Answer :
Answer:
[tex]p \geq 0[/tex]
Step-by-step explanation:
Given
[tex]R(p) = -16p^2 + 80p + 5[/tex]
Required
Determine the domain of the function
To do this, we need to solve for the vertex, p of the function
[tex]p= \frac{-b}{2a}[/tex]
Given the the general form of a quadratic function is:
[tex]y = ax^2 + bx + c[/tex]
By comparison, we have:
[tex]a = -16[/tex] [tex]b =80[/tex] [tex]c = 5[/tex]
So:
[tex]p= \frac{-b}{2a}[/tex]
[tex]p = \frac{-80}{-16 * 5}[/tex]
[tex]p = \frac{-80}{-80}[/tex]
[tex]p = 1[/tex]
Substitute 1 for p in [tex]R(p) = -16p^2 + 80p + 5[/tex]
[tex]R(1) = -16(1)^2 + 80(1) + 5[/tex]
[tex]R(1) = -16 + 80 + 5[/tex]
[tex]R(1) = 69[/tex]
This implies that
[tex](p, R) = (1,69)[/tex]
The interpretation of this is that;
For every value of p, there's a corresponding value of R.
However, because p indicates price and it's impossible to have a negative price, we can say that the minimum value of p is 0;
Hence, the domain is
[tex]p \geq 0[/tex]