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: Find critical points by differentiation
The first derivative of \( R(x) \) is:\[\frac{dR}{dx} = 450 - x.\]Set the first derivative to zero to find potential maxima or minima:\[450 - x = 0 \implies x = 450.\]Step 3: Confirm maximum 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 maximised at \( x = 450 \).Step 4: Conclusion
To maximise revenue, \( x = 450 \) units should be sold.