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Question

A city's main well was recently found to be contaminated with trichloroethylene, a cancer-causing chemical, as a result of an abandoned chemical dump leaching chemicals into the water. A proposal submitted to city council members indicates that the cost, measured in millions of dollars, of removing x% of the toxic pollutant is given by C(x)=0.5x100xC(x)=\frac{0.5 x}{100-x} C(x)=0.5x100xC(x)=\frac{0.5 x}{100-x}

Find the cost of removing 50%, 60%, 70%,80%, 90%, and 95% of the pollutant.

Solution

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Answered 1 year ago
Answered 1 year ago
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Given:

C(x)=0.5x100x with 0<x<100C(x)=\frac{0.5x}{100-x}\text{ with }0<x<100

Thus

C(50)=0.55010050=0.5C(60)=0.56010060=0.75C(70)=0.57010070=1.17C(80)=0.58010080=2C(90)=0.59010090=4.5C(95)=0.59510095=9.5\begin{align*} C(50)&=\frac{0.5\cdot 50}{100-50}=\color{#4257b2}{0.5}\\ C(60)&=\frac{0.5\cdot 60}{100-60}=\color{#4257b2}{0.75}\\ C(70)&=\frac{0.5\cdot 70}{100-70}=\color{#4257b2}{1.17}\\ C(80)&=\frac{0.5\cdot 80}{100-80}=\color{#4257b2}{2}\\ C(90)&=\frac{0.5\cdot 90}{100-90}=\color{#4257b2}{4.5}\\ C(95)&=\frac{0.5\cdot 95}{100-95}=\color{#4257b2}{9.5}\\ \end{align*}

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