# 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 produ - Mathematics and Statistics

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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 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.

#### Solution

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 number of chairs and tables respectively.
∴ They cannot be negative.
∴ x ≥ 0, y ≥ 0
Now, profit for one chair is ₹ 140 and profit for one table is ₹ 210
∴ Total profit (Z) = 140x + 210y
This is objective function to be maximized
∴ 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 ≤ 60 2x + 6y = 60 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) + (0) ≤ 60∴ 0 ≤ 60∴ Origin-side x ≥ 0 x = 0 – – R.H.S. of Y-axis y ≥ 0 y = 0 – – above X-axis

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 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(36/3, 10/3)
Here, the objective function is Z = 140x + 210y
Now, we will find 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) = 2310 C (10, 0) Z = 140(10) + 210(0) = 1400

∴ Z has maximum value 2310 at G (3, 9)
∴ Maximum profit is ₹ 2310, when x = number of chairs = 3, y = number of tables = 9.

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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 6 Linear Programming
Miscellaneous Exercise 6 | Q 4.08 | Page 104