D(x) = mx + c ; where x is selling price , c is constant, m is slope
given D = 21 when x = $16
21 = 16m + c ------------1) and also we know D = 24 unit, when x = $15,
24 = 15m + c ------------2)
subtract eqn 2) from eqn 1) we get
-3 = m , and plug this value of m in eqn 1) we get
21 = 16*(-3) + c => c = 69
hence demand function
D(x) = -3x + 69
Cost price is C = $1
Profit = Revenue - Cost price
P(x) = Demand*Selling price - Cost per unit* Demand
P(x) = (69 - 3x)*x - 1*(69 - 3x)
P(x) = 69x - 3x² - 69 + 3x ; or
P(x) = - 3x² + 72x - 69 -------------Answer
Differentiate profit function with respect to x (selling price)
dP/dx = -3*2x + 72 - 0
0 = 72 - 6x ; plug dP/dx = 0 for maxima minima
6x = 72
x = $12 per unit --------Answer
(Since d²P/dx² = -6 < 0 , hence profit will be maximum at x = $12 per unit
and Demand would be D(@12) = 33 unit
Revenue = 33*$12 = $396
Cost = $1*33 = $33
Max Profit = #396 - $33 = $363
