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

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!Step 1:    
 
!Step 1:    
 
|-
 
|-
|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  
 
|To do this, we set  
Line 43: Line 43:
 
\end{array}</math>
 
\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
 
|Hence, we have
 
|-
 
|-
|<math>\rho(x) = \left\{
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\rho(x) = \left\{
 
     \begin{array}{lr}
 
     \begin{array}{lr}
 
       -x^2+6x+16 &  \text{if }0\le x \le 8\\
 
       -x^2+6x+16 &  \text{if }0\le x \le 8\\
Line 59: Line 61:
 
</math>
 
</math>
 
|-
 
|-
|The graph of <math>\rho(x)</math> is displayed below.
+
|The graph of &nbsp;<math>\rho(x)</math>&nbsp; is displayed below.
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!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
 
|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
 
|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
 
|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>
 
|}
 
|}
  
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\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
|So there are approximately <math>251</math> trout.
+
|Thus, there are approximately &nbsp;<math>251</math>&nbsp; trout.
 
|}
 
|}
  
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!Final Answer: &nbsp;  
 
!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>''']]
 
[[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 12:28, 3 March 2017

The population density of trout in a stream is

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \rho (x)=|-x^{2}+6x+16|}

where    is measured in trout per mile and    is measured in miles.    runs from 0 to 12.

(a) Graph    and find the minimum and maximum.

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

Foundations:  
What is the relationship between population density    and the total populations?
       The total population is equal to  
       for appropriate choices of  


Solution:

(a)

Step 1:  
To graph    we need to find out when    is negative.
To do this, we set
       
So, we have
       
Hence, we get    and   
But,    is outside of the domain of  
Using test points, we can see that    is positive in the interval  
and negative in the interval  
Hence, we have
       
The graph of    is displayed below.
Step 2:  
We need to find the absolute maximum and minimum of  
We begin by finding the critical points of  
Taking the derivative, we have  
Solving    we get a critical point at  .
Now, we calculate  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \rho (0),\rho (3),\rho (12).}
We have
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \rho (0)=16,\rho (3)=25,\rho (12)=56.}
Therefore, the minimum of    is    and the maximum of    is  

(b)

Step 1:  
To calculate the total number of trout, we need to find
       
Using the information from Step 1 of (a), we have
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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
       
Thus, there are approximately  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 251}   trout.


Final Answer:  
    (a)     The minimum of  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho(x)}   is  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 16}   and the maximum of  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho(x)}   is  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 56.}   (See Step 1 for graph)
    (b)     There are approximately  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 251}   trout.

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