Answer:
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
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