A restaurant sells about 330 sandwiches each day at a price of $6 each. For each $0.25 decrease in price, 15 more sandwiches are sold per day. How much should the restaurant charge to maximize daily revenue? What is the maximum daily revenue?
To find the price that maximizes daily revenue, we need to determine the price at which the number of sandwiches sold multiplied by the price is the highest. Here's how we can solve it step by step:
1. Let's start with the initial price of $6 per sandwich. At this price, the restaurant sells about 330 sandwiches per day.
2. The problem states that for each $0.25 decrease in price, 15 more sandwiches are sold per day. So, for every $0.25 decrease from the initial price, the number of sandwiches sold will increase by 15.
3. To find the price that maximizes daily revenue, we need to consider both the price and the number of sandwiches sold. We can calculate the revenue for each potential price by multiplying the number of sandwiches sold by the price.
4. Let's create a table to track the number of sandwiches sold, the price, and the revenue for each potential price:
Price | Sandwiches Sold | Revenue
---------------------------------
$6 | 330 | $1,980
$5.75 | 345 | $1,988.75
$5.50 | 360 | $1,980
$5.25 | 375 | $1,968.75
$5.00 | 390 | $1,950
5. As we decrease the price by $0.25 in each row, the number of sandwiches sold increases by 15, and the revenue is calculated by multiplying the price by the number of sandwiches sold.
6. Based on the table, we can observe that the maximum revenue is achieved when the price is $5.75. At this price, 345 sandwiches are sold, resulting in a revenue of $1,988.75.
Therefore, the restaurant should charge $5.75 to maximize daily revenue, and the maximum daily revenue would be $1,988.75.
revenue = price * quantity, so if there are x price decreases, we have revenue is
y = (6-.25x)(330+15x)
= -3.75x^2 + 7.5x + 1980
Now, just find the vertex of that parabola to get the maximum revenue and its price. Recall that for
ax^2+bx+c
the vertex is at x = -b/2a