business analysis

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A company manufactures and sells two models of lamps, L1 and L2. To

manufacture each lamp, the manual work involved in model L1 is 20 minutes and for L2,

30 minutes. The mechanical (machine) work involved for L1 is 20 minutes and for L2, 10

minutes. The manual work available per month is 100 hours and the machine is limited

to only 80 hours per month. Knowing that the profit per unit is $15 and $10 for L1 and

L2, respectively, determine the quantities of each lamp that should be manufactured to

obtain the maximum benefit.

Step 1: Choose the unknowns.

x = number of lamps L1

y = number of lamps L2

Step 2: Write the objective function.

F(x, y) = 15x + 10y

Step 3: Write the constraints as a system of inequalities.

Convert the time from minutes to hours.

20 min = 1/3 h

30 min = 1/2 h

10 min = 1/6 h

L

1

L

2

Ti

m

e

Man

ual

1

/

3

1

/

2

10

0

Mac

hine

1

/

3

1

/

6

80

1/3x + 1/2y ≤ 100

1/3x + 1/6y ≤ 80

As the numbers of lamps are natural numbers, there are two more constraints:

x ≥ 0

y ≥ 0

Step 4: Find the set of feasible solutions that graphically represent the constraints.

Represent the constraints graphically.

As x ≥ 0 and y ≥ 0, work in the first quadrant.

Solve the in equation graphically: 1/3 x + 1/2 y ≤ 100; and take a point on the

plane, for example (0, 0).



1/3 · 0 + 1/2 · 0 ≤ 100

1/3 · 0 + 1/6 · 0 ≤ 80

The area of intersection of the

solutions of the inequalities would

be the solution to the system of

inequalities, which is the set of

feasible solutions.

Step 5: Calculate the coordinates of the vertices from the compound of feasible

solutions.

The optimal solution, if unique, is a vertex. These are the solutions to systems:

1/3x + 1/2y = 100; x = 0 (0, 200)

1/3x + 1/6y = 80; y = 0(240, 0)

1/3x + 1/2y = 100; 1/3x + 1/6y = 80(210, 60)

Step 6: Calculate the value of the objective function at each of the vertices to

determine which of them has the maximum or minimum values.

In the objective function, place each of the vertices that were determined in the

previous step.

F(x, y) = 15x + 10y

F(0, 200) = 15·0 + 10·200 = $2,000

F(240, 0 ) = 15·240 + 10·0 = $3,600

F(210, 60) = 15·210 + 10·60 = $3,750 Maximum

The optimum solution

is to manufacture 210

units of model L1 and 60 units of model L1 to obtain a benefit of $3,750.


With the start of school approaching, a store is planning on having a sale on

school materials. They have 600 notebooks, 500 folders and 400 pens in stock, and they

plan on packing it in two different forms. In the first package, there will be 2 notebooks,

1 folder and 2 pens, and in the second one, 3 notebooks, 1 folder and 1 pen. The price

of each package will be $6.50 and $7.00 respectively. How many packages should they

put together of each type to obtain the maximum benefit?


x = P1

y = P2


F(x, y) = 6.5x + 7y


P1 P2 Available

Notebooks 2 3 600

Folders 1 1 500

Pens 2 1 400

2x + 3y ≤ 600

x + y ≤ 500

2x + y ≤ 400

x ≥ 0

y ≥ 0



Step 5: Calculate the

coordinates of

the vertices from

the compound of

feasible solutions.

Step 6: Calculate the value of the objective function at each of the vertices to determine

which of them has the maximum or minimum values.

