Math Problem Statement

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

Solution

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

C(x)=62500+500x+x2C(x) = 62500 + 500x + x^2

Part a) The cost at the production level 1950

We calculate C(1950)C(1950) by substituting x=1950x = 1950 into the cost function:

C(1950)=62500+500(1950)+(1950)2C(1950) = 62500 + 500(1950) + (1950)^2

C(1950)=62500+975000+3802500=4840000C(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)AC(x) is given by:

AC(x)=C(x)xAC(x) = \frac{C(x)}{x}

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

AC(1950)=484000019502482.05AC(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)MC(x) is the derivative of the cost function C(x)C(x):

C(x)=500+2xC'(x) = 500 + 2x

Substitute x=1950x = 1950:

C(1950)=500+2(1950)=500+3900=4400C'(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)=C(x)xAC(x) = \frac{C(x)}{x}, we first find its derivative, set it to zero, and solve for xx.

  1. AC(x)=62500x+500+xAC(x) = \frac{62500}{x} + 500 + x

  2. Take the derivative:

AC(x)=62500x2+1AC'(x) = -\frac{62500}{x^2} + 1

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

62500x2+1=0-\frac{62500}{x^2} + 1 = 0

62500x2=1\frac{62500}{x^2} = 1

x2=62500x^2 = 62500

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

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


Part e) The minimal average cost

Substitute x=250x = 250 into the average cost function:

AC(250)=62500250+500+250=250+500+250=1000AC(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.

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Math Problem Analysis

Mathematical Concepts

Cost Functions
Marginal Cost
Average Cost
Optimization

Formulas

Cost Function: C(x) = 62500 + 500x + x^2
Average Cost: AC(x) = C(x)/x
Marginal Cost: MC(x) = dC(x)/dx

Theorems

Derivative to find Marginal Cost
Optimization using first-order conditions (setting derivative equal to zero)

Suitable Grade Level

College-level Economics/Mathematics