Question

(MEDAL + FAN!) the best highschool is running a fundriser, selling hats with their awesome logo. The hats have been donated by a local merchandise dealer. A survey determines that 50 students are willing to pay $5 for a hat. For every decrease of $.25 in the sale price, 10 more students are expected to purchase a hat. The school must raise at least $390 to meet their goal, but would prefer if as many students as possible purchase a hat. At what price should the hats be sold?

Answer 1

@mathmate @acxbox22

Answer 2

heres what i got:

Answer 3

let x= # of $.25 decreases in sale price
price= #students x price
(50+10x)(5-.25x) > 390 -----> the income

Answer 4

how do i account for the at least 390 and the most students possible?

Answer 5

i think the answer was around $3 if that helps

Answer 6

Hint:
You have correctly defined x as the \(number\) of 0.25 decreases.
You have correctly solved the maximum revenue would be obtained as 7.5 decreases.
Unfortunately you have interpreted 7.5 as the dollar amount.
Since x can only take on integer values, you need to verify the value of x (around 7.5) that gives the maximum revenue, and the maximum number of students.
Check those out, and post if necessary.

Btw OS rules stipulate that students should not ask for help from OS for school tests.

Answer 7

7.5 is the number of decreases?

Answer 8

Please note answer above:
"Since x can only take on integer values, you need to verify the value of x (around 7.5) that gives the maximum revenue, and the maximum number of students."

Answer 9

x=7 because when i plug it in, give gives 390

Answer 10

oh thank, i got $3.25

Answer 11

I suggest you read completely and understand answers to your posts. There is still time to reread my previous answer.