A park is open 16 weeks of the year. An equation that models the cost function in thousands is: c(x){5, 0<x<5 where x is the number of weeks since opening and x, x>5}
An equation that models the revenue function is r(x)=30(-5/x^2+1) for x>0
Determine the profit function for x>0
simple:
p(x) = r(x) - c(x)
To determine the profit function, we need to subtract the cost function from the revenue function.
Given:
Cost Function: c(x) = {5, 0 < x < 5; x, x > 5}
Revenue Function: r(x) = 30 * (-5/x^2 + 1), for x > 0
The profit function can be represented by the equation p(x) = r(x) - c(x).
Let's break it down into two parts:
1. For 0 < x < 5:
Since the cost function is constant 5 (in thousands), the profit function for this range will be:
p(x) = r(x) - c(x) = r(x) - 5
2. For x > 5:
The cost function is equal to x, so the profit function will be:
p(x) = r(x) - c(x) = r(x) - x
Therefore, the profit function for x > 0 is:
p(x) = {r(x) - 5, 0 < x < 5; r(x) - x, x > 5}
Note: The cost function is not defined for x = 5, so we consider two separate cases for x < 5 and x > 5.