The rate of growth of the profit is approximated by P'(x)=xe^(-x^2) where x represents time measured in years. The total profit in the third year that the new technology is in operation is $10,000.
1.Find the total profit function
2. what happens to the total amount of profit in the long run?
To find the total profit function, we need to integrate the rate of profit growth function, P'(x), with respect to x. Since P'(x) = xe^(-x^2), we can integrate it as follows:
∫(xe^(-x^2)) dx
To integrate this function, we can use a substitution. Let u = -x^2 and du = -2xdx. Rearranging, we get dx = du / -2x. Substituting these values in, we have:
∫(xe^(-x^2)) dx = ∫(-1/2)(e^u) du
Now, we can integrate e^u, which is a basic integral:
= (-1/2) ∫(e^u) du
= (-1/2) * e^u
= (-1/2) * e^(-x^2)
So, the total profit function, P(x), is given by:
P(x) = (-1/2) * e^(-x^2)
To find the total profit in the third year, we substitute x = 3 into the total profit function:
P(3) = (-1/2) * e^(-3^2)
= (-1/2) * e^(-9)
Given that the total profit in the third year is $10,000, we can set up the equation:
(-1/2) * e^(-9) = 10,000
Now we can solve for e^(-9):
e^(-9) = (10,000) / (-1/2)
e^(-9) = -20,000
Since exponential functions are always positive, there must be an error in the given information or calculations.
Regarding what happens to the total amount of profit in the long run, we can consider the behavior of the function as x approaches infinity.
As x approaches infinity, the term e^(-x^2) approaches 0 because e to any negative power approaches 0 as the exponent becomes increasingly negative. Therefore, the total amount of profit in the long run approaches zero. In other words, the profit growth slows down and eventually reaches a plateau.
If P'(x)=xe^(-x^2) , then
P(x) = (-1/2) e^(-x^2) + c
when t= 3 , P(3) = 10000
10000= (-1/2)e^(-9) + c
10000 = -.00006 + c
c = 10000.0000617
1. P(x) = (-1/2) e^(-x^2) + 10000.0000617
2. "in the long run" implies x --->∞
which means (-1/2) e^(-x^2) ---> 0
so P(∞) = 10000.0000617