Math Problem Statement

Find the marginal cost function and use it to estimate how fast the cost is increasing when x=10,000. C(x)=400,000+590x+0.001x^2

Solution

To find the marginal cost function, we first need to find the derivative of the cost function C(x)C(x), which represents the rate at which the cost is changing with respect to xx.

Given the cost function:

C(x)=400,000+590x+0.001x2C(x) = 400,000 + 590x + 0.001x^2

The marginal cost MC(x)MC(x) is the derivative of C(x)C(x) with respect to xx:

MC(x)=ddx(400,000+590x+0.001x2)MC(x) = \frac{d}{dx}\left(400,000 + 590x + 0.001x^2\right)

Now, let's differentiate each term:

  1. The derivative of 400,000400,000 is 0 because it's a constant.
  2. The derivative of 590x590x is 590590.
  3. The derivative of 0.001x20.001x^2 is 2×0.001×x=0.002x2 \times 0.001 \times x = 0.002x.

So, the marginal cost function is:

MC(x)=590+0.002xMC(x) = 590 + 0.002x

Now, to estimate how fast the cost is increasing when x=10,000x = 10,000, substitute x=10,000x = 10,000 into the marginal cost function:

MC(10,000)=590+0.002(10,000)=590+20=610MC(10,000) = 590 + 0.002(10,000) = 590 + 20 = 610

Thus, the rate at which the cost is increasing when x=10,000x = 10,000 is 610610.

Summary:

  • The marginal cost function is MC(x)=590+0.002xMC(x) = 590 + 0.002x.
  • When x=10,000x = 10,000, the cost is increasing at a rate of 610610 per unit increase in xx.

Would you like a more detailed explanation or any additional help with this? Here are some related questions you could explore:

  1. How do we interpret the marginal cost function in terms of business operations?
  2. How can we find the total cost at a given production level xx?
  3. What is the significance of the constant term (400,000) in the cost function?
  4. How would the marginal cost change if we had a different quadratic term in the cost function?
  5. How do we compute the average cost function from the given cost function?

Tip: The marginal cost function helps businesses understand the cost increase for producing one more unit. It’s particularly useful in pricing and production decision-making.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Marginal Cost
Derivative of Polynomial Functions

Formulas

Marginal Cost = d(C(x))/dx

Theorems

Power Rule for Differentiation

Suitable Grade Level

Grades 11-12