Chapter 1.4, Problem 65PS

Solution

To find: the price that the restaurant should charge to maximize daily revenue and find the maximum daily revenue.

To maximize revenue, each sandwich should be sold at $5.75 and the maximum daily revenue is $\$1,983.75$

Given information:

The restaurant sells about 330 sandwiches each day at price $6 each.

For each $0.25 decrease in price, about 15 more Sandwiches per month are sold.

Property Used:

Factoring and Zeros:

To find the maximum or minimum value of a quadratic function, first use factoring to write the function in intercept form $y=a\left(x-p\right)\left(x-q\right)$ . Because the function’s vertex lies on the axis of symmetry $x=\frac{p+q}{2}$ , the maximum or minimum occurs at the average of the zeros p and q .

Calculation:

Let x represent the price decrease and $R\left(x\right)$ represents the daily revenue.

Now, the required verbal model is:

Daily revenue(dollars) = Number of Sandwiches sold $\times$ price of each Sandwich

So, the quadratic equation is:

  $\begin{array}{l}R\left(x\right)=\left(330+15x\right)\times \left(6-0.25x\right)\\ R\left(x\right)=\left(15x+330\right)\left(-0.25x+6\right)\\ R\left(x\right)=15\left(x+22\right)\left[-0.25\left(x-24\right)\right]\\ R\left(x\right)=-3.75\left(x+22\right)\left(x-24\right)\end{array}$

Now, it is clear that here the zeros of the revenue function are -22 and 24.

The average of zeros is $\frac{24+\left(-22\right)}{2}=1$

So, to maximize revenue, each subscription should cost

  $\begin{array}{c}6-0.25x=6-0.25\times 1\\ =6-0.25\\ =\$5.75\end{array}$

The maximum revenue is:

  $\begin{array}{c}R\left(1\right)=-3.75\left(1+22\right)\left(1-24\right)\\ =-3.75\times 23\times \left(-23\right)\\ =\$1983.75\end{array}$