The annual total revenue for a product is given by \( R(x)=45,000 x-5 x^{2} \) dollars, where \( x \) is the number of units sold. To maximize revenue, how many units must be sold? What is the maximum possible annual revenue? To maximize revenue, \( \square \) units must be sold. (Simplify your answer.)

To maximize revenue, we need to find the value of \( x \) that maximizes the function \( R(x) = 45,000x - 5x^2 \). This is a quadratic function and opens downwards (since the coefficient of \( x^2 \) is negative), so its maximum occurs at the vertex. The vertex \( x \) can be found using the formula \( x = -\frac{b}{2a} \), where \( a = -5 \) and \( b = 45,000 \). Plugging in the values, we get: \[ x = -\frac{45,000}{2 \cdot (-5)} = \frac{45,000}{10} = 4,500 \] To find the maximum possible annual revenue, substitute \( x = 4,500 \) back into the revenue function: \[ R(4,500) = 45,000 \cdot 4,500 - 5 \cdot (4,500)^2 \] Calculating this gives: \[ R(4,500) = 202,500,000 - 101,250,000 = 101,250,000 \] Therefore, to maximize revenue, 4,500 units must be sold, leading to a maximum revenue of $101,250,000.