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Question
A chemical company produces a chemical containing three basic elements A, B, C, so that it has at least 16 litres of A, 24 litres of B and 18 litres of C. This chemical is made by mixing two compounds I and II. Each unit of compound I has 4 litres of A, 12 litres of B and 2 litres of C. Each unit of compound II has 2 litres of A, 2 litres of B and 6 litres of C. The cost per unit of compound I is ₹ 800 and that of compound II is ₹ 640. Formulate the problems as LPP and solve it to minimize the cost.
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Solution
Let the company buy x units of compound I and y units of compound II.
Then the total cost is z = ₹ (800x + 640y)
This is the objective function that is to be minimized.
The constraints are as per the following table:
| Compound I (x) |
Compound II (y) |
Minimum requirement |
|
| Element A | 4 | 2 | 16 |
| Element B | 12 | 2 | 24 |
| Element C | 2 | 6 | 18 |
From the table, the constraints are 4x + 2y ≥ 16, 12x + 2y ≥ 24, 2x + 6y ≥ 18.
Also, the number of units of compound I and compound II cannot be negative.
∴ x ≥ 0, y ≥ 0
∴ The mathematical formulation of given LPP is
Minimize z = 800x + 640y, subject to 4x + 2y ≥ 16, 12x + 2y ≥ 24, 2x + 6y ≥ 18, x ≥ 0, y ≥ 0.
First we draw the lines AB, CD and EF whose equations are 4x + 2y = 16, 12x + 2y = 24 and 2x + 6y = 18 respectively.
| Line | Equation | Points on the X-axis |
Points on the Y-axis |
Sign | Region |
| AB | 4x + 2y = 16 | A(4, 0) | B(0, 8) | ≥ | non-origin side of line AB |
| CD | 12x + 2y = 24 | C(2, 0) | D(0, 12) | ≥ | non-origin side of line CD |
| EF | 2x + 6y = 18 | E(9, 0) | F(0, 3) | ≥ | non-origin side of line EF |
The feasible region is shaded in the graph.
The vertices of the feasible region are E(9, 0), P, Q and D(0, 12).
P is the point of intersection of the lines
2x + 6y = 18 ...(1)
and 4x + 2y = 16 ...(2)
Multiplying equation (1) by 2, we get
4x + 12y = 36
Subtracting equation (2) from this equation, we get
10y = 20
∴ y = 2
∴ From (1), 2x + 6(2) = 18
∴ 2x = 6
∴ x = 3
∴ P = (3, 2)
Q is the point of intersection of the lines
12x + 2y = 24 ...(3)
and 4x + 2y = 16 ...(2)
On subtracting, we get
8x = 8
∴ x = 1
∴ From (2), 4(1) + 2y = 16
∴ 2y = 12
∴ y = 6
∴ Q = (1, 6)
The values of the objective function z = 800x + 640y at these vertices are
z(E) = 800(9) + 640(0) = 7200 + 0 = 7200
z(P) = 800(3) + 640(2) = 2400 + 1280 = 3680
z(Q) = 800(1) + 640(6) = 800 + 3840 = 4640
z(D) = 800(0) + 640(12) = 0 + 7680 = 7680
∴ The minimum value of z is 3680 at the point (3, 2).
Hence, the company should buy 3 units of compound I and 2 units of compound II to have the minimum cost of ₹ 3680.
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