Question

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?

Answer 1

C = Number of Dozens Chocolate Chip Cookies (2 flour 1 egg)
P = Number of Dozens Peanut Butter Cookies (3 flour 4 eggs)

90 cups of flour" 2C + 3P <= 90

80 eggs

C + 4P <= 80

Profit on the chocolate chip cookies is $1 per dozen and on the peanut butter is $1.50 per dozen

Profit C(1.00) + P(1.50)

Answer 2

The number of chocolate chip cookies are: 288 cookies.

Number of peanut butter cookies are: 168 cookies.

Explanation:

Let x denote the number of dozen chocolate chip cookies.

Let y denote the number of dozen peanut butter cookies.

Now is is given that:

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.

Also,

The club has 90 cups of flour and 80 eggs on hand.

Hence, the inequality that can be formed with the help of these information is:

2x+3y≤90

and x+4y≤80

Also, the objective function is:

Max. z=x+1.50y

where z is the profit function which is to be maximized.

Hence, on solving we get boundary point at:

(0,0) ,(45,0) , (24,14) and (0,20)

But as they want to offer both types of cookies this means that the case :

(0,0) ,(45,0) and (0,20) are discarded.

hence, we are left with the point (24,14)

Hence, the profit is:

z=24+1.50×14

z=$ 45

Hence, the maximum profit is: $ 45.

Hence, number of chocolate chip cookies=24 dozens.

As 1 dozen=12

Hence, 24 dozen=24×12=288 cookies.

Similarly Number of peanut butter cookies =14 dozen

Hence, 14 dozen=168 cookies.