Could someone explain this question please.

Suppose the population (in thousands of a certain species of insect after t months is described by the function p(t)=4t+ (100t)/(t^2+400) + 500. Determine the maximum population in the first three months.

when x is small (as in the interval [0,3]

100t/(t^2+400) is just basically t/4
so, p(t) looks a lot like 4t + t/4 + 500
that's just a line, so the max and min are at the ends of the interval.

If I factor it out, it'll be 17t/4 + 500.

then plug in 3

17(3)/4+500=512 x 1000

=512,750.

therefore the population would be 512,750?

To determine the maximum population in the first three months, we need to find the highest value of the population function p(t) within that time period.

First, we substitute t values from 0 to 3 months into the function p(t) and calculate the population for each month:

For t = 0:
p(0) = 4(0) + (100(0))/(0^2+400) + 500 = 0 + 0 + 500 = 500

For t = 1:
p(1) = 4(1) + (100(1))/(1^2+400) + 500 = 4 + 100/401 + 500 = 4 + 0.2494 + 500 = 504.2494

For t = 2:
p(2) = 4(2) + (100(2))/(2^2+400) + 500 = 8 + 0.4975 + 500 = 508.4975

For t = 3:
p(3) = 4(3) + (100(3))/(3^2+400) + 500 = 12 + 0.7463 + 500 = 512.7463

From these calculations, we observe that the maximum population occurs in the third month (t=3) with a population of 512.7463 thousand insects.

To determine the maximum population in the first three months, we need to find the maximum value of the population function p(t) within the range t = 0 to t = 3. Here's how you can find the answer:

1. Substitute the values of t = 0, 1, 2, and 3 into the population function p(t) and calculate the corresponding population values.

When t = 0:
p(0) = 4(0) + (100(0))/(0^2+400) + 500 = 0 + 0 + 500 = 500

When t = 1:
p(1) = 4(1) + (100(1))/(1^2+400) + 500 = 4 + 1/5 + 500 = 504.2

When t = 2:
p(2) = 4(2) + (100(2))/(2^2+400) + 500 = 8 + 1/6 + 500 = 508.167

When t = 3:
p(3) = 4(3) + (100(3))/(3^2+400) + 500 = 12 + 1/7 + 500 = 511.143

2. Compare the population values obtained for t = 0, 1, 2, and 3 and identify the highest value.

From the values obtained above, the maximum population in the first three months occurs at t = 3, with a population of 511.143 thousand insects.