F(x, y) = 6.5 · 200 + 7 · 0 = $1,300

F(x, y) = 6.5 · 0 + 7 · 200 = $1,400

F(x, y) = 6.5 · 150 + 7 · 100 = $1,675 Maximum

The optimum solution is to package 150 units of package 1 and 100 units

of package 2 to obtain $1,675

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On a chicken farm, the poultry is given a healthy diet to gain weight. The

chickens have to consume a minimum of 15 units of Substance A and another 15 units of

Substance B. In the market there are only two classes of compounds: Type X, with a

composition of one unit of A to five units of B, and another type, Y, with a composition

of five units of A to one of B. The price of Type X is $10 and Type Y, $30. What are the

quantities of each type of compound that have to be purchased to cover the needs of

the diet with a minimal cost?


x = X

y = Y


F(x, y) = 10x + 30y


X Y Minimal

A 1 5 15

B 5 1 15

x + 5y ≥ 15

5x + y ≥ 15

x ≥ 0

y ≥ 0





coordinates of

the vertices from the

compound of feasible solutions.


determine which of them has the maximum or minimum values. It must be taken

into account the possible non-existence of a solution if the compound is not

bounded.

F(0, 15) = 10 · 0 + 30 · 15 = 450

F(15, 0) = 10 · 15 + 30 · 0 = 150

F(5/2, 5/2) = 10 · 5/2 + 30 · 5/2 = 100 Minimum

The minimum cost is $100 for X = 5/2 and Y = 5/2.


There is only 600 milligrams of a certain drug that is needed to make both large

and small pills for small scale pharmaceutical distribution. The large tablets weigh 40

milligrams and the small ones, 30 milligrams. Consumer research determines that at

least twice the amount of the smaller tablets is needed than the large ones and there

needs to be at least three large tablets made. Each large tablet is sold for a profit of $2

and the small tablet, $1. How many tablets of each type have to be prepared to obtain

the maximum profit?


x = Large tablets

y = Small tablets


F(x, y) = 2x + y


40x + 30y ≤ 600

x ≥ 3

y ≥ 2x

x ≥ 0

y ≥ 0


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Step 5: Calculate the coordinates of

the vertices from the compound of feasible solutions.



F(x, y) = 2 · 3 + 16 = $22

F(x, y) = 2 · 3 + 6 = $12

F(x, y) = 2 · 6 + 12 = $24 Maximum

The maximum profit is $24, and is obtained by making 6 units of the large

tablets and 12 units of the small tablets.


Transport Company has two types of trucks, Type A and Type B. Type A has a

refrigerated capacity of 20 m 3 and a non-refrigerated capacity of 40 m 3 while Type B has

the same overall volume with equal sections for refrigerated and non-refrigerated stock.

A grocer needs to hire trucks for the transport of 3,000 m 3 of refrigerated stock and

4,000 m3 of non-refrigerated stock. The cost per kilometer of a Type A is $30 and $40

for Type B. How many trucks of each type should the grocer rent to achieve the

minimum total cost?


x = Type A trucks

y = Type B trucks


F(x,y) = 30x + 40y


A B Total

Refrigerated 20 30 3 000

Non-refrigerated 40 30 4 000

20x + 30y ≥ 3 000

40x + 30y ≥ 4 000

x ≥ 0


y ≥ 0



coordinates of

the vertices from the

compound of feasible solutions.



F(0, 400/3) = 30 · 0 + 40 · 400/3 = 5,333.332

F(150, 0) = 30 · 150 + 40 · 0 = 4,500

As x and y must be natural numbers round the value of y.

F(50, 67) = 30 · 50 + 40 ·67 = 4,180 Minimum

The minimum cost is $4,180. To achieve these 50 trucks of Type A and 67

trucks of Type B are needed.


A school is preparing a trip for 400 students. The company who is providing the

transportation has 10 buses of 50 seats each and 8 buses of 40 seats, but only has 9

drivers available. The rental cost for a large bus is $800 and $600 for the small bus.

Calculate how many buses of each type should be used for the trip for the least possible

cost.


x = small buses

y = big buses


F(x, y) = 600x + 800y


40x + 50y ≥ 400

x + y ≤ 9

x ≥ 0

y ≥ 0

Step 4: Find the set

of feasible solutions that

graphically represent the

constraints.



solutions.



