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Question
A carpenter makes chairs and tables. Profits are ₹ 140 per chair and ₹ 210 per table. Both products are processed on three machines: Assembling, Finishing and Polishing. The time required for each product in hours and availability of each machine is given by the following table:
| Product → | Chair (x) | Table (y) | Available time (hours) |
| Machine ↓ | |||
| Assembling | 3 | 3 | 36 |
| Finishing | 5 | 2 | 50 |
| Polishing | 2 | 6 | 60 |
Formulate the above problem as LPP. Solve it graphically
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Solution
Let the number of chairs and tables made by the carpenter be x and y respectively.
The profits are ₹ 140 per chair and ₹ 210 per table.
∴ total profit z = ₹ (140x + 210y)
This is the objective function which is to be maximized. The constraints are as per the following table:
| Chair (x) | Table (y) | Available time (hours) | |
| Assembling | 3 | 3 | 36 |
| Finishing | 5 | 2 | 50 |
| Polishing | 2 | 6 | 60 |
From the table, the constraints are
3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60.
The number of chairs and tables cannot be negative.
∴ x ≥ 0, y ≥ 0
Hence, the mathematical formulation of given LPP is:
Maximize z = 140x + 210y, subject to
3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60, x ≥ 0, y ≥ 0
We first draw the lines AB, CD and EF whose equations are 3x + 3y = 36, 5x + 2y = 50 and 2x + 6y = 60 respectively.
| Line | Equation | Points on the X-axis | Points on the Y-axis | Sign | Region |
| AB | 3x + 3y = 36 | A(12,0) | B(0,12) | ≤ | origin side of line AB |
| CD | 5x + 2y = 50 | C(10,0) | D(0,25) | ≤ | origin side of line CD |
| EF | 2x + 6y = 60 | E(30,0) | F(0,10) | ≤ | origin side of line EF |
The feasible region is OCPQFO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (10, 0), P, Q and F (0, 10).
P is the point of intersection of the lines
5x + 2y = 50 … (1)
and 3x + 3y = 36 … (2)
Multiplying equation (1) by 3 and equation (2) by 2, we get
15x + 6y = 150
6x + 6y = 72
On subtracting, we get
9x = 78 ∴ x = `26/3`
Substituting x = `26/3` in (2), we get
`3(26/3) + 3"y" = 36`
∴ 3y = 10
∴ y = `10/3`
∴ P is `(26/3, 10/3)`
Q is the point of intersection of the lines
3x + 3y = 36 ....(2)
and 2x + 6y = 60 ......(3)
Multiplying equation (2) by 2, we get
6x + 6y = 72
Subtracting equation (3) from this equation, we get
4x = 12 ∴ x = 3
Substituting x = 3 in (2), we get
3(3) + 3y = 36
∴ 3y = 27 ∴ y = 9
∴ Q is (3, 9).
Hence, the vertices of the feasible region are O (0, 0),
C(10, 0), P`(26/3, 10/3)`, Q(3,9) and F(0,10)
The values of the objective function z = 140x + 210y at these vertices are
z(O) = 140(0) + 210(0) = 0 + 0 = 0
z(C) = 140(10) + 210(0) = 1400 + 0 = 1400
z(P) = 140`(26/3) + 210(10/3) = (360 + 2100)/3 = 5740/3 = 1913.33`
z(Q) = 140(3) + 210(9) = 420 + 1890 = 2310
z (F) = 140(0) + 210(10) = 0 + 2100 = 2100
∴ z has maximum value 2310 when x = 3 and y = 9.
Hence, the carpenter should make 3 chairs and 9 tables to get the maximum profit of ₹ 2310.
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