Suppose the revenue \( R \) in thousands of dollars that a company receives from producing \( x \) thousand compact disc players is given by the formula \( R(x)=x(50-5 x) \). a. Evaluate \( R(3) \) and interpret the result. b. How many CD players should the company produce to maximize its revenue? c. What is the maximum revenue? a. Evaluate \( R(3) \). \( R(3)=\square \)

To evaluate \( R(3) \), plug \( x = 3 \) into the revenue function: \[ R(3) = 3(50 - 5 \cdot 3) = 3(50 - 15) = 3 \cdot 35 = 105. \] Interpreting this result, the company generates $105,000 in revenue when it produces 3,000 compact disc players. Quite a profitable venture, isn't it? To find the production level that maximizes revenue, we need to identify the vertex of the quadratic function \( R(x) = x(50 - 5x) \), which can be rewritten in standard form. The vertex occurs at \( x = -\frac{b}{2a} \). Here, \( a = -5 \) and \( b = 50 \), leading to: \[ x = -\frac{50}{2 \cdot -5} = 5. \] Thus, the company should produce 5,000 CD players to maximize revenue. To compute the maximum revenue, substitute \( x = 5 \) back into the function: \[ R(5) = 5(50 - 5 \cdot 5) = 5(50 - 25) = 5 \cdot 25 = 125. \] The maximum revenue is therefore $125,000. A winning strategy!