009B Sample Final 3, Problem 3
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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 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 |
| We have |
| Therefore, the minimum of is and the maximum of is |
(b)
| Step 1: |
|---|
| To calculate the total number of trout, we need to find |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{0}^{12}\rho (x)~dx.} |
| 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 |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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. |