Question

Aman has ₹ 1500 to purchase rice and wheat for his grocery shop. Each sack of rice and wheat costs ₹ 180 and Rupee ₹ 120 respectively. He can store a maximum number of 10 bags in his shop. He will earn a profit of ₹ 11 per bag of rice and ₹ 9 per bag of wheat.

1. Formulate a Linear Programming Problem to maximise Aman’s profit.
2. Calculate the maximum profit.

Answer 1

i. Let x be the number of rice sacks purchased and y be the number of wheat sacks purchased.

Cost of each sack of rice = ₹ 180

Cost of each sach of wheat = ₹ 120

Profit earned on each sack of rice = ₹ 11

Profit earned on each sack of wheat = ₹ 9

As a result, the problem can be expressed as LPP as follows:

Maximum P = 11x + 9y,

Subject to the constraints

180x + 120y ≤ 1500

or 3x + 2y ≤ 25  ...(i)

x + y ≤ 10  ...(ii)

x, y > 0

ii. Draw the lines 3x + 2y = 25 and x + y = 10 and shade the area bounded by the given inequations. The shaded region represents a bounded feasible region. Lines 3x + 2y = 25 and x + y = 10 intersect at (5, 5).

For equation (i)

3x + 2y = 25

x	0	`25/2`
y	`25/3`	0

Points: `(0, 25/2), (25/3, 0)`

For equation (ii)

x + y = 10

x	0	10
y	10	0

Points: (0, 10), (10, 0)

The four corner points of the feasible region OABC are O(0, 0), `A(25/3, 0)`, B(5, 5), C(0, 10)

At each corner point, we calculate Z = 11x + 9y.

Corner points	Value of objective function Z = 11x + 9y
O(0, 0)	Z = 11 × 0 + 9 × 0 = 0
`A(25/3, 0)`	Z = `11 xx 25/3 + 9 xx 0 = 275/3`
B(5, 5)	Z = 11 × 5 + 9 × 5 = 100
C(0, 10)	Z = 11 × 0 + 9 × 10 = 90

P is found to be greatest at B(5, 5), with a maximum value of P = 100.

Hence, Aman earns the greatest profit of ₹ 100 when purchasing 5 sacks of rice and 5 sacks of wheat.