Question

Solve the following problem :

A person makes two types of gift items A and B requiring the services of a cutter and a finisher. Gift item A requires 4 hours of cutter's time and 2 hours of finisher's time. B requires 2 hours of cutters time, 4 hours of finishers time. The cutter and finisher have 208 hours and 152 hours available times respectively every month. The profit of one gift item of type A is ₹ 75 and on gift item B is ₹ 125. Assuming that the person can sell all the items produced, determine how many gift items of each type should be make every month to obtain the best returns?

Answer 1

Let x gift items of type A and y gift items of type B be produced by the person.
∴ Total profit Z = 75x + 125y
This is the objective function to be maximized.
The given information can be tabulated as shown below:

	Type A (x)	Type B (y)	Total time available (in hours)
Cutter	4	2	208
Finisher	2	4	152

∴ The constraints are 4x + 2y ≤ 208, 2x + 4y ≤ 152, x ≥ 0, y ≥ 0
∴ Given problem can be formulated as
Maximize Z = 75x + 125y
Subject to, 4x + 2y ≤ 208, 2x + 4y ≤ 152, x ≥ 0, y ≥ 0
To draw feasible region, construct table as follows:

Inequality	4x + 2y ≤ 208	2x + 4y ≤ 152
Corresponding equation (of line)	4x + 2y = 208	2x + 4y = 152
Intersection of line with X-axis	(52, 0)	(76, 0)
Intersection of line with Y-axis	(0, 104)	(0, 38)
Region	Origin side	Origin side

Shaded portion OABC is the feasible region,
whose vertices are O ≡ (0, 0),
A ≡ (52, 0), B and C ≡ (0, 38).
B is the point of intersection of the lines 4x + 2y = 208 i.e. 2x + y = 104 and 2x + 4y = 152

Solving the above equations, we get B ≡ (44, 16)
Here, the objective function is Z = 75x + 125y
∴ Z at O(0, 0) = 75(0) + 125(0) = 0
Z at A(52, 0) = 75(52) + 125(0) = 3900
Z at B(44, 16) = 75(44) + 125(16) = 5300
Z at C(0, 38) = 75(0) + 125(38) = 4750
∴ Z has maximum value 5300 at B(44, 16)
∴ Z is maximum, when x = 44, y = 16
Thus, a person should make 44 gift items of type A and 16 gift items of type B every month to obtain the best returns of ₹ 5300.