Question

A chemical company based in Pahang makes two types of industrial solvents, S1 and S2. Each solvent is a mixture of three chemicals. Each kL of S1 requires 12L of chemical K, 9L of chemical L, and 30L of chemical M. Each kL of S2 requires 24L of chemical K, 5L of chemical L, and 30L of chemical M. The profit per kL of S1 is $100, and the profit per kL of S2 is $85. The inventory of the company shows 480 L of chemical K, 180 L of chemical L, and 720 L of chemical M. Assuming the company can sell all the solvent it makes, find the number of kL of each solvent that the company should make to maximize profit. {let x1=the # of kL of S1, and x2=the # of kL of S2}.

Answer 1

LPP Formulation:

Decision Variables: let

x₁=the # of kL of S1,

x₂=the # of kL of S2}.

Objective Function:

The objective is to maximize the total profit of the production mix.

Total profit = units of S1 x unit profit of S1 + units of S2 x unit profit of S2

Profit function is as follows:

Maximize Z = $100X₁ + $85X₂

Subject to:

Chemical K requirement and availability:

Total Chemical required for production <= available Liters of K

units of S1 x L of chemical K required per S1 + units of S2 x L of chemical K required per S2 <= available Liters of K

12X₁ + 24X₂ <= 480

Chemical L requirement and availability:

9X₁ + 5X₂ <= 180

Chemical L requirement and availability:

30X₁ + 30X₂ <= 720

Nonnegative Constraint:

X₁, X₂ >= 0

Graphical Method:

Constraint	Equality Equation	X-axis coordinate (X2 = 0)	Y-axis coordinate (X1 = 0)	Line and feasible region according to constraint type
1	12X1 + 24X2 = 480	X1 = 4 A (40 , 0)	X2 = 2 B (0, 20)	Towards origin
2	9X1 + 5X2 = 180	X1 = 20 C (20, 0)	X2 = 36 D (0, 36)	Towards origin
3	30X1 + 30X2 = 720	X1 = 24 E (24, 0)	X2 = 24 F (0, 24)	Towards origin

Plotting of equations:

According to graph feasible area is OFGDA.

The coordinates of G is obtained as follows:

At point G equation (2) and Equation (3) are intersecting, solve both equation by simultaneous methods.

Equation (3) – 6 x Equation (2)

30X₁ + 30X₂– 6*(9X₁ + 5X_2­) = 720 – 6*180

30X₁ + 30X₂– 54X₁ - 30X_2­) = 720 – 1080

-24 X₁ = - 360

X₁ = 15

X₁ = 15, submit in equation (3)

30*15 + 30X₂ = 720 or X₂ = 9

Coordinate of G are (15, 9)

At point D equation (1) and Equation (3) are intersecting, solve both equation by simultaneous methods. We get the coordinates as (8, 16).

By extreme point’s method:

Extreme Points	Objective function Value (z = $100X1 + $85X2)
O (0, 0)	100 x 0 + 85 x 0 = 0
F (20, 0)	100 x 20 + 85 x 0 = 2000
G (15, 9)	100 x 15 + 85 x 9 = 2265
D (8 , 16)	100 x 8 + 85 x 16 = 2160
A (0, 20)	100 x 0 + 85 x 20 = 1700

The highest objective function value is at point G (15,9)

Thus the optimal solution is obtained at point G, where,

X₁ = 15 and X₂ = 9

Z = 2265

The optimal solution is to produce 15 kL of S1 and 9 kL of S2 to realize the maximum profit of $2265.