Urban planning- Applied Optimization

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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?
johnny172
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Re: Urban planning- Applied Optimization

Hello, johnny172!

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: Select all

: - - - - - - - - - 18  - - - - - - - - - :
A *-------------o---------------------------* B
: - -  x  - - Q - - - -  18-x - - - - - - :

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

$\text{The smog from plant }A\text{ is: }\:\frac{80}{x}$
. . $\text{The smog from plant }B\text{ is: }\:\frac{720}{18\,-\,x}$

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

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

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

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

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

. . $9x^2 \:=\:324\,-\,36x\,+\,x^2 \qquad\qquad\Rightarrow\qquad\qquad 8x^2 \,+\,36x\,-\,324\;=\;0$

. . $2x^2 \,+\, 9x \,-\, 81 \:=\:0 \qquad\qquad\Rightarrow\qquad\qquad (2x\,-\,9)(x\,+\,9) \:=\:0$

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

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

soroban
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Location: Lexington, MA

Re: Urban planning- Applied Optimization

thank you so much that helps so much!!! i owe you one...
johnny172
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Posts: 12
Joined: Thu Oct 29, 2009 4:43 pm