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The given problem asks how many tractors per day the factory should produce to minimize costs. The cost function is:

\[
C(n) = n^2 - 4n + 803
\]

To minimize the cost, we need to find the value of \( n \) that minimizes \( C(n) \). This can be done by taking the derivative of \( C(n) \) with respect to \( n \), setting it equal to 0, and solving for \( n \). Let's go through the steps:

1. The derivative of \( C(n) \) is:
\[
C'(n) = 2n - 4
\]

2. Set the derivative equal to zero to find the critical points:
\[
2n - 4 = 0
\]

3. Solve for \( n \):
\[
2n = 4 \quad \Rightarrow \quad n = 2
\]

Thus, the factory should produce **2 tractors per day** to minimize costs.

Would you like more details or have any questions? Here are some related questions:

1. How do you determine if the critical point is a minimum or maximum?
2. What happens if the cost function has no real roots?
3. Can this method be applied to other cost functions?
4. How do changes in the coefficients of the quadratic function affect the minimum point?
5. What is the significance of the constant term in the cost function?

**Tip:** When optimizing quadratic functions, the vertex form is often useful in finding the minimum or maximum value directly.

A tractor factory produces n tractors per day. The cost of running the factory is C(n) = n^2 - 4n + 803. How many tractors per day should the factory produce to keep the costs to a minimum?

Learn how to minimize the cost function C(n) = n^2 - 4n + 803 for a tractor factory. This problem involves optimization using derivatives and is suitable for students in grades 10-12.

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