PackEmIn has another development in the works If it builds 3

Pack-Em-In has another development in the works. If it builds 30 houses in this development, it will be able to sell them at $250,000 each, but if it builds 80 houses, it will get only $200,000each. Obtain a linear demand equation. (Let p be the price of a house and q the number of houses.)

Determine how many houses it should build to get the largest revenue?
  houses

What is the largest possible revenue?
$

Solution

P(n) = An + B
P(30) = 250000 = 30A + B
P(80) = 200000 = 80A + B
50000 = - 50A
A = - 1000
B = 250000 + 30(1000) = 280000

The price of each house as a function of the number of houses:
P(n) = 280000 - 1000n
Revenue is the product of price per unit times the number of units
R(n) = nP(n) = 280000n - 1000n²

Maximum Revenue:
dR/dn = 280000 - 2000n = 0
n = 280/2 = 140

Maximum revenue will be at the 140 house level and is R(140) = 280000(140) - 1000(140²) = 19600000.