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Manufacturing Problems

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Manufacturing Problems - Lesson Summary

Manufacturing problems:
In these problems, we determine the number of units of different products that should be produced or sold by a firm when each product requires a fixed manpower, machine hours, labour hour per unit of product, warehouse space per unit of the output, etc., in order to make maximum profit.

Ex: A box manufacturer makes small and large boxes from a large piece of cardboard. The large boxes require 4 square feet of cardboard per box, while the small boxes require 3 square feet of cardboard per box. The manufacturer is required to make at least three large boxes and at least twice as many small boxes as large boxes. If 60 square feet of the cardboard is in stock, and if the profits on the large and small boxes are Rs 3 and Rs 2, respectively, how many of each should be made in order to maximise the total profit?

Sol:

Let x be the number of small boxes and y be the number of large boxes.

⇒ x ≥ 0, y ≥ 0 … (1)

Each small box requires 3 sq. ft. of cardboard and each large box requires 4 sq. ft. of cardboard.

The total cardboard required is (3x + 4y) sq. ft.

Total available card board = 60 sq. ft.

⇒ 3x + 4y £ 60 ... (2)

As at least three large boxes should be made, Þ y ³ 3 ... (3)

The number of small boxes is at least twice as that of the large boxes.

⇒ x ≥ 2y … (4)

The profit on one small box is Rs 2 and on one large box is Rs 3.

Therefore, total profit is Rs (2x + 3y).

Objective function Z = 2x + 3y

Z = 2x + 3y is maximised using the constraints given in (1) to (4).

x ≥ 0, y ≥ 0 … (1)

3x + 4y £ 60 ... (2)

y ³ 3 ... (3)

x ≥ 2y … (4)

3x + 4y £ 60

i) 3x + 4y = 60

    x

   20

   0

   y

   0

  15

The ordered pairs are (20, 0) and (0, 15).

x ≥ 2y

i) x = 2y

    x

    0

   12

    y

   0

    6

The ordered pairs are (0, 0), (12, 6).

y ³ 3

y = 3 is the line parallel to the X-axis.

The shaded region, ABC, is the feasible region.

The corner points of the feasible region are A (6, 3), B (12, 6) and C (16, 3).

    Corner points (x, y)

   Value of the objective function, Z = 2x + 3y

    A (6, 3)

    Z = 2 (6) + 3 (3) = 21

    B (12, 6)

    Z = 2 (12) + 3 (6) = 42

    C (16, 3)

    Z = 2 (16) + 3 (3) = 41



The maximum value of Z is 42, which occurs at the point, (12, 6).

To get the maximum profit, 12 small boxes and 6 large boxes should be made.

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