7. The cost per computer produced at a factory depends on how many computers the factory produces in a day: The cost function is modeled by \( C(n)=\frac{1}{500} n^{2}-n+200 \), where \( n \) is the number of computers produced in a day and \( C(n) \) is the unit cost, in dollars per computer. \( \begin{array}{ll}\text { (a) Calculate } C(50) \text { and give an interpretation of } & \text { (b) Explain why this function must have a minimum } \\ \text { value and not a maximum. } \\ \text { your answer in terms of the scenario described. } & \text { (d) Explain why this function cannot have any zeros } \\ \text { that are real numbers. } \\ \text { (c) What is the minimum value of the function? } \\ \text { (You may have to scroll through your table.) } & \text { ( }\end{array} \)

**(a) Calculation of \( C(50) \):** The cost function is \[ C(n)=\frac{1}{500}n^2 - n + 200. \] Substitute \( n = 50 \): \[ C(50)=\frac{1}{500}(50)^2 - 50 + 200. \] First, calculate \( (50)^2 \): \[ (50)^2 = 2500. \] Then, \[ \frac{1}{500} \times 2500 = 5. \] So, \[ C(50)=5 - 50 + 200 = 155. \] **Interpretation:** If the factory produces 50 computers in a day, the cost per computer is \(\$155\). --- **(b) Why the function must have a minimum (and not a maximum):** The cost function \[ C(n)=\frac{1}{500}n^2 - n + 200 \] is a quadratic function in \( n \). The quadratic term has the coefficient \( \frac{1}{500} \), which is positive. This means the parabola opens upwards, making the vertex the smallest point on the graph. Therefore, the function has a minimum value, not a maximum. In the scenario, as the production increases or decreases from the optimal number, the unit cost goes up, indicating that there is one optimal production level (the vertex) where the cost is minimized. --- **(c) Finding the minimum value of the function:** For a quadratic function in the form \[ C(n)=an^2+bn+c, \] the vertex occurs at \[ n = -\frac{b}{2a}. \] Here, \( a = \frac{1}{500} \) and \( b = -1 \). Thus, \[ n = -\frac{-1}{2\left(\frac{1}{500}\right)}=\frac{1}{\frac{2}{500}}=\frac{500}{2}=250. \] Now, compute \( C(250) \): \[ C(250)=\frac{1}{500}(250)^2 - 250 + 200. \] Calculate \( (250)^2 \): \[ (250)^2 = 62500. \] Then, \[ \frac{1}{500} \times 62500 = 125. \] Thus, \[ C(250)=125 - 250 + 200=75. \] So, the minimum cost per computer is \(\$75\) when 250 computers are produced in a day. --- **(d) Why the function cannot have any zeros that are real numbers:** To find the zeros, we set \[ \frac{1}{500}n^2 - n + 200 = 0. \] Multiply the entire equation by 500 to eliminate the fraction: \[ n^2 - 500n + 100000 = 0. \] The discriminant \( D \) of a quadratic equation \( n^2 - 500n + 100000 = 0 \) is given by \[ D = (-500)^2 - 4(1)(100000) = 250000 - 400000 = -150000. \] Since \( D < 0 \), there are no real solutions to the equation. This means the cost per computer never actually becomes \(\$0\) for any real number of computers produced.