**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 $.**

**Answer:**

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 + 0.5*x
Sold burritos = 220000 - 10000*x

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 + 0.5*x)*(220000-10000*x)

which operating leads to:

Daily revenue = (8*220000) + (0.5x*220000) - (8*10000x)-(0.5*10000x^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 + 0.5*x = 8 + 0.5*3 = 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=-5000*9+30000*3+1760000=1805000 $, which is the solution to part B