How can I determine the number of units x that results in the lowest average cost per unit based on the equation c = 0.001x³ + 5x + 250?

To find the number of units x that produces the minimum average cost per unit c given the equation c = 0.001x³ + 5x + 250, we will need to analyze the function and find its minimum point.

The provided equation is a polynomial equation, and the average cost per unit is given as a function of x. Our goal is to minimize this function. Here’s how you can do it step by step:

  1. Set up the equation: The function given is:
  2. c(x) = 0.001x³ + 5x + 250

  3. Find the derivative: To locate the minimum point, calculate the first derivative of the function with respect to x.
  4. c'(x) = 0.003x² + 5

  5. Set the derivative to zero: To find critical points, set the first derivative equal to zero and solve for x.
  6. 0.003x² + 5 = 0

    Since 0.003x² + 5 can never equal zero (as the term 5 ensures that it’s always positive), this means that there are no critical points that could give us a minimum based on the derivative test.

  7. Analyze the behavior of the function: Since the leading coefficient of is positive (0.001), the function will approach infinity as x increases indefinitely. Therefore, no local minima exists, but we can analyze the behavior of c(x) over a practical range to find a minimum average cost.
  8. Find the average cost over a range: You might want to compute the average cost per unit for specific values of x. For example, you could calculate:
  9. Average Cost (AC) = c(x) / x

As a strategy, you could compute this for different values of x (for example, from 1 to 100) to find a value that minimizes the average cost.

In conclusion, while the analytical method suggests no local minimum via the derivative, a numerical approach may yield a practical result within a range of units. You might employ numerical methods or use software tools to plot the average cost and visually identify the minimum point.

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