The population density of trout in a stream is
 $\rho (x)=x^{2}+6x+16$
where $\rho$ is measured in trout per mile and $x$ is measured in miles. $x$ runs from 0 to 12.
(a) Graph $\rho (x)$ and find the minimum and maximum.
(b) Find the total number of trout in the stream.
Foundations:

What is the relationship between population density $\rho (x)$ and the total populations?

The total population is equal to $\int _{a}^{b}\rho (x)~dx$

for appropriate choices of $a,b.$

Solution:
(a)
Step 1:

To graph $\rho (x),$ we need to find out when $x^{2}+6x+16$ is negative.

To do this, we set

$x^{2}+6x+16=0.$

So, we have

${\begin{array}{rcl}\displaystyle {0}&=&\displaystyle {x^{2}+6x+16}\\&&\\&=&\displaystyle {(x^{2}6x16)}\\&&\\&=&\displaystyle {(x+2)(x8).}\end{array}}$

Hence, we get $x=2$ and $x=8.$

But, $x=2$ is outside of the domain of $\rho (x).$

Using test points, we can see that $x^{2}+6x+16$ is positive in the interval $[0,8]$

and negative in the interval $[8,12].$

Hence, we have

$\rho (x)=\left\{{\begin{array}{ll}x^{2}+6x+16&{\text{if }}0\leq x\leq 8\\\,\,\,\,x^{2}6x16\qquad &{\text{if }}8<x\leq 12\end{array}}\right.$

The graph of $\rho (x)$ is displayed below.


Step 2:

We need to find the absolute maximum and minimum of $\rho (x).$

We begin by finding the critical points of

$x^{2}+6x+16.$

Taking the derivative, we get

$2x+6.$

Solving $2x+6=0,$ we get a critical point at

$x=3.$

Now, we calculate $\rho (0),~\rho (3),~\rho (12).$

We have

$\rho (0)=16,~\rho (3)=25,~\rho (12)=56.$

Therefore, the minimum of $\rho (x)$ is $16$ and the maximum of $\rho (x)$ is $56.$

(b)
Step 1:

To calculate the total number of trout, we need to find

$\int _{0}^{12}\rho (x)~dx.$

Using the information from Step 1 of (a), we have

$\int _{0}^{12}\rho (x)~dx=\int _{0}^{8}(x^{2}+6x+16)~dx+\int _{8}^{12}(x^{2}6x16)~dx.$

Step 2:

We integrate to get

${\begin{array}{rcl}\displaystyle {\int _{0}^{12}\rho (x)~dx}&=&\displaystyle {{\bigg (}{\frac {x^{3}}{3}}+3x^{2}+16x{\bigg )}{\bigg }_{0}^{8}+{\bigg (}{\frac {x^{3}}{3}}3x^{2}16x{\bigg )}{\bigg }_{8}^{12}}\\&&\\&=&\displaystyle {{\bigg (}{\frac {8^{3}}{3}}+3(8)^{2}+16(8){\bigg )}0+{\bigg (}{\frac {(12)^{3}}{3}}3(12)^{2}16(12){\bigg )}{\bigg (}{\frac {8^{3}}{3}}3(8)^{2}16(8){\bigg )}}\\&&\\&=&\displaystyle {8{\bigg (}{\frac {56}{3}}{\bigg )}+12{\bigg (}{\frac {12}{3}}{\bigg )}+8{\bigg (}{\frac {56}{3}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {752}{3}}.}\end{array}}$

Thus, there are approximately $251$ trout.

Final Answer:

(a) The minimum of $\rho (x)$ is $16$ and the maximum of $\rho (x)$ is $56.$ (See above for graph.)

(b) There are approximately $251$ trout.


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