the high school band wants to sell two types of cookies, chocolate chip and peanut butter, as a fundraiser. a dozen chocolate chip cookies requires 2 cups of flour and 1 egg. a dozen peanut butter cookies uses 3 cups of flour and 4 eggs. the club has 90 cups of flour and 80 eggs on hand. the profit on the chocolate chip cookies is $1 per dozen and on the peanut butter is $1.50 per dozen. if they want to offer both types of cookies, how many of each cookie should the club make to maximize profits? a. infeasible solutionsc.24 dozen chocolate chip14 dozen peanut butterb.30 dozen chocolate chip25 dozen peanut butterd.20 dozen chocolate chip18 dozen peanut butter

24 dozen chocolate chip

14 dozen peanut butter

Step-by-step explanation:

x = number of dozen chocolate chip cookies

y = number of dozen peanut butter cookies

Total number of cups of flour is:

2x + 3y ≤ 90

Total number of eggs is:

x + 4y ≤ 80

Total profit is:

P = x + 1.5y

Since this is multiple choice, one method would be to calculate the profit for each option, then choose the one that's largest.

But let's try solving this algebraically.  x and y are positive integers, and we want them to be as large as possible.

Let's start by assuming the solution is on the line 2x + 3y = 90.

x + 4y ≤ 80

2x + 8y ≤ 160

90 − 3y + 8y ≤ 160

5y ≤ 70

y ≤ 14

If y = 14, x = 24, and P = 45.

Now let's assume the solution is on the line x + 4y = 80.

2x + 3y ≤ 90

2 (80 − 4y) + 3y ≤ 90

160 − 8y + 3y ≤ 90

70 ≤ 5y

14 ≤ y

Therefore, to maximize the profit, they should bake 24 dozen chocolate chip cookies and 14 dozen peanut butter cookies.

the answer is c

hope this !

3 1/3 x 2 3/5 = 8 2/3

step-by-step explanation:

just multiply the first number by the second number to get your answer.