Answer:
To maximize the monthly rental profit, 90 units should be rented out.
The maximum monthly profit realizable is $38,200.
Step-by-step explanation:
Vertex of a quadratic function:
Suppose we have a quadratic function in the following format:
[tex]f(x) = ax^{2} + bx + c[/tex]
It's vertex is the point [tex](x_{v}, y_{v})[/tex]
In which
[tex]x_{v} = -\frac{b}{2a}[/tex]
[tex]y_{v} = -\frac{\Delta}{4a}[/tex]
Where
[tex]\Delta = b^2-4ac[/tex]
If a<0, the vertex is a maximum point, that is, the maximum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].
In this question:
Quadratic equation with [tex]a = -12, b = 2160, c = -59000[/tex]
To maximize the monthly rental profit, how many units should be rented out?
This is the x-value of the vertex, so:
[tex]x_{v} = -\frac{b}{2a} = -\frac{2160}{2(-12)} = \frac{2160}{24} = 90[/tex]
To maximize the monthly rental profit, 90 units should be rented out.
What is the maximum monthly profit realizable?
This is p(90). So
[tex]p(90) = -12(90)^2 + 2160(90) - 59000 = 38200[/tex]
The maximum monthly profit realizable is $38,200.