Difference between revisions of "009B Sample Final 3, Problem 3"

Revision as of 13:28, 3 March 2017

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
${\begin{array}{rcl}\displaystyle {\int _{0}^{12}\rho (x)~dx}&=&\displaystyle {{\bigg (}{\frac {-x^{3}}{3}}+3x^{2}+16x{\bigg )}{\bigg \|}_{0}^{8}+{\bigg (}{\frac {x^{3}}{3}}-3x^{2}-16x{\bigg )}{\bigg \|}_{8}^{12}}\\&&\\&=&\displaystyle {{\bigg (}{\frac {-8^{3}}{3}}+3(8)^{2}+16(8){\bigg )}-0+{\bigg (}{\frac {(12)^{3}}{3}}-3(12)^{2}-16(12){\bigg )}-{\bigg (}{\frac {8^{3}}{3}}-3(8)^{2}-16(8){\bigg )}}\\&&\\&=&\displaystyle {8{\bigg (}{\frac {56}{3}}{\bigg )}+12{\bigg (}{\frac {12}{3}}{\bigg )}+8{\bigg (}{\frac {56}{3}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {752}{3}}.}\end{array}}$
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 Step 1 for graph)
(b) There are approximately $251$ trout.

Return to Sample Exam

@@ Line 27: / Line 27: @@
 !Step 1: &nbsp;
 |-
-|To graph <math>\rho(x),</math> we need to find out when <math>-x^2+6x+16</math> is negative.
+|To graph &nbsp;<math>\rho(x),</math>&nbsp; we need to find out when &nbsp;<math>-x^2+6x+16</math>&nbsp; is negative.
 |-
 |To do this, we set
@@ Line 43: / Line 43: @@
 \end{array}</math>
 |-
-|Hence, we get <math>x=-2</math> and <math>x=8.</math> But, <math>x=-2</math> is outside of the domain of <math>\rho(x).</math>
+|Hence, we get &nbsp;<math>x=-2</math>&nbsp; and &nbsp;<math>x=8.</math>&nbsp;
 |-
-|Using test points, we can see that <math>-x^2+6x+16</math> is positive in the interval <math>[0,8]</math>
+|But, &nbsp;<math>x=-2</math>&nbsp; is outside of the domain of &nbsp;<math>\rho(x).</math>
 |-
-|and negative in the interval <math>[8,12].</math>
+|Using test points, we can see that &nbsp;<math>-x^2+6x+16</math>&nbsp; is positive in the interval &nbsp;<math>[0,8]</math>
+|-
+|and negative in the interval &nbsp;<math>[8,12].</math>
 |-
 |Hence, we have
 |-
-|<math>\rho(x) = \left\{
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\rho(x) = \left\{
       \begin{array}{lr}
         -x^2+6x+16 &  \text{if }0\le x \le 8\\
@@ Line 59: / Line 61: @@
 </math>
 |-
-|The graph of <math>\rho(x)</math> is displayed below.
+|The graph of &nbsp;<math>\rho(x)</math>&nbsp; is displayed below.
 |}
@@ Line 65: / Line 67: @@
 !Step 2: &nbsp;
 |-
-|We need to find the absolute maximum and minimum of <math>\rho(x).</math>
+|We need to find the absolute maximum and minimum of &nbsp;<math>\rho(x).</math>
 |-
-|We begin by finding the critical points of <math>-x^2+6x+16.</math>
+|We begin by finding the critical points of &nbsp;<math>-x^2+6x+16.</math>
 |-
-|Taking the derivative, we have <math>-2x+6.</math>
+|Taking the derivative, we have &nbsp;<math>-2x+6.</math>
 |-
-|Solving <math>-2x+6=0,</math> we get a critical point at <math>x=3</math>.
+|Solving &nbsp;<math>-2x+6=0,</math>&nbsp; we get a critical point at &nbsp;<math>x=3</math>.
 |-
-|Now, we calculate <math>\rho(0),\rho(3),\rho(12).</math>
+|Now, we calculate &nbsp;<math>\rho(0),\rho(3),\rho(12).</math>
 |-
 |We have
 |-
-|<math>\rho(0)=16,\rho(3)=25,\rho(12)=56.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\rho(0)=16,\rho(3)=25,\rho(12)=56.</math>
 |-
-|Therefore, the minimum of <math>\rho(x)</math> is <math>16</math> and the maximum of <math>\rho(x)</math> is <math>56.</math>
+|Therefore, the minimum of &nbsp;<math>\rho(x)</math>&nbsp; is &nbsp;<math>16</math>&nbsp; and the maximum of &nbsp;<math>\rho(x)</math>&nbsp; is &nbsp;<math>56.</math>
 |}
@@ Line 89: / Line 91: @@
 |To calculate the total number of trout, we need to find
 |-
-|<math> \int_0^{12} \rho(x)~dx.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math> \int_0^{12} \rho(x)~dx.</math>
 |-
 |Using the information from Step 1 of (a), we have
 |-
-|<math> \int_0^{12} \rho(x)~dx.=\int_0^8 (-x^2+6x+16)~dx+\int_8^{12} (x^2-6x-16)~dx.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math> \int_0^{12} \rho(x)~dx.=\int_0^8 (-x^2+6x+16)~dx+\int_8^{12} (x^2-6x-16)~dx.</math>
 |}
@@ Line 111: / Line 113: @@
 \end{array}</math>
 |-
-|So there are approximately <math>251</math> trout.
+|Thus, there are approximately &nbsp;<math>251</math>&nbsp; trout.
 |}
@@ Line 118: / Line 120: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The minimum of <math>\rho(x)</math> is <math>16</math> and the maximum of <math>\rho(x)</math> is <math>56.</math> (See Step 1 for graph)
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The minimum of &nbsp;<math>\rho(x)</math>&nbsp; is &nbsp;<math>16</math>&nbsp; and the maximum of &nbsp;<math>\rho(x)</math>&nbsp; is &nbsp;<math>56.</math>&nbsp; (See Step 1 for graph)
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; There are approximately <math>251</math> trout.
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; There are approximately &nbsp;<math>251</math>&nbsp; trout.
 |-
 |
 |}
 [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009B Sample Final 3, Problem 3"

Revision as of 13:28, 3 March 2017

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