By using the concept of maxima and minima, the result obtained is
Concentration is minimum at
[tex]x = \frac{d}{1 + 9^{\frac{1}{3}}}[/tex]
What is maxima and minima of a function?
Suppose there is a function and a set of range is given.
Maxima gives the maximum value of the function within the range and Minima gives the minimum value of the function within the range.
Distance between two smokestacks = d miles
Distance of deposit from one smokestack = x miles
Distance of deposit from other smoke stack = (d - x) miles
Concentration of combined deposit
[tex]S = \frac{c}{x^2}+\frac{k}{(d - x)^2}\\[/tex]
Here k = 9c
[tex]S = \frac{c}{x^2}+\frac{9c}{(d - x)^2}\\[/tex]
[tex]\frac{ds}{dx} = \frac{d}{dx}(\frac{c}{x^2}+\frac{9c}{(d-x)^2})\\[/tex]
[tex]= -\frac{2c}{x^3}+\frac{18c}{(d-x)^3}\\[/tex]
For minimum distance
[tex]\frac{ds}{dx}=0[/tex]
[tex]= -\frac{2c}{x^3}+\frac{18c}{(d-x)^3} = 0[/tex]
[tex]\frac{18c}{(d - x)^3} = \frac{2c}{x^3}\\\\9x^3 = (d - x)^3\\\\(\frac{d - x}{x})^3 = 9\\\frac{d - x}{x} = 9^{\frac{1}{3}}\\\\ \frac{d}{x} - 1 = 9^{\frac{1}{3}}\\\\\frac{d}{x} = 1 + 9^{\frac{1}{3}}\\\\ x = \frac{d}{1 + 9^{\frac{1}{3}}}[/tex]
For checking maximum or minimum, double derivative has to be checked.
[tex]\frac{d^2s}{dx^2} = \frac{d}{dx}(-\frac{2c}{x^3} + \frac{18c}{(d - x)^3})\\[/tex]
[tex]= \frac{6c}{x^4} + \frac{54c}{(d - x)^4} > 0[/tex]
Hence the concentration is minimum
Concentration is minimum at
[tex]x = \frac{d}{1 + 9^{\frac{1}{3}}}[/tex]
To learn more about maxima and minima, refer to the link -
https://brainly.com/question/27958412
#SPJ4
Complete Question
The complete question has been attached here
