Example  2
(Video in Development)

A company in Edmonton rents Segways to tour the river valley. The company charges \(\$60\) per day for each Segway rental. At this price, the company rents \(4\)8 Segways a week. An analyst predicts that for every \(\$5\) increase in rental price, \(2\) fewer Segways will be rented each week.

  1. Determine the rental price(s) that will give a weekly revenue of \(\$3 \thinspace 000.00\). Round to the nearest hundredth.

    Step 1: Write what you know.

    • Current: \(\$60\rm{/Segway}\) rental per day, \(48\) rentals/week
    • A rental increase of \(\$5\) each will decrease rental numbers by two per week.
    • Looking for the rental price(s) that will give a weekly revenue of \(\$3 \thinspace 000.00\).

    Step 2: Analyze how to solve the problem. Revenue = rental price \(\times\) number of rentals

    Both the rental price and the number of rentals can be represented in terms of the number of \(\$5\) price increases. Let \(p\) represent the number of \(\$5\) price increases. This means

    rental price = \(60 + 5p\)
    number of rentals = \(48 - 2p\)
    Revenue = \(3 \thinspace 000\)

    Using the new expressions for rental price and number of rentals, write the equation that represents the revenue after the increase in price. Simplify until the equation equals zero.

    \(\begin{align}
     {\rm{Revenue}}\thinspace &= \thinspace {\rm{rental \thinspace price}} \thinspace \times \thinspace {\rm{number \thinspace of \thinspace rentals}} \\
     3\thinspace 000 &= \left( {60 + 5p} \right)\left( {48 - 2p} \right) \\
     3\thinspace 000 &= 2\thinspace 880 - 120p + 240p - 10p^2  \\
     0 &= -120 + 120p - 10p^2  \\
     \end{align}\)

    Step 3: Solve for \(p\) using an appropriate method.

    First remove a GCF.

    \(\begin{align}
     0 &= -120 + 120p - 10p^2  \\
     0 &= -10\left( {p^2 - 12p + 12} \right) \\
     \end{align}\)

    The trinomial in the brackets is not easily factorable, so use the quadratic formula.

    \[\begin{align}
     a &= 1, b = -12, c = 12 \\
      p &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
      p &= \frac{{12 \pm \sqrt {\left( { - 12} \right)^2 - 4\left( 1 \right)\left( {12} \right)} }}{{2\left( 1 \right)}} \\
      p &= \frac{{12 \pm \sqrt {144 - 48} }}{2} \\
      p &= \frac{{12 \pm \sqrt {96} }}{2} \\
      p &= \frac{{12 \pm 4\sqrt 6 }}{2} \\
      p &= 6 \pm 2\sqrt 6  \\
      p &= 10.898...\thinspace {\rm{and} } \thinspace p = 1.101... \\
     \end{align}\]

    There are two possible rental prices.

    \(60 + 5\left( {6 - 2\sqrt 6 } \right) = 65.505...\) or \(60 + 5\left( {6 + 2\sqrt 6 } \right) = 114.494...\)

    The new price can be either \(\$65.51\) or \(\$114.49\) to give a revenue of \(\$3 \thinspace 000.00\).


  2. Will the price increase(s) make logical sense in a real-life context?

    While it does make logical sense to increase the price by just over \(\$5.00\), it is unlikely that a price increase of close to \(\$55.00\) is sustainable in real-life. For a new price of \(\$114.49\), the number of customers would decrease significantly. And, it is possible that the people of Edmonton may not be able to support this business in the long term with such a high price.