Chapter 3.1, Problem 57E

To determine

To calculate:

The maximize daily revenue and its charges

Answer to Problem 57E

Maximize daily revenue and its charges are $\$1983.75$ and $\$5.75$ respectively.

Explanation of Solution

Given information:

Price of sandwiches: $P=\$6$

Sell per day: $S=15$

Calculation:

The current revenue at the restaurant is estimated as:

  $R=\$6\times 330=\$1980$

Consider,

  $x$ : number of times the price to be dropped by $\$0.25$

So, the new price of the sandwich is calculated as:

  $P\text{'}=6-0.25x$

The new sell of sandwich per day is calculated as:

  $S\text{'}=330+15x$

The new revenue of the restaurant is:

  $R\text{'}=\left(6-0.25x\right)\left(330+15x\right)$

Here, need to find maximize the current revenue of the restaurant use the above new revenue equation to make it as complete square:

  $\begin{array}{l}R\text{'}=\left(6-0.25x\right)\left(330+15x\right)\\ R\text{'}=1980-82.5x+90x-3.75x^2\\ R\text{'}=1980-3.75\left(-2x+x^2\right)\\ R\text{'}=1980-3.75\left(1-1-2x+x^2\right)\\ R\text{'}=1980+3.75-3.75\left(x^2-2x-1\right)\end{array}$

  $R\text{'}=1983.75-3.75{\left(x-1\right)}^2$........(1)

Since, ${\left(x-1\right)}^2$ is always positive, need to minimize the second term to maximize the total revenue.

  $\begin{array}{l}{\left(x-1\right)}^2=0\\ x^2=1\\ x=1\end{array}$

From eq. (1):

  $\begin{array}{l}R\text{'}=1983.75-3.75\left(0\right)\\ R\text{'}=\$1983.75\end{array}$

Hence, the maximize daily revenue is calculated as $\$1983.75$

Now, the charges need to maximize daily revenue is calculated as:

  $\begin{array}{l}P\text{'}=6-0.25x\\ P\text{'}=6-0.25\left(x\right)\\ P\text{'}=6-0.25\\ P\text{'}=\$5.75\end{array}$