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
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.
2 answers
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
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
1 answer