Chapter 4.1, Problem 11CYU

To determine

To Find: The price that will maximize the income of the store.

Answer to Problem 11CYU

The maximum value of income is $3306.25$ dollars at the required price of $2.5.

Explanation of Solution

Given:

The store rents 1400 videos per week at $2.25 per videos.

The estimate is that 100 less videos are rented for each 0.25 dollar increase in the price.

Calculation:

Consider the number of price that increases the income is $x$ and

  $P\left(x\right)$ is the function of $x$ .

Then,

  $2.25+0.25x$

Since, the number of videos sold is $1400-100x$ then,

  $p(x)=\left(2.25+0.25x\right)\left(1400-100x\right)$

     $=3150-225x+350x-25x^2$

     $=-25x^2+125x+3150$

The above equation is the equation of the parabola with the coefficient of the $x^2$ as negative.

Consider the x coordinate is,

  $x=-\frac{b}{2a}$

   $=-\frac{125}{2\left(-25\right)}$

    $=2.5$

Then, the function is,

  $\begin{array}{c}p\left(2.5\right)=-25{\left(2.5\right)}^2+125\left(2.5\right)+3150\end{array}$

     $\begin{array}{c}=3306.25\end{array}$

Thus, the maximum value of income is $3306.25$ dollars at the required price of $2.5.