The population of a colony of bacteria is modeled by the function P(x) = 50(e^-x -e^-x^2)+10, for x is greater than or equal to 0, where population P is in thousands, x is in hours, and x=0 corresponds to the moment of introduction of a certain chemical into the colony's enviorment. At which time below is the rate of population growth the greatest?
A) x=1.2
B) x=2.2
c) x=3
D) x=.2
E) x=0.8
P' = 50(2x e^-x^2 - e^-x)
its rate of change is a maximum where its derivative is zero
P" = 50((-4x^2+2) e^-x^2 + e^-x) = 0
x = 0.84843
dP/dx = 50 [ -e^-x - (e^-x^2) * -2x) ] = 50 [ -e^-x + 2 x e^-x^2 ]
zero when
e^-x = 2 x e^-x^2
1/2 = x e^(x-x^2)
try x = 3
3 e^(-6) way too small , needs to be 0.5
try x = 0.8
.8 e^(0.16 )= 9.4 , too big
try 1.2
etc
To find the time at which the rate of population growth is the greatest, we need to determine the derivative of the population function and evaluate it at each given time.
Given the population function: P(x) = 50(e^-x - e^(-x^2)) + 10
Step 1: Find the derivative of the population function.
P'(x) = 50(-e^-x + 2x*e^(-x^2))
Step 2: Evaluate the derivative at each given time.
To determine the time at which the rate of population growth is the greatest, we are looking for the maximum value of the derivative function.
For option A) x = 1.2:
P'(1.2) = 50(-e^-1.2 + 2 * 1.2 * e^(-1.2^2))
For option B) x = 2.2:
P'(2.2) = 50(-e^-2.2 + 2 * 2.2 * e^(-2.2^2))
For option C) x = 3:
P'(3) = 50(-e^-3 + 2 * 3 * e^(-3^2))
For option D) x = 0.2:
P'(0.2) = 50(-e^-0.2 + 2 * 0.2 * e^(-0.2^2))
For option E) x = 0.8:
P'(0.8) = 50(-e^-0.8 + 2 * 0.8 * e^(-0.8^2))
Now, we can calculate the values of P'(x) for each of the given options.
Compare the values of P'(x) for each option and determine which one has the highest value. The option with the highest derivative value represents the time at which the rate of population growth is the greatest.