Suppose that a when a cell phone store charges 16$ for a car charger, it sells 92 units. when it drops the price to 15$ it sells 96 units. assume that demand is a linear function of price.
if each phone charger costs 1$ to make, what price should the store charge to maximize its profit?
if x is the number of times the price is reduced by one dollar.
find a function for total profit with respect to x. a negative value for x will mean the price is increased.

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