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 October 29th, 2009, 03:00 PM #1 Newbie   Joined: Oct 2009 Posts: 12 Thanks: 0 Urban planning- Applied Optimization Two industrial plants, A and B, are located 18 miles apart, and each day, respectively emit 80 ppm (parts per million) and 720 ppm of particulate matter. Plant A is surounded b a restricted area of radius 1 mile, while the restricted area around plant B has a radius of 2 miles. The concentration of particulate matter arriving at any other point Q from each plant decreases with the reciprocal of the distance between that plant and Q Where should a house be located on a road joining the two plants to minimize the total concentration of particulate matter arriving from both plants?
October 29th, 2009, 09:09 PM   #2
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Re: Urban planning- Applied Optimization

Hello, johnny172!

Quote:
 Two industrial plants, A and B, are located 18 miles apart. Each day, respectively emit 80 ppm (parts per million) and 720 ppm of smog. The concentration of smog arriving at any other point Q from each plant decreases with the reciprocal of the distance between that plant and Q. Where should a house be located on a road joining the two plants to minimize the total concentration of smog arriving from both plants?
Code:
      : - - - - - - - - - 18  - - - - - - - - - :
A *-------------o---------------------------* B
: - -  x  - - Q - - - -  18-x - - - - - - :

$\text{The plants are located at }A\text{ and }B:\;AB= 18$
$\text{The house is located at }Q.$
$\text{Let: }\,x \:=\: AQ,\;18\,-\,x \:=\:QB$

$\text{The smog from plant }A\text{ is: }\:\frac{80}{x}$
[color=beige]. . [/color]$\text{The smog from plant }B\text{ is: }\:\frac{720}{18\,-\,x}$

$\text{The total smog at }Q\text{ is: }\;S \;=\;80x^{-1} \,+\, 720(18\,-\,x)^{-1}$

$\text{Differentiate and equate to zero:}$

[color=beige]. . [/color]$S' \;=\;-80x^{-2} \,-\, 720(18\,-\,x)^{-2}(-1) \;=\;0$

[color=beige]. . [/color]$-\frac{80}{x^2} \,+\,\frac{720}{(18\,-\,x)^2} \;=\;0 \qquad\qquad\Rightarrow\qquad\qquad \frac{720}{(18\,-\,x)^2} \;=\;\frac{80}{x^2}$

$\text{Cross-multiply: }\;720x^2 \;=\;80(18\,-\,x)^2 \qquad\qquad\Rightarrow\qquad\qquad 9x^2 \;=\;(18\,-\,x)^2$

[color=beige]. . [/color]$9x^2 \:=\:324\,-\,36x\,+\,x^2 \qquad\qquad\Rightarrow\qquad\qquad 8x^2 \,+\,36x\,-\,324\;=\;0$

[color=beige]. . [/color]$2x^2 \,+\, 9x \,-\, 81 \:=\:0 \qquad\qquad\Rightarrow\qquad\qquad (2x\,-\,9)(x\,+\,9) \:=\:0$

$\text{Hence: }\;x\:=\:\frac{9}{2}\qquad (x \,=\,-9)$

$\text{Therefore, the house should be located }4\frac{1}{2}\text{ miles from plant }A.$

 October 29th, 2009, 09:26 PM #3 Newbie   Joined: Oct 2009 Posts: 12 Thanks: 0 Re: Urban planning- Applied Optimization thank you so much that helps so much!!! i owe you one...

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