At time t (in days), the size S of a population of butterflies is given by the formula s= 600/1+49(0.6)^t.
When does the maximum growth rate occur?
Well, the maximum growth rate occurs when the butterflies discover an all-you-can-eat buffet of their favorite flowers while simultaneously avoiding their natural predators, who have taken a vacation. But since we are talking about a mathematical formula here, and not the whims of the butterfly community, let's find the real answer.
To find the maximum growth rate, we need to determine the point where the population size, S, reaches its highest value. This point occurs when the derivative of S with respect to time, t, is equal to zero.
Now, let's find that derivative:
dS/dt = 600 * (-49) * (0.6)^t * ln(0.6)
Setting this derivative equal to zero:
0 = 600 * (-49) * (0.6)^t * ln(0.6)
We can ignore the constants since ln(0.6) is nonzero, so we are left with:
0 = (0.6)^t * ln(0.6)
To solve this equation, we realize that ln(0.6) is negative and (0.6)^t is always positive, so there is no real value of t that can satisfy this equation.
So, in the context of this formula, there is no specific time when the population experiences maximum growth. But don't worry, the butterflies will continue to flourish in their own whimsical ways!
To find when the maximum growth rate occurs, we need to determine where the rate of change of the population size is highest. This can be found by taking the derivative of the population size formula with respect to time.
Let's first find the derivative of the population size formula with respect to t:
s = 600 / (1 + 49(0.6)^t)
To simplify the calculation, let's use the substitution u = 0.6^t:
s = 600 / (1 + 49u)
We can rewrite the equation as:
s = 600(1 + 49u)^-1
Taking the derivative of s with respect to u:
ds/du = -600 * (1 + 49u)^-2 * (49)
We want to find when the derivative is equal to zero, so let's set ds/du = 0 and solve for u:
0 = -600 * (1 + 49u)^-2 * (49)
Since the derivative is always negative (because of the minus sign and the squared term), we don't need to consider it. Therefore, we can ignore the first part of the equation:
0 = (1 + 49u)^-2
Taking the square root of both sides:
1 + 49u = 0
49u = -1
u = -1/49
Since we substituted u = 0.6^t, we can solve for t:
0.6^t = -1/49
To get rid of the negative sign, let's take the reciprocal of both sides:
1/0.6^t = -49
0.6^t = -1/49
Since 0.6^t is always positive and the right-hand side of the equation is negative, there are no solutions for t in this case. This means that there is no maximum growth rate for the given population size formula.
To find out when the maximum growth rate occurs in this population, we need to analyze the given formula and understand its components. The formula to calculate the population size at any given time t is:
S = 600 / (1 + 49 * (0.6)^t)
Let's break down the formula:
- The formula uses an exponential function, (0.6)^t, where t represents time in days. This function determines how the population grows over time.
- The exponential term inside the parentheses, (0.6)^t, represents the growth factor. As t increases, the growth factor determines how fast the population size expands.
- The denominator of the equation, 1 + 49 * (0.6)^t, calculates the effective growth rate. The larger this value, the slower the growth rate becomes.
- The numerator, 600, represents the initial population size or the carrying capacity of the environment. It is the maximum population size that can be sustained under ideal conditions.
To find when the maximum growth rate occurs, we need to determine the time t value that results in the largest possible growth factor, (0.6)^t.
Since the base of the exponential term is 0.6, which is less than 1, the growth factor will decrease as time increases. Hence, the maximum growth rate will occur at the beginning, when t is close to zero.
Therefore, the maximum growth rate occurs at or near the start of the population observation, or when t is approximately zero days.
I will assume you meant:
s = 600/(1 + 49(.6)^t)
= 600(1 + 49(.6)^t)^-1
then growth rate = s'
= -600(1+49(.6)^t)^-2 (49ln.6)(.6)^t
now you want the max of growth rate, so
you need s'' and set that equal to zero and solve
I suggest taking the product rule, and carefully doing the algebra. It's going to be messy!