Chapter 2.4, Problem 34E

To determine

The lowest charge of the two parks you have to spend for staying four or more nights.

Explanation of Solution

Given information:

Park A charges a $\$100$ membership fee plus $\$35$ per night.

Park B charges a $\$20$ membership fee plus $\$55$ per night.

Calculations:

Let the number of days spend be $x$ .

Therefore charges for park A is

  $\begin{array}{l}=100+(x\times 35)\\ =35x+100\end{array}$

Similarly charges for park B is

  $\begin{array}{l}=20+(x\times 55)\\ =55x+20\end{array}$

Now if,

  $\begin{array}{l}35x+100=55x+20\\ 20x=80\\ x=4\end{array}$

Therefore we can see that if a person stays for four days then he has to spend equal amount for both the parks in which he stays.

If a person stays for less than four days then he has to spend more money if he stays in park A than park B.

If a person stays for more than four days then he has to less more money if he stays in park A than park B.

Let us check,

Let the number of days spend be $3$ .

Therefore charges for park A is

  $\begin{array}{l}=100+(3\times 35)\\ =205\end{array}$

Similarly charges for park B is

  $\begin{array}{l}=20+(3\times 55)\\ =185\end{array}$

So, charges for park A is more than charges for park B for $3$ days spend.

Let the number of days spend be $4$ .

Therefore charges for park A is

  $\begin{array}{l}=100+(4\times 35)\\ =240\end{array}$

Similarly charges for park B is

  $\begin{array}{l}=20+(4\times 55)\\ =240\end{array}$

So, charges for park A are equal to charges for park B for $4$ days spend.

Let the number of days spend be $5$ .

Therefore charges for park A is

  $\begin{array}{l}=100+(5\times 35)\\ =275\end{array}$

Similarly charges for park B is

  $\begin{array}{l}=20+(5\times 55)\\ =295\end{array}$

So, charges for park A are less than charges for park B for $5$ days spend.

Thus it is checked.