Question

A Gift for You makes floral arrangements and fruit baskets. The small business has a maximum of 40 hours per week available in the assembly department and a maximum of 10 hours per week in the packaging department. Each floral arrangement takes 20 minutes to assemble and 6 minutes to package. Each fruit basket takes 15 minutes to assemble and 2 minutes to package. The profit for each floral arrangement is $50 and the profit for each fruit basket is $35. The company wants to maximize their profit.a. Set up the linear programming problem by writing the objective function as well as the system of constraints. b. How many floral arrangements and fruit baskets must be sold to maximize profit? c. What is the maximum profit? Let x the number of floral arrangements and y the number of fruit baskets.Give the objective function a. Max P= 50x+ 35y b. Max P= 20x + 6y c. Min P= 15x+2yd. Max P= 35x +20ye. Min P= 35x+20yGive a constrainta. 20x+ 6y <=40b. 20x+15y<=40c. 6x+2y<=10d. 20x+ 15y<=24,00e. 6x+2y>=600How many of each should they sell to maximize their profit?a. 84 Floral Arrangements and 48 Fruit Baskets b. 0 Floral Arrangements and 160 Fruit Baskets c. 0 Floral Arrangements and 300 Fruit Baskets d. 100 Floral Arrangements and O Fruit Baskets e. 48 Floral Arrangements and 84 Fruit Baskets

Answer 1

z(max) = 5000

x = 100

y = 0

Explanation:

Let call

x floral arrangements and

y fruit baskets

Then Objective function is according to profits in each gift

z = 50*x + 35*y

Constraints:

1.- Hours available in Assembly department 40 in minutes is 2400 minutes

20*x + 15*y ≤ 2400

2.- Hours available in packaging department 10 in minutes is 600

6*x + 2*y ≤ 600

3.- x and y must be x ≥ 0 y ≥ 0

Then the system is:

z - 50*x - 50*y = 0 To maximize subject to:

20*x + 15*y ≤ 2400

6*x + 2*y ≤ 600

x ≥ 0 y ≥ 0

Simplex Method:

z x y s₁ s₂ Cte

1 -50 -35 0 0 = 0

0 20 15 1 0 = 2400

0 6 2 0 1 = 600

First iteration: 6 is a pivot we dvede R3 by 6

z x y s₁ s₂ Cte

1 0 15 0 50/6 = 5000

0 0 -50/6 -1 20/6 = -400

0 20 15 1 0 = 2400

0 1 2/6 0 1/6 = 100

We have done no negative number in the objective function we stop iteration and

z(max) = 5000

x = 100

y = 0

to add to R1 50*R3 [ 0 50 100/6 0 50/6 5000

to add to R2 20*R3 [ 0 20 40/6 0 20/6 2000