F(0, 8) = 600 · 0 + 800 · 8 = $6,400

F(0, 9) = 600 · 0 + 800 · 9 = $7,200

F(5, 4) = 6 00 · 5 + 800 · 4 = $6,200 € Minimum

The minimum cost is $6,200. This is achieved with 4 large and 5 small

buses.

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A store wants to liquidate 200 of its shirts and 100 pairs of pants from last

season. They have decided to put together two offers, A and B. Offer A is a package of

one shirt and a pair of pants which will sell for $30. Offer B is a package of three shirts

and a pair of pants, which will sell for $50. The store does not want to sell less than 20

packages of Offer A and less than 10 of Offer B. How many packages of each do they

have to sell to maximize the money generated from the promotion?


x = number of packages of Offer A

y = number of packages of Offer B


F(x, y) = 30x + 50y


A B

Mi

ni

m

al



S

h

ir

t

s

1 3

2

0

0

P

a

n

t

s

1 1

1

0

0

x + 3y ≤ 200

x + y ≤ 100

x ≥ 20

y ≥ 10



solutions.



F(x, y) = 30 · 20 + 50 · 10 = $1,100

F(x, y) = 30 · 90 + 50 · 10 = $3,200

F(x, y) = 30 · 20 + 50 · 60 = $3,600

F(x, y) = 30 · 50 + 50 · 50 = $4,000 Maximum

50 packages of each offer generate a maximum amount of $4,000 in sales.

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A calculator company produces a scientific calculator and a graphing calculator. Long-term

projections indicate an expected demand of at least 100 scientific and 80 graphing calculators each

day. Because of limitations on production capacity, no more than 200scientific and 170 graphing

calculators can be made daily. To satisfy a shipping contract, a total of at least 200 calculators much

be shipped each day.

If each scientific calculator sold results in a $2 loss, but each graphing calculator produces

a $5 profit, how many of each type should be made daily to maximize net profits?

The question asks for the optimal number of calculators, so my variables will stand for that:

x: number of scientific calculators produced

y: number of graphing calculators produced

Since they can't produce negative numbers of calculators, I have the two

constraints, x > 0 and y > 0. But in this case, I can ignore these constraints, because I already have

that x > 100 and y > 80. The exercise also gives maximums: x < 200 and y < 170. The minimum

shipping requirement gives me x + y > 200; in other words, y > –x + 200. The revenue relation will be

my optimization equation: R = –2x + 5y. So the entire system is:

R = –2x + 5y, subject to:

100 < x < 200

80 < y < 170

y > –x + 200

The feasibility region graphs as: Copyright © Elizabeth Stapel 2006-2011 All Rights Reserved

When you test the corner points at (100, 170), (200, 170), (200, 80), (120, 80),

and (100, 100), you should obtain the maximum value of R = 650 at (x, y) = (100,

170). That is, the solution is "100scientific calculators and 170 graphing calculators".


You need to buy some filing cabinets. You know that Cabinet X costs $10 per unit, requires six

square feet of floor space, and holds eight cubic feet of files. Cabinet Y costs $20 per unit, requires

eight square feet of floor space, and holds twelve cubic feet of files. You have been given $140 for this

purchase, though you don't have to spend that much. The office has room for no more than 72 square

feet of cabinets. How many of which model should you buy, in order to maximize storage volume?

The question ask for the number of cabinets I need to buy, so my variables will stand for that:

x: number of model X cabinets purchased

y: number of model Y cabinets purchased

Naturally, x > 0 and y > 0. I have to consider costs and floor space (the "footprint" of each

unit), while maximizing the storage volume, so costs and floor space will be my constraints, while

volume will be my optimization equation.

