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 number of units \( x \) that maximizes the revenue function: \[ R(x) = 45{,}000x - 5x^2 \] This is a quadratic function of the form \( R(x) = ax^2 + bx + c \), where: - \( a = -5 \) (since the coefficient of \( x^2 \) is negative, the parabola opens downward, indicating a maximum point) - \( b = 45{,}000 \) The number of units \( x \) that maximizes the revenue can be found using the vertex formula for a parabola: \[ x = -\frac{b}{2a} \] Plugging in the values: \[ x = -\frac{45{,}000}{2(-5)} = \frac{45{,}000}{10} = 4{,}500 \] So, **4,500 units** must be sold to maximize revenue. Next, to find the **maximum possible annual revenue**, substitute \( x = 4{,}500 \) back into the revenue function: \[ R(4{,}500) = 45{,}000 \times 4{,}500 - 5 \times (4{,}500)^2 \] Calculate each term: 1. \( 45{,}000 \times 4{,}500 = 202{,}500{,}000 \) 2. \( 5 \times (4{,}500)^2 = 5 \times 20{,}250{,}000 = 101{,}250{,}000 \) Subtract the second term from the first: \[ R(4{,}500) = 202{,}500{,}000 - 101{,}250{,}000 = 101{,}250{,}000 \text{ dollars} \] **Summary:** - **To maximize revenue, 4,500 units must be sold.** - **The maximum possible annual revenue is \$101,250,000.**