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 given by: \[ R(x) = x \cdot p(x) = x \cdot \left( 450 - \frac{x}{2} \right) = 450x - \frac{x^2}{2}. \] Step 2: Differentiate to find critical points
The first derivative of \( R(x) \) is: \[ \frac{dR}{dx} = 450 - x. \] For maximum or minimum, 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: \[ \frac{d^2R}{dx^2} = -1<0. \] Since \( \frac{d^2R}{dx^2}<0 \), \( R(x) \) is maximum when \( x = 450 \). 
Step 4: Final result
The number of units that should be sold to maximise revenue is \( x = 450 \).

Answer 2

Step 1: Calculate the price at \( x = 450 \)
The price is given by: \[ p = 450 - \frac{x}{2} = 450 - \frac{450}{2} = 225. \] Step 2: Compute the rebate
The original price is Rs. 350. The rebate is: \[ {Rebate} = 350 - 225 = 125 \, {(Rs. per calculator)}. \] Step 3: Final result
The rebate required to maximise the revenue is Rs. 125 per calculator.