Cost: 10x + 20y < 140, or y < –( 1/2 )x + 7

Space: 6x + 8y < 72, or y < –( 3/4 )x + 9

Volume: V = 8x + 12y

This system (along with the first two constraints) graphs as:

When you test the corner points at (8, 3), (0, 7), and (12, 0), you should obtain a

maximal volume of100 cubic feet by buying eight of model X and three of model Y.



In order to ensure optimal health (and thus accurate test results), a lab technician needs to

feed the rabbits a daily diet containing a minimum of 24 grams (g) of fat, 36 g of carbohydrates, and 4

g of protein. But the rabbits should be fed no more than five ounces of food a day.

Rather than order rabbit food that is custom-blended, it is cheaper to order Food X and Food Y,

and blend them for an optimal mix. Food X contains 8 g of fat, 12 g of carbohydrates, and 2 g of

protein per ounce, and costs $0.20 per ounce. Food Y contains 12 g of fat, 12 g of carbohydrates, and

1 g of protein per ounce, at a cost of $0.30 per ounce.

What is the optimal blend?

Since the exercise is asking for the number of ounces of each food required for the optimal

daily blend, my variables will stand for the number of ounces of each:

x: number of ounces of Food X

y: number of ounces of Food Y

Since I can't use negative amounts of either food, the first two constrains are the usual ones:

x > 0 and y > 0. The other constraints come from the grams of fat, carbohydrates, and protein per

ounce:

Fat: 8x + 12y > 24

Carbs: 12x + 12y > 36

Protein: 2x + 1y > 4

Also, the maximum weight of the food is five ounces, so:

x + y < 5

The optimization equation will be the cost relation C = 0.2x + 0.3y, but this time I'll be finding

the minimum value, not the maximum.

After rearranging the inequalities, the system graphs as:

When you test the corners at (0, 4), (0, 5), (3, 0), (5, 0), and (1, 2), you should get a

minimum cost of sixty cents per daily serving, using three ounces of Food X only.



Given the following constraints, maximize and minimize the value of z = –0.4x + 3.2y.

First I'll solve the fourth and fifth constraints for easier graphing:

The feasibility region looks like this:

From the graph, I can see which lines cross to form the corners, so I know which lines to pair

up in order to verify the coordinates. I'll start at the "top" of the shaded area and work my way

clockwise around the edges:

y = –x + 7

y = x + 5

y = –x + 7

x = 5

x = 5

y = 0

–x + 7 = x + 5

2 = 2x

1 = x

y = (1) + 5 = 6

y = –(5) + 7 = 2 [nothing to do]

corner at (1, 6) corner at (5, 2) corner at (5, 0)


Now I'll plug each corner

point into the

optimization

equation, z = –0.4x + 3.2y:

(1, 6): z = –0.4(1) + 3.2(6) = –0.4 + 19.2 = 18.8

(5, 2): z = –0.4(5) + 3.2(2) = –2.0 + 6.4 = 4.4

(5, 0): z = –0.4(5) + 3.2(0) = –2.0 + 0.0 = –2.0

(4, 0): z = –0.4(4) + 3.2(0) = –1.6 + 0.0 = –1.6

(0, 2): z = –0.4(0) + 3.2(2) = –0.0 + 6.4 = 6.4

(0, 5): z = –0.4(0) + 3.2(5) = –0.0 + 16.0 = 16.0

Then the maximum is 18.8 at (1, 6) and the minimum is –2 at (5, 0).

y = 0

y = –( 1/2 )x + 2

y = –( 1/2 )x + 2

x = 0

x = 0

y = x + 5

–( 1/2 )x + 2 = 0

2 = (1/2)x

4 = x

y = –( 1/2 )(0) + 2

y = 0 + 2

y = 2

y = (0) + 5 = 5

corner at (4, 0) corner at (0, 2) corner at (0, 5)

Linear Programming

Problems

Submitted to:

Submitted by:Dumngalon, norvin john d.

business analysis

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y step

x y minimal

type y

price of type x

p2 step

compound of feasible

previous step

objective function