SOLUTION: Oh my gosh, I need help again. Here is the problem. The annual cost in dollars for removing p% of the toxic chemicals froma town's water supply is given by this formula. C(p) 500

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Oh my gosh, I need help again. Here is the problem. The annual cost in dollars for removing p% of the toxic chemicals froma town's water supply is given by this formula. C(p) 500      Log On


   

Question 178411This question is from textbook Elementary and Intermediate Algebra
: Oh my gosh, I need help again. Here is the problem.
The annual cost in dollars for removing p% of the toxic chemicals froma town's water supply is given by this formula. C(p) 500,000/100-p.
A. Estimate the cost for removing 90% and 95% of the toxic chemicals.
B. Use the formula to find C(99.5) and C(99.9).
C. What happens to the cost as the percentage of pollutants removed approaches 100% This question is from textbook Elementary and Intermediate Algebra

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

a)

C%28p%29=%28500000%29%2F%28100-p%29 Start with the given function


C%2890%29=%28500000%29%2F%28100-90%29 Plug in p=90


C%2890%29=%28500000%29%2F%2810%29 Subtract


C%2890%29=50000 Divide


So it costs $50,000 to remove 90% of the toxic chemicals.

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C%28p%29=%28500000%29%2F%28100-p%29 Start with the given function


C%2895%29=%28500000%29%2F%28100-95%29 Plug in p=95


C%2895%29=%28500000%29%2F%285%29 Subtract


C%2895%29=100000 Divide


So it costs $100,000 to remove 95% of the toxic chemicals.


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b)


C%28p%29=%28500000%29%2F%28100-p%29 Start with the given function


C%2899.5%29=%28500000%29%2F%28100-99.5%29 Plug in p=99.5


C%2899.5%29=%28500000%29%2F%280.5%29 Subtract


C%2899.5%29=1000000 Divide


So it costs $1,000,000 to remove 99.5% of the toxic chemicals.


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C%28p%29=%28500000%29%2F%28100-p%29 Start with the given function


C%2899.9%29=%28500000%29%2F%28100-99.9%29 Plug in p=99.9


C%2899.9%29=%28500000%29%2F%280.1%29 Subtract


C%2899.9%29=5000000 Divide


So it costs $5,000,000 to remove 99.9% of the toxic chemicals.


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c)

As the percentage approaches 100, the cost increases dramatically. Notice how the cost increased tenfold (ie multiplied by 10) as "p" changed from 95 to 99.5

Also, take note that "p" CANNOT be equal to 100. If p was equal to 100, then there would be a division by zero (which is undefined). So this means that we CANNOT completely remove the toxic chemicals.