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.

ExpandFoundations:
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)

ExpandStep 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
$\rho (x)=\left\{{\begin{array}{ll}-x^{2}+6x+16&{\text{if }}0\leq x\leq 8\\\,\,\,\,x^{2}-6x-16\qquad &{\text{if }}8<x\leq 12\end{array}}\right.$
The graph of  $\rho (x)$  is displayed below.

ExpandStep 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)

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

ExpandStep 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.

ExpandFinal 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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