A manufacturer finds that the revenue generated by selling \( x \) units of a certain commodity is given by the function \( R(x)=100 x-0.2 x^{2} \), where the revenue \( R(x) \) is measured in dollars. What is the maximum revenue, and how many units should be manufactured to obtain this maximum? \( \$ \square \) units Need Help? Watch it Master lt

We are given the revenue function \[ R(x) = 100x - 0.2x^2. \] **Step 1. Find the derivative of \( R(x) \).** Differentiate \( R(x) \) with respect to \( x \): \[ R'(x) = \frac{d}{dx} (100x - 0.2x^2) = 100 - 0.4x. \] **Step 2. Set the derivative equal to zero to find the critical point.** \[ 100 - 0.4x = 0. \] Solving for \( x \): \[ 0.4x = 100 \quad \Rightarrow \quad x = \frac{100}{0.4} = 250. \] **Step 3. Confirm that this is a maximum.** The revenue function is a quadratic function with a negative coefficient for \( x^2 \) (i.e., \(-0.2\)), which means the graph is a downward-opening parabola. Therefore, the critical point \( x = 250 \) represents a maximum. **Step 4. Calculate the maximum revenue.** Substitute \( x = 250 \) back into \( R(x) \): \[ R(250) = 100(250) - 0.2(250)^2. \] Calculate each term: \[ 100(250) = 25000, \] \[ (250)^2 = 62500, \] \[ 0.2 \times 62500 = 12500. \] So, \[ R(250) = 25000 - 12500 = 12500. \] **Answer:** The maximum revenue is \(\$12500\) and it is achieved by manufacturing \(\boxed{250}\) units.