The question that we’re really asking is to find the absolute extrema of \(P\left( t \right)\) on the interval \(\left[ {0,4} \right]\). Since this function is continuous everywhere we know we can do this.

Let’s start with the derivative.

\[P'\left( t \right) = 3 + 4\cos \left( {4t} \right)\]

We need the critical points of the function. The derivative exists everywhere so there are no critical points from that. So, all we need to do is determine where the derivative is zero.

\[\begin{align*}3 + 4\cos \left( {4t} \right) & = 0\\ \cos \left( {4t} \right) & = - \frac{3}{4}\end{align*}\]

The solutions to this are,

\[\begin{array}{*{20}{c}}{4t = 2.4189 + 2\pi n,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots }\\{4t = 3.8643 + 2\pi n,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots }\end{array}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{array}{*{20}{c}}{t = 0.6047 + \displaystyle \frac{{\pi n}}{2},\,\,\,\,n = 0, \pm 1, \pm 2, \ldots }\\{t = 0.9661 + \displaystyle \frac{{\pi n}}{2},\,\,\,\,n = 0, \pm 1, \pm 2, \ldots }\end{array}\]

So, these are all the critical points. We need to determine the ones that fall in the interval \(\left[ {0,4} \right]\). There’s nothing to do except plug some \(n\)’s into the formulas until we get all of them.

\(n = 0\) :

\[t = 0.6047\hspace{1.0in}t = 0.9661\]

We’ll need both of these critical points.

\(n = 1\)

\[t = 0.6047 + \frac{\pi }{2} = 2.1755\hspace{1.0in}t = 0.9661 + \frac{\pi }{2} = 2.5369\]

We’ll need these.

\(n = 2\)

\[t = 0.6047 + \pi = 3.7463\hspace{1.0in}t = 0.9661 + \pi = 4.1077\]

In this case we only need the first one since the second is out of the interval.

There are five critical points that are in the interval. They are,

\[0.6047,\,\,\,0.9661,\,\,\,2.1755,\,\,\,2.5369,\,\,\,3.7463\]

Finally, to determine the absolute minimum and maximum population we only need to plug these values into the function as well as the two end points. Here are the function evaluations.

\[\begin{align*}P\left( 0 \right) & = 100.0\hspace{1.0in} & P\left( 4 \right) & = 111.7121\\
P\left( {0.6047} \right) & = 102.4756\hspace{1.0in} & P\left( {0.9661} \right) & = 102.2368\\
P\left( {2.1755} \right) & = 107.1880\hspace{1.0in} & P\left( {2.5369} \right) & = 106.9492\\
P\left( {3.7463} \right) & = 111.9004 & & \end{align*}\]

From these evaluations it appears that the minimum population is 100,000 (remember that \(P\) is in thousands…) which occurs at \(t = 0\) and the maximum population is 111,900 which occurs at \(t = 3.7463\).