Question

Suppose the demand for a new pharmaceutical drug, on which the manufacturer has a patent monopoly, is given by \( Q = (100 - P)A^{0.5 \), where \( Q \) is output, \( P \) is the price and \( A \) is advertising expenditure. Production cost of the patented drug is given by \( C(Q) = 60Q \). At the firm’s optimal choices, the ratio of advertising expenditure to sales revenue for the pharmaceutical product will be 1: _________ (in integer).}

Hint:

For optimal pricing and advertising expenditure, use the revenue and profit maximization conditions.

Answer 1

Correct Answer: 8

The revenue from sales is given by: \[ R = P \times Q \] Substitute the demand function \( Q = (100 - P)A^{0.5} \) into this: \[ R = P \times (100 - P)A^{0.5} \] Now, differentiate the revenue with respect to \( P \) to find the price that maximizes revenue: \[ \frac{dR}{dP} = (100 - 2P)A^{0.5} \] Set the derivative equal to zero to find the optimal price: \[ 100 - 2P = 0 \quad \Rightarrow \quad P = 50 \] Substitute \( P = 50 \) into the demand function to find \( Q \): \[ Q = (100 - 50)A^{0.5} = 50A^{0.5} \] The cost of producing \( Q \) units is: \[ C(Q) = 60Q = 60 \times 50A^{0.5} = 3000A^{0.5} \] The firm's profit is: \[ \text{Profit} = R - C(Q) = 50 \times 50A^{0.5} \times A^{0.5} - 3000A^{0.5} = 2500A - 3000A^{0.5} \] Maximize profit with respect to \( A \). Differentiate the profit function: \[ \frac{d(\text{Profit})}{dA} = 2500 - 1500A^{-0.5} \] Set the derivative equal to zero: \[ 2500 - 1500A^{-0.5} = 0 \quad \Rightarrow \quad A^{0.5} = \frac{5}{3} \] Squaring both sides: \[ A = \left( \frac{5}{3} \right)^2 = \frac{25}{9} \] The ratio of advertising expenditure to sales revenue is: \[ \frac{A}{R} = \frac{\frac{25}{9}}{2500} = 0.0011 \] Thus, the ratio of advertising expenditure to sales revenue is \( 1: 900 \).