find the area of the trapezium whose parallel sides 24 cm and 20 cm and the distance between them is 15 cm.
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We presume your cost function is
c(p) = 124p/((10 +p)(100 -p))
This can be rewritten as
c(p) = (124/11)*(10/(100 -p) -1/(10 +p))
The average value of this function over the interval [50, 55] is given by the integral
This evaluates to
(-124/55)*(ln(65/60)+10ln(45/50)) ≈ 2.19494
The average cost of removal of 50-55% of pollutants is about
$2.19 hundred thousand = $219,000
c(p) = 124p/((10 +p)(100 -p))
This can be rewritten as
c(p) = (124/11)*(10/(100 -p) -1/(10 +p))
The average value of this function over the interval [50, 55] is given by the integral
This evaluates to
(-124/55)*(ln(65/60)+10ln(45/50)) ≈ 2.19494
The average cost of removal of 50-55% of pollutants is about
$2.19 hundred thousand = $219,000