Question

If demand function is given as the following: 

Qz = 230 -2.75 Pz + 0.5 I + 1.2 Pm + 0.6A 

Where Qz is quantity of Good z sold, Pz is price of Good z per unit, I is per capita income, Pm is price of competitor and A is the amount of advertising spent. 

Current values:  Pz= RM 55 I= RM 9000    Pm= RM 50    A =RM 12,000

Determine the total revenue maximizing price and quantity.

Answer 1

At current values $Qz = 230 - 2.75×55 + 0.5×9,000 + 1.2×50 + 0.6×12,000 = 11,838.75$units.

Qz = 11,990 - 2.75Pz,

Pz = 4,360 - Qz/2.75,

$TR = Pz×Qz = 4,360Qz - Qz^2/2.75.$

Total revenue is maximized when TR'(Qz) = 0.

4,360 - 2Qz/2.75 = 0,

2Qz = 11,990,

Qz = 5,995 units,

Pz = 4,360 - 5,995/2.75 = 2,180.