Question

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. 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 $.

Answer 1

Answer:

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

$200000-10000x$

Revenue after price change $= (6+0.5x) (200000-10000x)\\= 1200000+100000x-60000x-5000x^2\\= 1200000+40000x-5000x^2$

Now we can use calculus to find maximum

Let R(x) = $1200000+40000x-5000x^2$

$R'(x) = 40000-10000x\\R"(x) = -10000$

Equate first derivative to 0

we get x = 4

So maximum revenue is when 4 times price increased.

New price = $6+4(0.5) = 8$

Sales = $200000-40000 =160000$

Max revenue = $160000*8=1280000$.