Management Science Mid-Term examination
AY 2022-2023 (1st Semester)
I have created these 5 questions to access your knowledge and practice your skills on
computation and practical applications.
Write your answer in word document of if computer is not accessible, use a piece of paper,
picture it and upload it. I need solutions computations, and business analysis of the problem,
including graphical presentation.
20 points each problem
Problem 1:
A prestige chocolatier shop has two products: its flagship assortment of triangular chocolates,
called Toblerox, and the more decadent and deluxe Toblerox Nuit. How much of each should it
produce to maximize profits? Every box of Toblerox has a a profit of PhP 10. Every box of Toblerox
Nuit has a profit of $60. • The daily demand is limited to at most 2000 boxes of Toblerox and 3000
boxes of Toblerox Nuit. The current workforce can produce a total of at most 4000 boxes of
chocolate per day. Let x1 be # of boxes of Pyramide, x2 be # of boxes of Nuit.
Answer:
Objective Function: P = 10x1 + 60 x2
Constraints:
X1 ≤ 2000
x2 ≤ 3000
x1+x2 ≤ 4000
Non-Negativity Constraints: x1, x2 ≥ 0
(0,3000) = 10 (0) + 60(3000) = 180000
(1000,3000) = 10 (1000) + 60 (3000) = 190000
(2000,2000) = 10 (2000) + 60(2000) = 140000
(2000,0) 10(2000) + 60 (0) = 20000
Business Analysis of the Problem:
The coordinate is (1000,3000) will give the highest value of Php 190000. The decision is to create
1000 boxes of Pyramide and 3000 boxes of nuit.
Problem 2:
Let’s say you hold two part-time jobs, Job I and Job II. You never want to work more than a total
of 12 hours a week. You determined that for every hour you work at Job I, you need 2 hours of
preparation time, and for every hour you works at Job II, you need one hour of preparation time,
and You cannot spend more than 16 hours for preparation.
If You make PhP 400 an hour at Job I, and PhP 300 an hour at Job II, how many hours should you
work per week at each job to maximize your income?
Answer:
Objective Function: P = 400x1 + 300x2
Constraints:
x1 + x2 ≤ 12
2x1 + x2 ≤ 16
Non-Negativity Constraints: x1, x2 ≥ 0
X1 + x2 = 12
Let x1 = 0
x1 = 12
Let x2= 0
X1= 12
2x1 + x2 = 16
Let x1 = 0
X1= 16
Let x2 = 0
2x = 16
Divide bot side by 2.
X2= 8
X1 + x2 = 12
2x1 + x2 = 16
To eliminate multiply both equation to a certain number to cancel 1 variable.
(-1) X1 + x2 = 12
(1) 2x1 + x2 = 16
-1x1 = -12
2x1 = 16
X1 = 4
Substitute the value to one equation:
4 + x2 = 12
X2 = 8
(0,12) = 400 (0) + 300 (12) = 3600
(4,8) = 400 (4) + 300 (8) = 4000
(8,0) = 400 (8) + 300 (0) = 3200
Business Analysis of the Problem:
The coordinate is (4,8) will give the highest value of Php 4000. The decision is to work 4 hours
for Job I and 8 hours to Job II.
Problem 3
A gadget and accessory factory manufacture two types of gadgets, regular and premium. Each
gadget requires the use of two operations, assembly and finishing, and there are at most 12 hours
available for each operation. A regular gadget requires 1 hour of assembly and 2 hours of finishing,
while a premium gadget needs 2 hours of assembly and 1 hour of finishing. Due to other
restrictions, the company can make at most 7 gadgets a day. If a profit of PhP20 is realized for
each regular gadget and PhP30 for a premium gadget, how many of each should be manufactured
to maximize profit?
Answer:
Let x = be the number of regular gadget; and Let y = be the number of the premium gadget
Objective Function: P= 20x +30y
Constraints:
X+y≤7
x +2 y ≤ 12
2x + y ≤ 12
Non-Negativity Constraints: x, y ≥ 0
x+y = 7
Let x = 0
Y=7
Let y=0
X= 7
X+2y =12
Let x = 0
2y=12
Divide both side by 2
Y=6
Let y=0
X= 12
2x + y ≤ 12
Let x = 0
Y=12
Let x = 0
2x=12
Divide both side by 2
X=6
(0,6) = 20 (0) + 30 (6) = 180
(2,5) = 20 (2) + 30(5) = 190
(5,2) = 20 (5) + 30(2) = 160
(6,0) = 20 (6) + 30 (0) = 120
Business Analysis of the Problem:
The coordinate is (2.5) will give the highest value of Php 190. The decision is create 2 regular
gadget and 5 premium gadget.
Problem 4:
One kind of cake requires 200g of flour and 25g of fat, and another kind of cake requires 100g of
flour and 50g of fat. Formulate this problem as a linear programming problem to find the
maximum number of cakes that can be made from 5kg of flour and 1 kg of fat assuming that there
is no shortage of the other ingredients used in making the cakes.
Answer:
Let x= be the number of one kind of cake; and let y= be the number of another kind of cake.
Objective Function: P = x + y
Constraint =
200x+100y≤5000
25x+50y≤1000
Non-Negativity Constraints: x, y ≥ 0
(0,20) = 1 (0) +1 (20) = 20
(20,10) = 1 (20) + 1(10) = 30
(25,0) = 1 (25) + 1(0) = 25
Business Analysis of the Problem:
The coordinate is (20,10) will give the highest number of cakes of 30 pcs. The decision is create
20 one kind of cake and 10 another kind of cake.
Problem 5:
You’re on a special diet and know that your daily requirement of five nutrients is 60 milligrams of
vitamin C, 1,000 milligrams of calcium, 18 milligrams of iron, 20 milligrams of niacin, and 360
milligrams of magnesium. You have two supplements to choose from: Vega Vita and Happy Health.
Vega Vita costs 20 cents per tablet, and Happy Health costs 30 cents per tablet. Vega Vita contains
20 milligrams of vitamin C, 500 milligrams of calcium, 9 milligrams of iron, 2 milligrams of niacin,
and 60 milligrams of magnesium. Happy Health contains 30 milligrams of vitamin C, 250 milligrams
of calcium, 2 milligrams of iron, 10 milligrams of niacin, and 90 milligrams of magnesium. How
many of each tablet should you take each day to meet your minimum requirements while
spending the least amount of money?
A good way to organize this is to make a chart or table listing the requirements, costs, and amount
of nutrients in each tablet.
With all the information organized into the table, it’s time to solve for the number of tablets that
will minimize your cost using linear programming.
Answer:
Let x = be the number of vega vital and let y= be the number of happy health.
Objective Function: Z = 0.20x + 0.30y
20x + 30y ≥ 60
500x + 250y ≥ 1000
9x + 2y ≥ 18
2x + 10y ≥ 20
60x+90y≥ 360
Non-Negativity Constraints: x, y ≥ 0
Round up to whole number:
(2,4) = 0.20 (2) + 0.30 (4) = $1.60
(5.2) = 0.20 (5) + 0.30 (2) = $1.60
(0.9) = 0.20 (0) + 0.30 (9) = $2.70
(10,0) = 0.20 (10) + 0.30 (0) = $2.00
Business Analysis of the Problem:
There are two coordinate that satisfy the problem, (2,4) and (5,2) will give the minimize cost of
$1.60 . The decision is created 2 vega vita and 4 happy health or 5 vega vita and 2 happy health.