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Ncert Solutions Maths class 12th Maths Class 12th Linear Programming

27. Two godowns, A and B, have grain capacities of 100 quintals and 50 quintals, respectively. They supply to 3 ration shops, D, E and F, whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops is given in the following table:

Transportation Cost per Quintal (in Rs)

From/To

A

B

D

6

4

E

3

2

F

2.50

3

How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

Let godown A supply x and y quintals of grain to shops D and E.

So, $(100 - x - y)$ will be supplied to shop F.

Since x quintals are transported from godown A, the requirement at shop D is 60 quintals. Hence, the remaining (60 – x) quintals will be transported from godown B.

Similarly, (50 – y) quintals and $40 - (100 - x - y) = (x + y - 60)$ quintals will be transported from godown B to shop E and F.

The given problem can be represented diagrammatically as given below:

$x \geq 0, y \geq 0, a n d 100 - x - y \geq 0$

Then, $x \geq 0, y \geq 0, a n d x + y \leq 100$

$\begin{array}{l} 60 - x \geq 0, 50 - y \geq 0, a n d x + y - 60 \geq 0 \end{array}$

$\begin{array}{l} T h e n, x \leq 60, y \leq 50, a n d x + y \geq 60 \end{array}$

Total transportation cost z is given by,

$\begin{array}{l} z = 6 x + 3 y + 2.5 (100 - x - y) + 4 (60 - x) + 2 (50 - y) + 3 (x + y - 60) \end{array}$

$\begin{array}{l} = 6 x + 3 y + 250 - 2.5 x - 2.5 y + 240 - 4 x + 100 - 2 y + 3 x + 3 y - 180 \end{array}$

$\begin{array}{l} = 2.5 x + 1.5 y + 410 \end{array}$

The given problem can be formulated as given below:

$M i n i m i s e z = 2.5 x + 1.5 y + 410 \dots \dots \dots \dots . (i)$

Subject to the constraints,

$\begin{array}{l} \end{array} x + y \leq 100 \dots \dots \dots . . (i i)$

$x \leq 60 \dots \dots \dots . . (i i i)$

$y \leq 50 \dots \dots \dots . (i v)$

$x + y \geq 60 \dots \dots \dots (v)$

$x, y \geq 0 \dots \dots \dots \dots . . (v i)$

The feasible region determined by the system of constraints is

...more

Let godown A supply x and y quintals of grain to shops D and E.

So, $(100 - x - y)$ will be supplied to shop F.

Since x quintals are transported from godown A, the requirement at shop D is 60 quintals. Hence, the remaining (60 – x) quintals will be transported from godown B.

Similarly, (50 – y) quintals and $40 - (100 - x - y) = (x + y - 60)$ quintals will be transported from godown B to shop E and F.

The given problem can be represented diagrammatically as given below:

$x \geq 0, y \geq 0, a n d 100 - x - y \geq 0$

Then, $x \geq 0, y \geq 0, a n d x + y \leq 100$

$\begin{array}{l} 60 - x \geq 0, 50 - y \geq 0, a n d x + y - 60 \geq 0 \end{array}$

$\begin{array}{l} T h e n, x \leq 60, y \leq 50, a n d x + y \geq 60 \end{array}$

Total transportation cost z is given by,

$\begin{array}{l} z = 6 x + 3 y + 2.5 (100 - x - y) + 4 (60 - x) + 2 (50 - y) + 3 (x + y - 60) \end{array}$

$\begin{array}{l} = 6 x + 3 y + 250 - 2.5 x - 2.5 y + 240 - 4 x + 100 - 2 y + 3 x + 3 y - 180 \end{array}$

$\begin{array}{l} = 2.5 x + 1.5 y + 410 \end{array}$

The given problem can be formulated as given below:

$M i n i m i s e z = 2.5 x + 1.5 y + 410 \dots \dots \dots \dots . (i)$

Subject to the constraints,

$\begin{array}{l} \end{array} x + y \leq 100 \dots \dots \dots . . (i i)$

$x \leq 60 \dots \dots \dots . . (i i i)$

$y \leq 50 \dots \dots \dots . (i v)$

$x + y \geq 60 \dots \dots \dots (v)$

$x, y \geq 0 \dots \dots \dots \dots . . (v i)$

The feasible region determined by the system of constraints is given below:

A (60, 0), B (60, 40), C (50, 50) and D (10, 50) are the corner points.

The values of z at these corner points are given below:

Corner Point	z = 2.5x + 1.5y + 410
A (60, 0)	560
B (60, 40)	620
C (50, 50)	610
D (10, 50)	510	Minimum

The minimum value of z is 510 at (10, 50).

Hence, the amount of grain transported from A to D, E, and F, is 10 quintals, 50 quintals and 40 quintals, respectively, and from B to D, E and F are 50 quintals, 0 quintals, 0 quintals, respectively.

Thus, the minimum cost is Rs. 510.

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Transportation Cost per Quintal (in Rs)
From/To	A	B
D	6	4
E	3	2
F	2.50	3