Question

College student makes decorative Styrofoam Lions and Husky's which she sells each year. Each one uses 8 oz of Styrofoam and takes 35 minutes to make. Each husky uses 12 oz of Styrofoam and takes 20 minutes to me. Furthermore, each line brings in a profit of $2.50 while each husky brings in a profit of $3. If she only has 3840 ounces of Styrofoam to work with and has 9650 minutes, how many of each animal should she make in order to maximize profits? What is the maximum profit possible?

Answer 1

Answer:

She should make 122 Styrofoam lions and 268 Styrofoam Husky's

The maximum profit, she makes is  $1109

Step-by-step explanation:

The given information are;

The volume required to make each lion = 8 oz

The time it takes to make each lion = 35 minutes

The amount of profit each lion brings = $2.50

The volume required to make each Husky = 12 oz

The time it takes to make each Husky = 20 minutes

The amount of profit each Husky brings = $3

Let the number of lion Styrofoam cups she makes = x

Let the number of Husky Styrofoam cups she makes = y

Therefore, we have;

x × 35 + y × 20 = 9650..........(1)

x × 8 + y × 12  = 3840.............(2)

Dividing equation (2) by 8 and making x the subject, we get;

x × 8/8 + y × 12/8  = 3840/8

x + 4/3·y = 480

x = 480 - 4/3·y

Substituting the value of x in equation (1) gives;

(480 - 4/3·y) × 35 + y × 20 = 9650

16800 - 80/3·y = 9650

y = 268.125

x = 480 - 4/3·y  = 480 - 4/3×268.125 = 122.5

Therefore, she can make 122 Styrofoam lions and 268 Styrofoam Husky's

The maximum profit, she makes is therefore given by the following equation;

The maximum profit = 122 × 2.5 + 268 × 3 = $1109.