SOLUTION: A restaurant sells 330 sandwiches each day. For each $0.25 decrease in price, the restaurant sells about 15 more sandwiches. How much should the restaurant charge to maximize daily

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: A restaurant sells 330 sandwiches each day. For each $0.25 decrease in price, the restaurant sells about 15 more sandwiches. How much should the restaurant charge to maximize daily      Log On


   

Question 995488: A restaurant sells 330 sandwiches each day. For each $0.25 decrease in price, the restaurant sells about 15 more sandwiches. How much should the restaurant charge to maximize daily revenue? What is the maximum daily revenue? (All sandwiches are $6).
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +n+ = the number of $.25 decreases in price
Let +R+ = daily revenue in dollars
given:
+R+=+%28+330+%2B+15n+%29%2A%28+6+-+.25n+%29+
+R+=+1980+%2B+90n+-+82.5n+-+3.75n%5E2+
+R+=+-3.75n%5E2+%2B+7.5n+%2B+1980+
+R%5Bmax%5D+ is at +n+=+-b%2F%282a%29+ where
+b+=+7.5+
+a+=+-3.75+
---------------
+-b%2F%282a%29+=+-7.5%2F%28+2%2A%28-3.75%29+%29+
+-b%2F%282a%29+=+1+
When +n+=+1+, +R%28n%29+ is maximum
+R%28n%29+=+-3.75n%5E2+%2B+7.5n+%2B+1980+
+R%281%29+=+-3.75%2A1%5E2+%2B+7.5%2A1+%2B+1980+
+R%281%29+=+-3.75+%2B+7.5+%2B+1980+
+R%281%29+=+1983.75+
The maximm revenue is $1,983.75
with 1 decrease in price
---------------------
check:
Here's the plot:
+graph%28+400%2C+400%2C+-30%2C+30%2C+-300%2C+2200%2C+-3.75x%5E2+%2B+7.5x+%2B+1980+%29+
----------------------
To further check:
If +n+=+2+, then +R%282%29+ should be less
than +R%281%29+
--------------
+R%28n%29+=+-3.75n%5E2+%2B+7.5n+%2B+1980+
+R%282%29+=+-3.75%2A2%5E2+%2B+7.5%2A2+%2B+1980+
+R%282%29+=+-15+%2B+15+%2B+1980+
+R%282%29+=+1980+
OK