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1 Solution Activity 1

The factory must produce the maximum number of units of models A and B to maximize profits, subject to maximum production constraints. The oil company must minimize transportation costs of gasoline to city 1 from 3 refineries, subject to limits on the amount transported from each one. The television company must maximize profits by producing 27" and 20" televisions, subject to restrictions on production hours and estimated demand.
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0% found this document useful (0 votes)
21 views4 pages

1 Solution Activity 1

The factory must produce the maximum number of units of models A and B to maximize profits, subject to maximum production constraints. The oil company must minimize transportation costs of gasoline to city 1 from 3 refineries, subject to limits on the amount transported from each one. The television company must maximize profits by producing 27" and 20" televisions, subject to restrictions on production hours and estimated demand.
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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SOLUTION ACTIVITY 1 – UNIT 3

1.- A factory produces two models A and B of a product. The profit generated by
Model A is $40,000/unit and Model B is $60,000/unit. Daily production
cannot exceed 4000 units of model A or 3000 of model B due to the
production conditions of the plant. The marketing department informs that the
demand according to the received orders is 600 units How many
units of each model the factory must produce to achieve the maximum
benefit?
Objective: maximize profit.

Model A = X1
Model B = X2

$40,000
Value B = $60,000

Zmax (X1,X240000x1+ 60000X2

Restrictions:
X1≤ 4000 units
X2≤ 3000 units
X1+ X2>= 600 units
X1, X2≥ 0

A petroleum company has three refineries to supply gasoline to the city.


The first is in city M, the second in city N and the third
In the city Q. From the plant of M, 15,000 liters of gasoline can be transported to
day, while from N 20,000 and from Q 35,000. The cost per liter of gasoline
The transportation from M is $1.50, from N is $2.00, and from Q is $2.20. If the
the demand for gasoline in city 1 is 28,000 liters per day, what is the
combination that minimizes transportation costs?

Variables:
X115,000 liters per day = $1.50 per liter
X2= 20,000 liters per day = $2.00 per liter
X3= 35,000 liters per day -= $2.20 per liter
City 1 = 28,000 liters per day.
X1+ X2+ X328000

Z min (X12,X3) = 1.5 X12 X2+2.2 X3

Restrictions:
X1≤ 15000
X2≤ 20000
X3≤ 35000
X1, X2, X3≥ 0

3.-Ápex Television must decide the number of 27" and 20" televisions.
produced in one of its factories, market research indicates sales to the
more than 40 televisions of 27" and 10 of 20" each month. The maximum number of hours-
The available man is 500 per month, a 27” television requires 20 hours.
one man and 20" requires 10 man-hours, each 27" television produces a
profit of $120 and each one of 20" gives a profit of $80. A distributor is
agrees to buy all the televisions produced as long as not
exceeds the maximum indicated by the market study.

a) formulate the linear programming model.


b) Use the graphical method to solve the model.

Variables:
X1TV 27”
X2Quantity of 20” TV

Zmax (X1, X2) = 120 X1+ 80 X2

Restrictions:
20 X1+ 10 X2<= 500
X1≤ 40
X2≤ 10
4.- The company Trim-Look Company manufactures several lines of skirts, dresses and
sports jackets. Recently, a consulting firm proposed that the company
will reevaluate its South Islander line and will allocate its resources to products
capable of maximizing the contribution to profits and overhead expenses.
Each product requires the same fabric of polyester and has to go through the
cutting and sewing departments. The following data was collected for this
study:
The cutting department has 100 hours of capacity, the sewing department has
180 hours of capacity and has 60 yards of material. Each skirt contributes
with $5 for utilities and general expenses; each dress, for $17; and each
sports jacket, with $30.

Specify the objective function and the constraints for this problem.
b) Use some computer program to solve the problem.
Variables:
Skirts1
Dresses -> X2
Jackets3

Z max (X1, X2, X3) = 5 X1+ 17 X2+ 30 X3

Restrictions:
X1+ 3 X2+ 4 X3less than or equal to 100
X1+ 4 X2+ 6 X3≤ 180 (SEWING)
X1+ X2+ 4 X3≤ 60 (MATERIAL)

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