Suppose that the population (in thousands) of a certain kind of insect after t months is given by the following formula.

P(t) = 3t + sin(4t) + 100

Determine the minimum and maximum population in the first 4 months.

**Solution:**

The question that we’re really asking is to find the absolute extrema of P(t) on the interval [0,4].

**Step 1:** Find the derivative

P'(t) = 3 + 4cos(4t)

**Step 2:** Isolate t

3 + 4cos(4t) = 0

4cos(4t) = -3

cos(4t) = -3/4

4t = arccos(-3/4)

4t = 2.4189 + 2πn, n = 0, ±1, ±2, ... -> I get why it's 2.4189 and the degree is 138.59, but I don't get why it had + 2πn, n - 0, ±1, ±2, ...

4t = 3.8643 + 2πn, n= 0, ±1, ±2, ... -> I don't know where 3.8643 came from and I want to know it's degree too.

t = 2.4189/4 + 2πn, n= 0, ±1, ±2, ...

t = 3.8643/4 + 2πn, n= 0, ±1, ±2, ...

t = 0.6047 + πn/2, n= 0, ±1, ±2, ...

t = 0.9661 + πn/2, n= 0, ±1, ±2, ...

**Step 3:** Find the critical points

at n = 0 -> t = 0.6047, t = 0.9661 -> i don't get why we had to use positive n=0,1,2

at n = 1 -> t = 0.6047 + π/2 = 2.1755, t = 0.9661 + π/2 = 2.5369

at n = 2 -> t = 0.6047 + π = 3.7463, t = 0.9661 + π = 4.1077 -> because [0,4] we don't need this

**Step 4:** Plug critical points into the original function

P(t) = 3t + sin(4t) + 100

P(0) = 100

P(0.6047) = 102.4756

P(0.9661) = 102.2368

P(2.1755) = 107.1880

P(2.5369) = 106.9492

P(3.7463) = 111.9004

P(4) = 111.7121

* remember that P is in thousands

**Answer:**

Absolute Minimum Population is 100,000 which occurs at t = 0 or (0, 100)

Absolute Maximum Population is 111,900 which occurs at t = 3.7463 or (3.7463, 111.9004)