009B Sample Final 3, Problem 3

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
${\begin{array}{rcl}\displaystyle {0}&=&\displaystyle {-x^{2}+6x+16}\\&&\\&=&\displaystyle {-(x^{2}-6x-16)}\\&&\\&=&\displaystyle {-(x+2)(x-8).}\end{array}}$
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
$\rho (x)=\left\{{\begin{array}{lr}-x^{2}+6x+16&{\text{if }}0\leq x\leq 8\\x^{2}-6x-16&{\text{if }}8<x\leq 12\end{array}}\right.$
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 have $-2x+6.$
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
$\int _{0}^{12}\rho (x)~dx.=\int _{0}^{8}(-x^{2}+6x+16)~dx+\int _{8}^{12}(x^{2}-6x-16)~dx.$

Step 2:
We integrate to get
${\begin{array}{rcl}\displaystyle {\int _{0}^{12}\rho (x)~dx}&=&\displaystyle {{\bigg (}{\frac {-x^{3}}{3}}+3x^{2}+16x{\bigg )}{\bigg |}_{0}^{8}+{\bigg (}{\frac {x^{3}}{3}}-3x^{2}-16x{\bigg )}{\bigg |}_{8}^{12}}\\&&\\&=&\displaystyle {{\bigg (}{\frac {-8^{3}}{3}}+3(8)^{2}+16(8){\bigg )}-0+{\bigg (}{\frac {(12)^{3}}{3}}-3(12)^{2}-16(12){\bigg )}-{\bigg (}{\frac {8^{3}}{3}}-3(8)^{2}-16(8){\bigg )}}\\&&\\&=&\displaystyle {8{\bigg (}{\frac {56}{3}}{\bigg )}+12{\bigg (}{\frac {12}{3}}{\bigg )}+8{\bigg (}{\frac {56}{3}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {752}{3}}.}\end{array}}$
So there are approximately $251$ trout.

Final Answer:
(a)     The minimum of $\rho (x)$ is $16$ and the maximum of $\rho (x)$ is $56.$ (See Step 1 for graph)
(b)     There are approximately $251$ trout.

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