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.

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: \[ \frac{dR}{dx} = 450 - x. \] To find extrema, set \( \frac{dR}{dx} = 0 \): \[ 450 - x = 0 \implies x = 450. \] Step 3: Verify using the second derivative
The second derivative of \( R(x) \) is: \[ \frac{d^2R}{dx^2} = -1<0. \] As \( \frac{d^2R}{dx^2}<0 \), \( R(x) \) attains a maximum at \( x = 450 \).
Step 4: Final result
The optimal number of units to sell for maximum revenue is \( x = 450 \).

Answer 2

Step 1: Calculate the price at \( x = 450 \)
The price function is given by:\[p = 450 - \frac{x}{2}.\]Substituting \( x = 450 \), we get:\[p = 450 - \frac{450}{2} = 225.\]Step 2: Compute the rebate
Given an original price of Rs. 350, the rebate is calculated as:\[{Rebate} = 350 - 225 = 125 \, {(Rs. per calculator)}.\]Step 3: Final result
The rebate that maximises revenue is Rs. 125 per calculator.