Arestaurant chain sells 200,000 burritos each day when it charges $6.00 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? b. the maximum daily revenue is $.

1280000

Step-by-step explanation:

Given that a restaurant chain sells 200,000 burritos each day when it charges $6.00 per burrito.

For each $0.50 increase in price, the restaurant chain sells 10,000 less burritos.

If x is the number of times price increased then sales would be

Revenue after price change

Now we can use calculus to find maximum

Let R(x) =

Equate first derivative to 0

we get x = 4

So maximum revenue is when 4 times price increased.

New price =

Sales =

Max revenue = .

50x5=$250

step-by-step explanation:

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116