Chapter 2, Problem 125CR

a.

To determine

To determine the cost of seizing $25\%,50\%\text{ and }75\%$ of illegal drugs.

Answer to Problem 125CR

  $86$ (In million$), $86$ (In million$) and $774$ (In million$)

Explanation of Solution

Given:

  $C=\frac{258p}{100-p}\text{, }0\le p<100$ , where $p$ denotes seizing $p\%$ of drugs.

Calculation:

Given representation of the illegal drugs cost,

  $\Rightarrow C=\frac{258p}{100-p}\text{, }0\le p<100$

Now, for seizing $25\%$ of drugs,

  $\Rightarrow C=\frac{258\times 25}{100-25}\text{, }0\le 25<100$$\left[\because C=\frac{258p}{100-p}0\le p<100\right]$

  $\Rightarrow C=\frac{258\times 25}{100-25}\text{, }0\le 25<100$

  $\Rightarrow C=\frac{6450}{75}\text{, }0\le 25<100$

  $\Rightarrow C=86$ (In million$)

Now, for seizing $75\%$ of drugs,

  $\Rightarrow C=\frac{258\times 75}{100-75}\text{, }0\le 75<100$$\left[\because C=\frac{258p}{100-p}0\le p<100\right]$

  $\Rightarrow C=\frac{258\times 75}{25}\text{, }0\le 75<100$

  $\Rightarrow C=\frac{19350}{25}\text{, }0\le 25<100$

  $\Rightarrow C=774$ (In million$)

Now, for seizing $50\%$ of drugs,

  $\Rightarrow C=\frac{258\times 50}{100-50}\text{, }0\le 50<100$$\left[\because C=\frac{258p}{100-p}0\le p<100\right]$

  $\Rightarrow C=\frac{258\times 50}{50}\text{, }0\le 75<100$

  $\Rightarrow C=\frac{12900}{50}\text{, }0\le 25<100$

  $\Rightarrow C=258$ (In million$)Conclusion:

Therefore, the cost of seizing $25\%,50\%\text{ and }75\%$ of illegal drugs are $86$ (In million$), $86$ (In million$) and $774$ (In million$) respectively.

b.

To determine

To determine the graphical unit of the function and choosing an appropriate viewing window.

Explanation of Solution

Given:

  $C=\frac{258p}{100-p}\text{, }0\le p<100$ , where $p$ denotes seizing $p\%$ of drugs.

Calculation for graph:

Function: $C=\frac{258p}{100-p}\text{, }0\le p<100$

Consider $C=\frac{258p}{100-p},$

Values of p	Values of C
0	0
25	86
50	258

By taking different values of p , the graph can be plotted.

Graph:

  

The chosen window gives good description about the values of the graph. Here if one take a look the value of $C$ its $0$ same as value of $p$ .Interpretation:

Therefore, the discussed graphical unit of the function is having the appropriate viewing window.

c.

To determine

To determine possibility to seize $100\%$ of drugs.

Answer to Problem 125CR

No

Explanation of Solution

Given:

  $C=\frac{258p}{100-p}\text{, }0\le p<100$ , where $p$ denotes seizing $p\%$ of drugs.

Calculation:

Given representation of the illegal drugs cost,

$\Rightarrow C=\frac{258p}{100-p}\text{, }0\le p<100$

Now, for seizing $100\%$ of drugs, At first value range of $p$ is $0\le p<100$

So, $p\ne 100$ or $\ge 100$ .

Now calculating the cost

  $\Rightarrow C=\frac{258\times 100}{100-100}$$\left[\because C=\frac{258p}{100-p}0\le p<100\right]$

  $\Rightarrow C=\frac{25800}{100-100}$

Now here denominator becomes zero.

Thus it is not a real value of $C$ .

Conclusion:

Therefore, it is not possibility to seize $100\%$ of drugs.