 + x \)
   
2. Take the derivative:

\[
AC'(x) = -\frac{62500}{x^2} + 1
\]

3. Set the derivative to zero to find the critical point:

\[
-\frac{62500}{x^2} + 1 = 0
\]

\[
\frac{62500}{x^2} = 1
\]

\[
x^2 = 62500
\]

\[
x = \sqrt{62500} = 250
\]

Thus, the production level that minimizes the average cost is **250**.

---

### Part 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**.

---

Would you like further explanation or details? Here are some related questions for you:

1. How does marginal cost relate to average cost?
2. Why do we take the derivative of the average cost function to find the production level that minimizes it?
3. What is the significance of the minimal average cost in production?
4. Can a higher production level always lower the average cost?
5. How would changes in fixed costs (like 62500) affect the production decisions?

**Tip:** The marginal cost shows the additional cost of producing one more unit, while the average cost provides the overall cost per unit. Both play key roles in optimizing production decisions.

For the given cost function C(x) = 62500 + 500x + x^2, find: 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

Learn how to calculate cost, average cost, marginal cost, and minimize the average cost using the given cost function C(x) = 62500 + 500x + x^2 at the production level of 1950. This solution covers core concepts in economics and mathematics including derivatives, optimization, and cost analysis.

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