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?

dp/dt = t e^-t^2

if p = c - .5e^-t^2
p' = t e^-t^2 =

when t = 3
10,000 = c - .5e^-9
c = 10,000 + .5 e^-9
c = 10,000 + .0000617
so
p = 10,000 - .5/ e^(t^2) essentially

It becomes 10,000 as t---> oo

To find the total profit function, we integrate the rate of growth of the profit function, P'(x).

1. Total profit function:
We start by integrating the function P'(x) = xe^(-x^2) with respect to x:

∫ P'(x) dx = ∫ xe^(-x^2) dx

To solve this integral, we can use the substitution method:
Let u = -x^2, then du/dx = -2x, and dx = -du/(2x).

Now substituting the values into the integral:

∫ P'(x) dx = ∫ xe^(-x^2) dx = -∫ (1/2) e^u du

= -(1/2) ∫ e^u du = -(1/2) e^u + C

Since u = -x^2, we substitute it back into the equation:

Total profit function: P(x) = -(1/2) e^(-x^2) + C

Now, to find the value of the constant C, we use the information given that the total profit in the third year is $10,000.

When x = 3 (third year), P(x) = $10,000:

P(3) = -(1/2) e^(-3^2) + C = 10,000

Solving for C:

C = 10,000 + (1/2) e^(-9)

Therefore, the total profit function is:

P(x) = -(1/2) e^(-x^2) + 10,000 + (1/2) e^(-9)

2. What happens to the total amount of profit in the long run?

To determine what happens to the total amount of profit in the long run, we need to examine the behavior of the exponential function as x approaches infinity.

When x approaches infinity, e^(-x^2) becomes very close to zero. Therefore, the total profit function approaches:

P(x) ≈ -(1/2) * 0 + 10,000 + (1/2) * 0

P(x) ≈ 10,000

Hence, in the long run, the total amount of profit approaches $10,000.

To find the total profit function, we need to integrate the rate of growth function with respect to time.

1. Integrating the rate of growth function P'(x) = xe^(-x^2) with respect to x gives us the total profit function P(x):

P(x) = ∫(P'(x)) dx

To evaluate this integral, we need to use the substitution method. Let u = -x^2, then du/dx = -2x, and dx = -du/(2x). Substituting these values into the integral, we have:

P(x) = ∫(xe^(-x^2)) dx
= ∫(-e^u) (-du/(2x))
= (1/2) ∫(e^u) du
= (1/2) e^u + C

Since u = -x^2, we can substitute back to get:

P(x) = (1/2) e^(-x^2) + C

To find the value of C, we can use the given information that the total profit in the third year is $10,000. Since x represents time measured in years, we substitute x = 3 into P(x) and equate it to $10,000:

P(3) = (1/2) e^(-3^2) + C = 10,000

Simplifying this equation allows us to solve for C.

2. To find what happens to the total amount of profit in the long run, we need to consider the behavior of the exponential term e^(-x^2) as x approaches infinity. When x is large, the exponential term approaches 0, which means that the total profit will approach a constant value in the long run. This suggests that the rate of growth of profit will decrease until it levels off, resulting in a steady, constant total amount of profit.