A restaurant chain sells 220,000 burritos each day when it charges $8.00 per burrito. For each $0.50 increase in price, the restaurant chain sells 10,000 less burritos.
a. How much should the restaurant chain charge to maximize daily revenue?
The restaurant chain should charge $ per burrito.
b. What is the maximum daily revenue?
The maximum daily revenue is $.
To solve this problem you need to focus on the fact that for each increment of the prize by 0.5 $ the restaurant sells 10000 less burritos. Thus, if x is the number of increments done, we can write:
Burrito’s prize = 8 + 0.5x
Sold burritos = 220000 - 10000x
Then, if you consider that the daily revenue of the restaurant is calculated by multiplying the sold burritos by their prize, we can express it as:
Daily revenue = (8 + 0.5x)(220000-10000*x)
which operating leads to:
Daily revenue = (8220000) + (0.5x220000) - (810000x)-(0.510000x^2)
Daily revenue = 1760000+110000x-80000x-5000x^2
Daily revenue = -5000x^2+30000x+1760000
In order to calculate the value of x for which the daily revenue is maximum, we need to establish it’s derivative value and then make it equal to 0
Considering that the derivative value of the daily revenue respect to x is;
d(Daily revenue)/dx = -10000x+30000
If we make this equal to 0, we find that the daily revenue is maximum for x=3 since:
-10000 x + 30000 = 0
-x + 3 = 0
x = 3
Thus, if we substitute the value calculated for x in the formula for the prize of the burritos we find:
Burrito’s prize = 8 + 0.5x = 8 + 0.53 = 9.5 $ which is the solution to part A
On the other hand, if we substitute the value calcualted for x in the formula for the daily revenue we obtain:
Daily revenue = -5000x^2+30000x+1760000=-50009+300003+1760000=1805000 $, which is the solution to part B