My Notes Ask Your Teacher 3. -3 points WaneFMAC6 9.1.028 Pack-Em-In hi another d

Question

My Notes Ask Your Teacher 3. -3 points WaneFMAC6 9.1.028 Pack-Em-In hi another development in the works. If it builds 60 houses in this development, it will be able to sell them at $400,000 each, but if as it builds 70 houses, it will get only $380,000 each. Obtain a linear demand equation. (Let p be the price of a house and q the number of houses p(q) Determine how many houses it should build to get the largest revenue? houses What is the largest possible revenue? Need Help? Read It Practice It Talk to a Tutor

Explanation / Answer

essentially, yes.

P(q) = Aq + B
P(60) = 400000 = 60A + B
P(70) = 380000 = 70A + B
20000 = - 10A
A = - 2000
B = 400000 + 60(2000) = 520000

The price of each house as a function of the number of houses:
P(q) = 520000 - 2000q = 1000(520-2q)

Revenue is the product of price per unit times the number of units
R(q) = qP(q) = 520000q - 2000q²

Maximum Revenue:
dR/dq = 520000 - 4000q
q = 520/4 = 130

Maximum revenue will be at the 130 house level and is R(130) = 520000(130) - 2000(130²) = 33,800,000.

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