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Question
A chemist has a compound to be made using 3 basic elements X, Y, Z so that it has at least 10 litres of X, 12 litres of Y and 20 litres of Z. He makes this compound by mixing two compounds (I) and (II). Each unit compound (I) had 4 litres of X, 3 litres of Y. Each unit compound (II) had 1 litre of X, 2 litres of Y and 4 litres of Z. The unit costs of compounds (I) and (II) are ₹ 400 and ₹ 600 respectively. Find the number of units of each compound to be produced so as to minimize the cost
Solution
Let the chemist produce x units of compound I and y units of compound II.
Since x and y cannot be negative, x ≥ 0, y ≥ 0
The unit costs of compounds I and II are ₹ 400 and ₹ 600 respectively.
Total Cost = Z = 400x + 600y
We construct a table with constraints of X, Y and Z as follows:
| Compound I | Compound II | Least value | |
| X | 4 | 1 | 10 |
| Y | 3 | 2 | 12 |
| Z | – | 4 | 20 |
From the table, the constraints are
4x + y ≥ 10
3x + 2y ≥ 12
4y ≥ 20
∴ Given problem can be formulated as follows:
Minimize Z = 400x + 600y
Subject to 4x + y ≥ 10
3x + 2y ≥ 12
4y ≥ 20, x ≥ 0, y ≥ 0
To draw the feasible region, construct table as follows:
| Inequality | 4x + y ≥ 10 | 3x + 2y ≥ 12 | 4y ≥ 20 |
| Corresponding equation (of line) | 4x + y = 10 | 3x + 2y = 12 | 4y = 20 |
| Intersection of line with X-axis | `(5/2, 0)` | (4, 0) | – |
| Intersection of line with Y-axis | (0, 10) | (0, 6) | (0, 5) |
| Region | Non-Origin side | Non-Origin side | Non-Origin side |
Shaded portion EABY is the feasible region, whose vertices are A and B(0, 10).
A is the point of intersection of the lines 4y = 20 and 4x + y = 10,
Solving the above equations, we get
x = `5/4`, y = 5
∴ A ≡ `(5/4, 5)`
Here, the objective function is
Z = 400x + 600y
∴ Z at A`(5/4, 5) = 400(5/4) + 600(5)`
= 500 + 3000
= 3500
Z at B (0, 10) = 400(0) + 600(10)
= 6000
∴ Z has minimum value 3500 at x = `5/4` and y = 5.
∴ The chemist should produce `5/4` units of compound I and 5 units of compound II to minimize the cost.
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