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

Answer 1

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.