Given the cost of removing p percent of pollutant from a river is Smith County in dollars as
[tex]C(p)=\frac{71000p}{100-p}[/tex]
To find the cost of removing the pollutant for a particular percentage, we will substitute the value of the pollutant in the cost formula above.
Thus, for p equal to 20%
[tex]\begin{gathered} C(20)=\frac{71000\times20}{100-20}=\frac{1420000}{80} \\ C(20)=\text{ \$}17,750 \end{gathered}[/tex]
Hence, the cost of removing 20% of the pollutant is $17,750
The cost of removing half of the pollutants is equivalent to the cost of removing 50%, thus, p in percentage is 50%
[tex]\begin{gathered} C(50)=\frac{71000\times50}{100-50}=\frac{3550000}{50} \\ C(50)=\text{ \$}71,000 \end{gathered}[/tex]
Hence, the cost of removing half of the pollutants is $71,000
The cost of removing all but 5% of the pollutant is equivalent to the cost of removing 95% of the pollutants. Hence, p is 95
[tex]\begin{gathered} C(95)=\frac{71000\times95}{100-95}=\frac{6745000}{5} \\ C(95)=\text{ \$}1,349,000 \end{gathered}[/tex]
Hence, the cost of removing all but 5% of the pollutants is $1,349,000
