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

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|Solving &nbsp;<math style="vertical-align: -4px">-2x+6=0,</math>&nbsp; we get a critical point at &nbsp;<math style="vertical-align: 0px">x=3.</math>
 
|Solving &nbsp;<math style="vertical-align: -4px">-2x+6=0,</math>&nbsp; we get a critical point at &nbsp;<math style="vertical-align: 0px">x=3.</math>
 
|-
 
|-
|Now, we calculate &nbsp;<math style="vertical-align: -5px">\rho(0),\rho(3),\rho(12).</math>
+
|Now, we calculate &nbsp;<math style="vertical-align: -5px">\rho(0),~\rho(3),~\rho(12).</math>
 
|-
 
|-
 
|We have
 
|We have
 
|-
 
|-
|&nbsp; &nbsp; &nbsp; &nbsp;<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 &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">16</math>&nbsp; and the maximum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">56.</math>
 
|Therefore, the minimum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">16</math>&nbsp; and the maximum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">56.</math>
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|Using the information from Step 1 of (a), we have
 
|Using the information from Step 1 of (a), we have
 
|-
 
|-
|&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>
+
|&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>
 
|-
 
|-
|Thus, there are approximately &nbsp;<math>251</math>&nbsp; trout.
+
|Thus, there are approximately &nbsp;<math style="vertical-align: -1px">251</math>&nbsp; trout.
 
|}
 
|}
  
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|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The minimum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">16</math>&nbsp; and the maximum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">56.</math>
 
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The minimum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">16</math>&nbsp; and the maximum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">56.</math>
 
|-
 
|-
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; There are approximately &nbsp;<math>251</math>&nbsp; trout.
+
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; There are approximately &nbsp;<math style="vertical-align: -1px">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 12:38, 3 March 2017

The population density of trout in a stream is

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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \rho (x),}   we need to find out when    is negative.
To do this, we set
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -x^{2}+6x+16=0.}
So, we have
        Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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    and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=8.}  
But,    is outside of the domain of  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \rho (x).}
Using test points, we can see that    is positive in the interval  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle [0,8]}
and negative in the interval  
Hence, we have
       Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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    is displayed below.
Step 2:  
We need to find the absolute maximum and minimum of  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \rho (x).}
We begin by finding the critical points of  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -x^{2}+6x+16.}
Taking the derivative, we have  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -2x+6.}
Solving  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -2x+6=0,}   we get a critical point at  
Now, we calculate  
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  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.}

(b)

Step 1:  
To calculate the total number of trout, we need to find
       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 \int_0^{12} \rho(x)~dx.}
Using the information from Step 1 of (a), we have
       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 \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
        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 \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  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.}
    (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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