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Standard XII

Mathematics

Objective Function

A factory use...

Question

A factory uses three different resources for the manufacture of two different products, 20 units of the resources A, 12 units of B and 16 units of C being available. 1 unit of the first product requires 2, 2 and 4 units of the respective resources and 1 unit of the second product requires 4, 2 and 0 units of respective resources. It is known that the first product gives a profit of 2 monetary units per unit and the second 3. Formulate the linear programming problem. How many units of each product should be manufactured for maximizing the profit? Solve it graphically.

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Solution

Let x units of first product and y units of second product be manufactured.

Therefore, $x, y \geq 0$

The given information can be tabulated as follows:

Product Resource A Resource B Resource C

First(x) 2 2 4

Second(y) 4 2 0

Availability 20 12 16

Therefore, the constraints are

$2 x + 4 y \leq 20 2 x + 2 y \leq 12 4 x + 0 y \leq 16 or 4 x \leq 16$

It is known that the first product gives a profit of 2 monetary units per unit and the second 3. Therefore, profit gained from x units of first product and y units of second product is 2x monetary units and 4y monetary units respectively.

Total profit = Z = $2 x + 3 y$ which is to be maximised

Thus, the mathematical formulation of the given linear programmimg problem is

Max Z = $2 x + 3 y$

subject to

$2 x + 4 y \leq 20 2 x + 2 y \leq 12 4 x + 0 y \leq 16 or 4 x \leq 16$
$x, y \geq 0$

First we will convert inequations into equations as follows :
2x + 4y = 20, 2x + 2y = 12, 4x = 16, x = 0 and y = 0

Region represented by 2x + 4y ≤ 20:
The line 2x + 4y = 20 meets the coordinate axes at A₁(10, 0) and B₁(0, 5) respectively. By joining these points we obtain the line 2x + 4y = 20.Clearly (0,0) satisfies the 3x + 2y = 210. So,the region which contains the origin represents the solution set of the inequation 2x + 4y ≤ 20.

Region represented by 2x + 2y ≤ 12:
The line 2x +2y =16 meets the coordinate axes at C₁(6, 0) and D₁(0, 6) respectively. By joining these points we obtain the line 2x + 2y = 12.Clearly (0,0) satisfies the inequation 2x + 2y ≤ 12. So, the region which contains the origin represents the solution set of the inequation 2x + 2y ≤ 12.

Region represented by 4x ≤ 16:
The line 4x =16 or x = 4 is the line passing through the point E₁(4, 0) and is parallel to Y axis.The region to the left of the line x = 4 would satisfy the inequation 4x ≤ 16.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 2x + 4y ≤ 20, 2x + 2y ≤ 12, 4x ≤ 16, x ≥ 0 and y ≥ 0 are as follows

The corner points are O(0, 0), B₁(0, 5), G₁ $(2, 4)$ , F₁(4, 2) and E₁(4, 0).

The values of Z at these corner points are as follows

Corner point Z= 2x + 3y

O 0

B₁ 15

G₁ 16

F₁ 14

E₁ 8

The maximum value of Z is 16 which is attained at G₁ $(2, 4)$ .

Thus, the maximum profit is 16 monetary units obtained when 2 units of first product and 4 units of second product were manufactured.

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Product	Resource A	Resource B	Resource C
First(x)	2	2	4
Second(y)	4	2	0
Availability	20	12	16