The manager of a large apartment complex knows from experience that 120 units will be occupied if the rent is 450 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 5 dollar increase in rent. Similarly, one additional unit will be occupied for each 5 dollar decrease in rent. What rent should the manager charge to maximize revenue

To maximize revenue, the manager should charge $525

Step-by-step explanation:

Given.

Rent = $450.

Unit = 120

Yes assume our function to be F(x).

To maximise F(x), the function will be based how much of the available units (x) is less than the initial 120 units with a corresponding increase in revenue dollars of 5x above the initial revenue.

Initial Revenue = 450 * 120 = $54,000

F(x) = (120 - x)(450 + 5x)

Converting this into standard quadratic form;

F(x) = 54000 + 600x - 450x - 5x²

F(x) = 54000 + 150x - 5x²

F(x) = -5x² + 150x + 54000

Using axis of symmetry formula

(x = -b/2a) for a parabola to determine the x coordinate of the function's vertex (maximum point):

F(x) = -5x² + 150x + 54,000

Where a = -5

b = 150

x = -b/2a

x = -150/2(-5)

x = -150/-10

x = 15 units less than 120

Next is to calculate the corresponding value of F(x) .

By Substitution

F(x) = -5(15)² + 150(15) + 54000

F(x) = -1125 + 2250 + 54000

F(x) = 55,125 --- Maximum Revenue

Calculating the optimum monthly rent based upon the maximum revenue dollars divided by by the previously determined number of units (15) less than the initial 120 units:

Monthly Rent = 55,125/(120 - 15)

Monthly Rent = 55,125/105

Monthly Rent = $525

To maximize revenue, the manager should charge $525

y=-2.5x+12.5

step-by-step explanation:

5y=2x+10

y=0.4x+2

slope=-2.5

5=-2.5*3+b

b=12.5