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प्रश्न
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 the availability of each machine is given by the following table.
| Product/Machines | Chair (x) |
Table (y) |
Available time (hours) |
| Assembling | 3 | 3 | 36 |
| Finishing | 5 | 2 | 50 |
| Polishing | 2 | 6 | 60 |
Formulate and solve the following Linear programming problems using graphical method.
उत्तर
Let x be the number of chairs and y be the number of tables.
∴ The constraints are
3x + 3y ≤ 36
5x + 2y ≤ 50
2x + 6y ≤ 60
Since x and y are the numbers of chairs and tables, respectively.
∴ They cannot be negative.
∴ x ≥ 0, y ≥ 0
Now, the profit for one chair is ₹ 140 and the profit for one table is ₹ 210
∴ Total profit (Z) = 140x + 210y
This is an objective function to be maximized
∴ The given problem can be formulated as
Maximize Z = 140x + 210y
Subject to 3x + 3y ≤ 36
5x + 2y ≤ 50
2x + 6y ≤ 60
x ≥ 0, y ≥ 0
To find the graphical solution, construct the table as follows:
| Inequation | Equation | Double intercept form | Points (x, y) | Region |
| 3x + 3y ≤ 36 | 3x + 3y = 36 | `x/(2) + y/(12)` = 1 | A (12, 0) B (0, 12) |
3(0) + 3(0) ≤ 36 ∴ (0) ≤ 36 ∴ Origin-side |
| 5x + 2y ≤ 50 | 5x + 2y = 50 | `x/(10) + y/(25)` = 1 | C 10, 0) D (0, 25) |
5(0) + 2(0) ≤ 50 ∴ 0 ≤ 50 ∴ Origin-side |
| 2x + 6y ≤60 | 2x + 6y = 60 | `x/(30) + y/(10)` = 1 | E (30, 0) F (0, 10 |
2(0) + 6(0) ≤ 60 ∴ 0 ≤ 60 ∴ Origin-side |
| x ≥ 0 | x = 0 | – | – | R.H.S. of Y-axis |
| y ≥ 0 | y = 0 | – | – | above X-axis |
The shaded portion OFG HC is the feasible region,
Whose vertices are O (0, 0), F (0, 10), G, H and C (10, 0)
G is the point of intersection of lines.
2x + 6y = 60
i.e., x + 3y = 30 …(i)
and 3x + 3y = 36
i.e., x + y = 12 …(ii)
∴ By (i) – (ii), we get
x + 3y = 30
x + y = 12
– – –
2y = 18
∴ y = 9
Substituting y = 9 in (ii), we get
x + 9 = 12
∴ x = 12 – 9
∴ x = 3
∴ G = (3, 9)
H is the point of intersection of lines.
3x + 3y = 36
i.e., x + y = 12 …(ii)
5x + 2y = 50 …(iii)
∴ By 2 x (ii) – (iii), we get
2x + 2y = 24
5x + 2y = 50
– – –
– 3x – 26
∴ x = `(26)/(3)`
Substituting x = `(26)/(3)` in (ii), we get
`(26)/(3) + y` = 12
∴ y = `12 - (26)/(3) = (36 - 26)/(3)`
∴ y = `(10)/(3)`
∴ H`(26/3, 10/3)`
Here, the objective function is Z = 140x + 210y
Now, we will find the maximum value of Z as follows:
| Feasible Points | The value of Z = 140x + 210y |
| O (0, 0) | Z = 140(0) + 210(0) = 0 |
| F (0, 10) | Z = 140(0) + 210(10) = 2100 |
| G (3, 9) | Z = 140(3) + 210(9) = 420 + 1890 = 2310 |
| H`(36/3, 10/3)` | Z = `140(26/3) + 210(10/3) = (3640)/(3) + (2100)/(3)` = 1913.33 |
| C (10, 0) | Z = 140(10) + 210(0) = 1400 |
∴ Z has a maximum value of 2310 at G (3, 9)
∴ Maximum profit is ₹ 2310, when x = number of chairs = 3, y = number of tables = 9.
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