# 009B Sample Final 3, Problem 3

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The population density of trout in a stream is

${\displaystyle \rho (x)=|-x^{2}+6x+16|}$

where  ${\displaystyle \rho }$  is measured in trout per mile and  ${\displaystyle x}$  is measured in miles.  ${\displaystyle x}$  runs from 0 to 12.

(a) Graph  ${\displaystyle \rho (x)}$  and find the minimum and maximum.

(b) Find the total number of trout in the stream.

Foundations:
What is the relationship between population density  ${\displaystyle \rho (x)}$  and the total populations?
The total population is equal to  ${\displaystyle \int _{a}^{b}\rho (x)~dx}$
for appropriate choices of  ${\displaystyle a,b.}$

Solution:

(a)

Step 1:
To graph  ${\displaystyle \rho (x),}$  we need to find out when  ${\displaystyle -x^{2}+6x+16}$  is negative.
To do this, we set
${\displaystyle -x^{2}+6x+16=0.}$
So, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {0}&=&\displaystyle {-x^{2}+6x+16}\\&&\\&=&\displaystyle {-(x^{2}-6x-16)}\\&&\\&=&\displaystyle {-(x+2)(x-8).}\end{array}}}$
Hence, we get  ${\displaystyle x=-2}$  and  ${\displaystyle x=8.}$
But,  ${\displaystyle x=-2}$  is outside of the domain of  ${\displaystyle \rho (x).}$
Using test points, we can see that  ${\displaystyle -x^{2}+6x+16}$  is positive in the interval  ${\displaystyle [0,8]}$
and negative in the interval  ${\displaystyle [8,12].}$
Hence, we have
${\displaystyle \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
The graph of  ${\displaystyle \rho (x)}$  is displayed below.
Step 2:
We need to find the absolute maximum and minimum of  ${\displaystyle \rho (x).}$
We begin by finding the critical points of
${\displaystyle -x^{2}+6x+16.}$
Taking the derivative, we get
${\displaystyle -2x+6.}$
Solving  ${\displaystyle -2x+6=0,}$  we get a critical point at
${\displaystyle x=3.}$
Now, we calculate  ${\displaystyle \rho (0),~\rho (3),~\rho (12).}$
We have
${\displaystyle \rho (0)=16,~\rho (3)=25,~\rho (12)=56.}$
Therefore, the minimum of  ${\displaystyle \rho (x)}$  is  ${\displaystyle 16}$  and the maximum of  ${\displaystyle \rho (x)}$  is  ${\displaystyle 56.}$

(b)

Step 1:
To calculate the total number of trout, we need to find
${\displaystyle \int _{0}^{12}\rho (x)~dx.}$
Using the information from Step 1 of (a), we have
${\displaystyle \int _{0}^{12}\rho (x)~dx=\int _{0}^{8}(-x^{2}+6x+16)~dx+\int _{8}^{12}(x^{2}-6x-16)~dx.}$
Step 2:
We integrate to get
${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{12}\rho (x)~dx}&=&\displaystyle {{\bigg (}{\frac {-x^{3}}{3}}+3x^{2}+16x{\bigg )}{\bigg |}_{0}^{8}+{\bigg (}{\frac {x^{3}}{3}}-3x^{2}-16x{\bigg )}{\bigg |}_{8}^{12}}\\&&\\&=&\displaystyle {{\bigg (}{\frac {-8^{3}}{3}}+3(8)^{2}+16(8){\bigg )}-0+{\bigg (}{\frac {(12)^{3}}{3}}-3(12)^{2}-16(12){\bigg )}-{\bigg (}{\frac {8^{3}}{3}}-3(8)^{2}-16(8){\bigg )}}\\&&\\&=&\displaystyle {8{\bigg (}{\frac {56}{3}}{\bigg )}+12{\bigg (}{\frac {12}{3}}{\bigg )}+8{\bigg (}{\frac {56}{3}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {752}{3}}.}\end{array}}}$
Thus, there are approximately  ${\displaystyle 251}$  trout.

(a)     The minimum of  ${\displaystyle \rho (x)}$  is  ${\displaystyle 16}$  and the maximum of  ${\displaystyle \rho (x)}$  is  ${\displaystyle 56.}$ (See above for graph.)
(b)     There are approximately  ${\displaystyle 251}$  trout.