Respuesta :
Using a linear and a quadratic function, it is found that the largest possible revenue is of $9,800,000.
The demand function has a linear format, given by:
[tex]P = mh + b[/tex]
In which:
- m is the slope, that is, the rate of change.
- b is the intercept, that is, the value of P when h = 0.
40 houses can be sold for $200,000 each, thus, the first point of the linear function is: (40, 200000).
60 houses can be sold for $160,000 each, thus, the second point is: (60, 160000).
The slope is given by change in the output divided by change in the input, thus:
[tex]m = \frac{160000 - 200000}{60 - 40} = -2000[/tex]
Thus:
[tex]P = -2000h + b[/tex]
Point (40, 200000) means that when [tex]x = 40, y = 200000[/tex]. This is used to find b.
[tex]P = -2000h + b[/tex]
[tex]200000 = -2000(40) + b[/tex]
[tex]b = 280000[/tex]
Then
[tex]P = -2000h + 280000[/tex]
The revenue function is:
[tex]R = hp = -2000h^2 + 280000h[/tex]
It is a quadratic equation with [tex]a = -2000, b = 280000[/tex]
Since the quadratic equation is concave down, the maximum value is:
[tex]R_{MAX} = -\frac{b^2 - 4ac}{4a} = -\frac{280000^2}{4(-2000)} = 9800000[/tex]
The largest possible revenue is of $9,800,000.
A similar problem is given at https://brainly.com/question/14469903
