A supply company manufactures copy machines. The unit cost $C$ (the cost in dollars to make each copy machine) depends on the number of machines made. If $x$ machines are made, then the unit cost is given by the function $C(x)=0.6 x^{2}-120 x+25,602$. How many machines must be made to minimize the unit cost? Do not round your answer. Number of copy machines: $\times \quad 5$

Step 1 ：The problem is asking for the number of machines that must be made to minimize the unit cost. This is a calculus problem, where we need to find the minimum of the function \(C(x)=0.6 x^{2}-120 x+25,602\). The minimum of a function occurs where its derivative is zero. So, we need to find the derivative of the function, set it equal to zero, and solve for \(x\).

Step 2 ：The derivative of the function \(C(x)=0.6 x^{2}-120 x+25,602\) is \(C'(x)=1.2x - 120\). Setting this equal to zero gives us the critical points of the function.

Step 3 ：Solving \(1.2x - 120 = 0\) gives us the critical point \(x = 100\). This means that the unit cost is minimized when 100 machines are made.

Step 4 ：However, we need to verify that this is indeed a minimum and not a maximum or a point of inflection. We can do this by taking the second derivative of the function and checking its sign at \(x = 100\). If the second derivative is positive, then the function has a minimum at \(x = 100\).

Step 5 ：The second derivative of the function \(C(x)=0.6 x^{2}-120 x+25,602\) is \(C''(x)=1.2\). The second derivative at the critical point \(x = 100\) is positive, which means that the function has a minimum at \(x = 100\). Therefore, the unit cost is minimized when 100 machines are made.

Step 6 ：Final Answer: The number of machines that must be made to minimize the unit cost is \(\boxed{100}\).