Assume that the revenue, $R$, for a certain product is determined by the following equation: $R=100 x-0.02 x^{2}$
a.) Find the number of units, $x$, that produces a maximum revenue.
b.) Find the maximum revenue.

Step 1 ：Assume that the revenue, \(R\), for a certain product is determined by the following equation: \(R=100x-0.02x^{2}\)

Step 2 ：To find the number of units, \(x\), that produces a maximum revenue, we need to find the maximum point of the function \(R=100x-0.02x^{2}\). This is a quadratic function, and the maximum point of a quadratic function \(ax^{2}+bx+c\) is given by \(-\frac{b}{2a}\). In this case, \(a=-0.02\) and \(b=100\), so we can substitute these values into the formula to find \(x\).

Step 3 ：Substituting \(a=-0.02\) and \(b=100\) into the formula \(-\frac{b}{2a}\), we get \(x = 2500.0\).

Step 4 ：The number of units, \(x\), that produces a maximum revenue is \(\boxed{2500}\).

Step 5 ：To find the maximum revenue, we substitute \(x=2500\) into the equation \(R=100x-0.02x^{2}\).

Step 6 ：Substituting \(x=2500\) into the equation \(R=100x-0.02x^{2}\), we get \(R = 125000.0\).

Step 7 ：The maximum revenue is \(\boxed{125000}\).