MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Objective Function

A small firm ...

Question

A small firm manufacturers items A and B. The total number of items A and B that it can manufacture in a day is at the most 24. Item A takes one hour to make while item B takes only half an hour. The maximum time available per day is 16 hours. If the profit on one unit of item A be Rs 300 and one unit of item B be Rs 160, how many of each type of item be produced to maximize the profit? Solve the problem graphically.

Open in App

Solution

Let the firm manufacture x items of A and y items of B per day.
Number of items cannot be negative.
Therefore, x ≥0 and y ≥0.

It is given that the total number of items manufactured per day is at most 24.

∴ x + y ≤ 24

Item A takes 1 hour to make and item B takes 0.5 hour to make.

The maximum number of hours available per day is 16 hours.

∴ x + 0.5y ≤ 16

If the profit on one unit of item A be Rs 300 and one unit of item B be Rs 160.Therefore, profit gained on x items of A and y items of B is
Rs 300x and Rs 160y respectively.

∴ Total profit, Z = 300x + 160y

Thus, the mathematical formulation of the given problem is:

Maximise Z = 300x + 160y

subject to the constraints

x+ y ≤ 24

x + 0.5y ≤ 16

x, y ≥0

First we will convert inequations into equations as follows:
x + y = 24, x + 0.5y = 16, x = 0 and y = 0

Region represented by x + y ≤ 24:
The line x + y = 24 meets the coordinate axes at A₁(24, 0) and B₁(0, 24) respectively. By joining these points we obtain the line x + y = 24. Clearly (0,0) satisfies the x + y = 24. So, the region which contains the origin represents the solution set of the inequation
x + y ≤ 24.

Region represented by x + 0.5y ≤ 16:
The line x + 0.5y = 16 meets the coordinate axes at C₁(16, 0) and D₁(0, 32) respectively. By joining these points we obtain the line x + 0.5y = 16. Clearly (0,0) satisfies the inequation x + 0.5y ≤ 16. So,the region which contains the origin represents the solution set of the inequation x + 0.5y ≤ 16.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.

The feasible region determined by the system of constraints x + y ≤ 24, x + 0.5y ≤ 16, x ≥ 0 and y ≥ 0 are as follows.

The corner points are O(0, 0) ,C₁(16, 0), E₁(8, 16) and B₁(0, 24).

The value of Z at these corner points are as follows:

Corner point Z = 300x + 160y

O(0, 0) 0

C₁(16, 0) 4800

E₁(8, 16) 4960

B₁(0, 24) 3840

Thus, the maximum value of Z is 4960 at E₁(8, 16).

Thus, 8 units of item A and 16 units of item B should be manufactured per day to maximise the profits.

Suggest Corrections

thumbs-up

0

Similar questions

Q. A firm makes items

A

B

and the total number of items it can make in a day is

24

It takes one hour to make an item of

A

and only half an hour to make an item of

B

. The maximum time available per day is

16

hours. The profit on an item of

A

R s . 300

and on one item of

B

R s . 160

. How many times of each type should be produced to maximize the profit? Solve the problem graphically.

Q. A manufacturer has employed

5

skilled men and

10

semi-skilled men and makes two models

A

B

of an article. The making of one item of model

A

2

hours work by a skilled men and

2

hours work by a semi-skilled man. One item of model

B

1

hour by a skilled man and

3

hours by a semi-skilled man. No man is expected to work more than

8

hours per day. The manufacturer's profit on an item of model

A

15

and on an item of model

B

10.

How many of items of each model should be made per day in order to produce maximize daily profit ? Formulate the above LPP and solve it graphically and find the maximum profit.

Q. A manufacturer has three machines installed in his factory. machines I and II are capable of being operated for at most 12 hours whereas Machine III must operate at least for 5 hours a day. He produces only two items, each requiring the use of three machines. The number of hours required for producing one unit each of the items on the three machines is given in the following table:

Item	Number of hours required by the machine
A B	I	II	III
A B	1 2	2 1	1 5/4

He makes a profit of Rs 6.00 on item A and Rs 4.00 on item B. Assuming that he can sell all that he produces, how many of each item should he produces so as to maximize his profit? Determine his maximum profit. Formulate this LPP mathematically and then solve it.

Q. A manufacturer has employed

5

skilled men and

10

semi-skilled men and makes two models

A

B

of an article. The making of one item of model

A

2

hours work by a skilled men and

2

hours work by a semi-skilled man. One item of model

B

1

hour by a skilled man and

3

hours by a semi-skilled man. No man is expected to work more than

8

hours per day. The manufacturer's profit on an item of model

A

15

and on an item of model

B

10,

then the total number of items of each model should be made per day in order to maximize daily profit.

[

n (A) =

number of items of artical

A

n (B) =

number of items of artical

A]

Q. A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is Rs 100 and that on a bracelet is Rs 300. Formulate on L.P.P. for finding how many of each should be produced daily to maximize the profit?
It is being given that at least one of each must be produced.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Terminology

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Objective Function

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon