Difference between revisions of "009B Sample Final 3, Problem 3"
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|Solving <math style="vertical-align: -4px">-2x+6=0,</math> we get a critical point at <math style="vertical-align: 0px">x=3.</math> | |Solving <math style="vertical-align: -4px">-2x+6=0,</math> we get a critical point at <math style="vertical-align: 0px">x=3.</math> | ||
|- | |- | ||
| − | |Now, we calculate <math style="vertical-align: -5px">\rho(0),\rho(3),\rho(12).</math> | + | |Now, we calculate <math style="vertical-align: -5px">\rho(0),~\rho(3),~\rho(12).</math> |
|- | |- | ||
|We have | |We have | ||
|- | |- | ||
| − | | <math>\rho(0)=16,\rho(3)=25,\rho(12)=56.</math> | + | | <math>\rho(0)=16,~\rho(3)=25,~\rho(12)=56.</math> |
|- | |- | ||
|Therefore, the minimum of <math style="vertical-align: -5px">\rho(x)</math> is <math style="vertical-align: -1px">16</math> and the maximum of <math style="vertical-align: -5px">\rho(x)</math> is <math style="vertical-align: -1px">56.</math> | |Therefore, the minimum of <math style="vertical-align: -5px">\rho(x)</math> is <math style="vertical-align: -1px">16</math> and the maximum of <math style="vertical-align: -5px">\rho(x)</math> is <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 | ||
|- | |- | ||
| − | | <math> \int_0^{12} \rho(x)~dx | + | | <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 <math>251</math> trout. | + | |Thus, there are approximately <math style="vertical-align: -1px">251</math> trout. |
|} | |} | ||
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| '''(a)''' The minimum of <math style="vertical-align: -5px">\rho(x)</math> is <math style="vertical-align: -1px">16</math> and the maximum of <math style="vertical-align: -5px">\rho(x)</math> is <math style="vertical-align: -1px">56.</math> | | '''(a)''' The minimum of <math style="vertical-align: -5px">\rho(x)</math> is <math style="vertical-align: -1px">16</math> and the maximum of <math style="vertical-align: -5px">\rho(x)</math> is <math style="vertical-align: -1px">56.</math> | ||
|- | |- | ||
| − | | '''(b)''' There are approximately <math>251</math> trout. | + | | '''(b)''' There are approximately <math style="vertical-align: -1px">251</math> 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. |