Question

Determine the number of units (\( x \)) that should be sold to maximise the revenue \( R(x) = x \cdot p(x) \). Also verify the result.

Hint:

To maximise revenue or profit, always verify the nature of the critical point using the second derivative test.

Answer 1

Step 1: Express revenue as a function of \( x \)
Revenue is calculated as: \[ R(x) = x \cdot p(x) = x \cdot \left( 450 - \frac{x}{2} \right) = 450x - \frac{x^2}{2}. \]
Step 2: Differentiate to find critical points
The first derivative of \( R(x) \) is: \[ \frac{dR}{dx} = 450 - x. \] Setting the first derivative to zero to find extrema: \[ 450 - x = 0 \implies x = 450. \] 
Step 3: Verify using the second derivative
The second derivative of \( R(x) \) is: \[ \frac{d^2R}{dx^2} = -1 < 0. \] As the second derivative is negative, \( R(x) \) is maximized at \( x = 450 \). 
Step 4: Final result
The revenue is maximized when \( x = 450 \) units are sold.