Question

A wholesale dealer deals in two kinds of mixtures A and B of nuts. Each kg of mixture A contains 60 grams of almonds, 30 grams of cashew and 30 grams of hazel nuts. Each kg of mixture B contains 30 grams of almonds, 60 grams of cashew and 180 grams of hazel nuts. A dealer is contemplating to use mixtures A and B to make a bag which will contain at least 240 grams of almonds, 300 grams of cashew and 540 grams of hazel nuts. Mixture A costs ₹ 8 and B costs ₹ 12 per kg. How many kgs of each mixture should he use to minimize the cost of the kgs

Answer 1

Let the dealer use x kg of mixture A and y kg of mixture B.

Since x and y cannot be negative, x ≥ 0, y ≥ 0

Mixture A costs ₹ 8 and Mixture B costs ₹ 12 per kg.

∴ Total cost = Z = 8x + 12y

We construct a table with the constraints of Almond, Cashew and Hazelnut as follows:

	Mixture A	Mixture B	Least value
Almond	60	30	240
Cashew	30	60	300
Hazelnut	30	180	540

From the table, the constraints are

60x + 30y ≥ 240

30x + 60y ≥ 300

30x + 180y ≥ 540

∴ Given problem can be formulated as follows:

Minimize Z = 8x + 12y

Subject to 60x + 30y ≥ 240

30x + 60y ≥ 300

30x + 180y ≥ 540, x ≥ 0, y ≥ 0

To draw the feasible region, construct table as follows:

Inequality	60x +30y ≥ 240	30x+ 60y ≥ 300	30x+ 180y ≥ 540
Corresponding equation (of line)	60x+ 30y = 240	30x + 60y = 300	30x+ 180y = 540
Intersection of line with X-axis	(4, 0)	(10, 0)	(18, 0)
Intersection of line with Y-axis	(0, 8)	(0, 5)	(0, 3)
Region	Non-origin side	Non-origin side	Non-origin side

Shaded portion XABCDY is the feasible region, whose vertices are A(18, 0), B, C and D(0, 8).

B is the point of intersection of the lines 30x + 180y = 540 and 30x + 60y = 300.

Solving the above equations, we get

x = 6, y = 2

∴ B ≡ (6, 2)

C is the point of intersection of the lines 60x + 30y = 240 and 30x + 60y = 300.

Solving the above equations, we get

x = 2, y = 4

∴ C ≡ (2, 4)

Here, the objective function is

Z = 8x + 12y

∴ Z at A(18, 0) = 8(18) + 12(0)

= 144

Z at B(6, 2) = 8(6) + 12(2)

= 48 + 24

= 72

Z at C(2, 4) = 8(2) + 12(4)

= 16 + 48

= 64

Z at D(0, 8) = 8(0) + 12(8)

= 96

∴ Z has minimum value 64 at x = 2 and y = 4.

∴ 2 kgs of mixture A and 4 kgs of mixture B should be used to minimize the cost of the kgs.