Question

(1 pt) The manager of a large apartment complex knows from experience that 90 units will be occupied if the rent is 420 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 3 dollar increase in rent. Similarly, one additional unit will be occupied for each 3 dollar decrease in rent. What rent should the manager charge to maximize revenue?

Answer 1

Answer:

Monthly rent of $345 would maximize revenue

Explanation:

Revenue = Price * Quantity

Quantity depends on price. We need to work out the relationship between price and quantity (that is, the demand function)

When the rent is $420, quantity demanded is 90 units:

When P = 420 we have Q = 90

Let x be the change in price. For every 3 dollar increase (decrease) in price demanded quantity will decrease (increase) 1 unit:

P = 420 + x (a) we have Q = 90 - x/3 (b)

To find the relationship between P and Q we seek to eliminate x.

Multiply both sides of (b) with 3 we have: 3Q = 270 - x (b')

From (a) and (b') we have: P + 3Q = 420 + x + 270 - x

=> P = 690 - 3Q

Revenue R = P * Q = (690 - 3Q) * Q = 690Q - 3Q^2

To find maximum set derivative of R to 0:

dR = 690 - 6Q = 0

=> Q = 690/6 = 115

To lease 115 the price should be P = 690 - 3Q = 690 - 3*115 = 345