A company produces two types of pens A and B. Pen A is of superior qua

1.	A company produces two types of pens A and B. Pen A is of superior quality and pen B is of lower quality. Profits on pens A and B are Rs 5 and Rs 3 per pen respectively. Raw materials required for each pen A is twice as that of pen B. The supply of raw material is sufficient only for 1000 pens per day. Pen A requires a special clip and only 400 such clips are available per day. For pen B, only 700 clips are available per day. Formulate this problem as a linear programming problem.
Answer» (i) Variables: Let x₁ and x₂ denotes the number of pens in type A and type B. (ii) Objective function: Profit on x₁ pens in type A is = 5x₁ Profit on x₂ pens in type B is = 3x₂ Total profit = 5x₁ + 3x₂ Let Z = 5x₁ + 3x₂, which is the objective function. Since the B total profit is to be maximized, we have to maximize Z = 5x₁ + 3x₂ (iii) Constraints: Raw materials required for each pen A is twice as that of pen B. i.e., for pen A raw material required is 2x₁ and for B is x₂. Raw material is sufficient only for 1000 pens per day ∴ 2x₁ + x₂ ≤ 1000 Pen A requires 400 clips per day ∴ x₁ ≤ 400 Pen B requires 700 clips per day ∴ x₂ ≤ 700 (iv) Non-negative restriction: Since the number of pens is non-negative, we have x₁ > 0, x₂ > 0. Thus, the mathematical formulation of the LPP is Maximize Z = 5x₁ + 3x₂ Subj ect to the constrains 2x₁ + x₂≤ 1000, x₁ ≤ 400, x₂ ≤ 700, x₁, x₂ ≥ 0

A company produces two types of pens A and B. Pen A is of superior quality and pen B is of lower quality. Profits on pens A and B are Rs 5 and Rs 3 per pen respectively. Raw materials required for each pen A is twice as that of pen B. The supply of raw material is sufficient only for 1000 pens per day. Pen A requires a special clip and only 400 such clips are available per day. For pen B, only 700 clips are available per day. Formulate this problem as a linear programming problem.

Answer»

(i) Variables:

Let x₁ and x₂ denotes the number of pens in type A and type B.

(ii) Objective function:

Profit on x₁ pens in type A is = 5x₁

Profit on x₂ pens in type B is = 3x₂

Total profit = 5x₁ + 3x₂

Let Z = 5x₁ + 3x₂, which is the objective function.

Since the B total profit is to be maximized, we have to maximize Z = 5x₁ + 3x₂

(iii) Constraints:

Raw materials required for each pen A is twice as that of pen B.

i.e., for pen A raw material required is 2x₁ and for B is x₂.

Raw material is sufficient only for 1000 pens per day

∴ 2x₁ + x₂ ≤ 1000

Pen A requires 400 clips per day

∴ x₁ ≤ 400

Pen B requires 700 clips per day

∴ x₂ ≤ 700

(iv) Non-negative restriction:

Since the number of pens is non-negative, we have x₁ > 0, x₂ > 0.

Thus, the mathematical formulation of the LPP is

Maximize Z = 5x₁ + 3x₂

Subj ect to the constrains

2x₁ + x₂≤ 1000, x₁ ≤ 400, x₂ ≤ 700, x₁, x₂ ≥ 0