he rate of growth of the profit from an invention is approximated by P'(x)=xe^-x^2, where x represents time measured in years. The total profit in year 1 that the invention is in operation is $15,000. Find the total profit function.
I know the answer is +.199 but how? I set up the problem as 15000=-.5e^-(1)^2 + c but I can't get the correct answer.
if dP/dx = x e^-x^2
then p = -(1/2)e^-x^2 + C
15,000 = -.5 /e + C agree
so
C = 15,000 + .1839 = 15,000.1839
so
p = -(1/2)e^-x^2 + 15,000.18
It looks like your profit function is correct as
p(x) = (-1/2) e^(-x^2) + c
15000 = (-1/2)(e^-1) + c
15000 = -1/(2e) + c
c = 15000.184
so p(x) = (-1/2) e^(-x^2) + 15000.184
strange result.
Since the exponential term always has a negative value and approaches zero
the profit is basically steady at 15,000
To find the total profit function, you need to integrate the rate of growth of profit function, P'(x), with respect to x.
The rate of growth of profit function provided is P'(x) = xe^(-x^2).
Integrating P'(x) will give you the total profit function, denoted as P(x):
∫P'(x) dx = ∫(xe^(-x^2)) dx
To solve this integral, you can use a substitution. Let u = -x^2, then du = -2x dx.
The integral becomes:
∫(xe^(-x^2)) dx = ∫(1/(-2)) e^u du
Simplifying and changing the limits:
= -(1/2) ∫ e^u du
= -(1/2) e^u + C
Now substitute back -x^2 in for u:
= -(1/2) e^(-x^2) + C
Since the total profit in year 1 is $15,000, you can solve for C as follows:
P(1) = -(1/2) e^(-1^2) + C = 15000
Simplifying:
-(1/2) e^(-1) + C = 15000
C = 15000 + (1/2) e^(-1)
C ≈ 15000 + 0.199
Therefore, the total profit function is approximately:
P(x) = -(1/2) e^(-x^2) + 15000 + 0.199
To find the total profit function, we need to integrate the given rate of growth of profit function, P'(x), with respect to x. This will give us the total profit function, P(x).
The rate of growth of profit function is given as P'(x) = xe^(-x^2).
We can integrate P'(x) with respect to x to find P(x). The integral of xe^(-x^2) can be evaluated using integration techniques, such as substitution or integration by parts.
Using integration by parts, let's set u = x and dv = e^(-x^2)dx.
Differentiating u, we get du = dx, and integrating dv, we get v = -0.5e^(-x^2).
We can now use the formula for integration by parts: ∫u dv = uv - ∫v du.
∫xe^(-x^2) dx = -0.5xe^(-x^2) - ∫(-0.5e^(-x^2)) dx
= -0.5xe^(-x^2) + 0.5∫e^(-x^2) dx
The integral ∫e^(-x^2) dx does not have an elementary function representation, meaning it cannot be expressed using standard functions. It is a special function called the error function, denoted by erf(x).
So, the indefinite integral of xe^(-x^2) dx is given by:
∫xe^(-x^2) dx = -0.5xe^(-x^2) + 0.5erf(x) + C
Now, we can find the constant of integration, C, by using the given information.
In the first year of operation (x = 1), the total profit is $15,000. So, P(1) = 15,000. Plugging this into our expression for P(x):
15,000 = -0.5e^(-1^2) + 0.5erf(1) + C
15,000 = -0.5e^(-1) + 0.5erf(1) + C
Now, we need to evaluate -0.5e^(-1) + 0.5erf(1) + C to find the value of C.
Using a calculator, we find that erf(1) ≈ 0.8427.
15,000 = -0.5e^(-1) + 0.5(0.8427) + C
15,000 = -0.5e^(-1) + 0.42135 + C
15,000 ≈ -0.5(0.3679) + 0.42135 + C
15,000 ≈ -0.18395 + 0.42135 + C
15,000 ≈ 0.2374 + C
Now, we solve for C:
C ≈ 15,000 - 0.2374
C ≈ 14,999.7626
Therefore, the total profit function, P(x), is approximately:
P(x) ≈ -0.5xe^(-x^2) + 0.5erf(x) + 14,999.7626
To find the value of P'(1), we can differentiate P(x) with respect to x and evaluate it at x = 1:
P'(x) ≈ -0.5e^(-x^2) + 0.5(2/sqrt(pi))e^(-x^2)
P'(1) ≈ -0.5e^(-1) + 0.5(2/sqrt(pi))e^(-1)
P'(1) ≈ -0.5(0.3679) + 0.5(2/sqrt(pi))(0.3679)
P'(1) ≈ -0.18395 + (1.13198/sqrt(pi))
P'(1) ≈ 0.94803
Hence, the rate of growth of profit in year 1 is approximately 0.94803.
The answer you provided, +0.199, seems to be incorrect.