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The given cost function is \( C(x) = 28x^2 - 371x + 3125 \), where \( C \) represents the cost in dollars, and \( x \) is the number of cell phones produced (in thousands).

To find the number of units that minimizes the cost, we need to calculate the value of \( x \) that minimizes this quadratic function.

### Steps:

1. The cost function is a quadratic equation in the form:
   \[
   C(x) = ax^2 + bx + c
   \]
   where \( a = 28 \), \( b = -371 \), and \( c = 3125 \).

2. For any quadratic function \( ax^2 + bx + c \), the value of \( x \) that minimizes (or maximizes) the function is given by:
   \[
   x = -\frac{b}{2a}
   \]

3. Substituting the values of \( a \) and \( b \):
   \[
   x = -\frac{-371}{2(28)} = \frac{371}{56} \approx 6.63
   \]

Thus, the company should produce approximately **6.63 thousand** cell phones (or 6630 units) to minimize the cost.

Would you like more details, or do you have any further questions?

Here are 5 related questions for deeper understanding:
1. What is the significance of the coefficients in the cost function?
2. How can you verify whether this value of \( x \) truly minimizes the cost function?
3. What is the cost at the minimum production level?
4. How does the second derivative test confirm the minimum?
5. Can this model be extended to predict costs for different production levels?

**Tip:** The vertex formula \( x = -\frac{b}{2a} \) is useful for finding the minimum or maximum of any quadratic function.

A company that produces cell phones has a cost function C = 28x^2 - 371x + 3125, where C is the cost in dollars and x is the number of cell phones produced (in thousands). How many units of cell phones (in thousands) minimizes the cost function?

Learn how to minimize the cost function C(x) = 28x^2 - 371x + 3125 for a company producing cell phones. This problem covers key concepts in algebra, such as quadratic functions and optimization, and is suitable for students in Grades 10-12.

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