A company manufactures two types of cardigans: type A and type B. It costs ₹ 360 to make a type A cardigan and ₹ 120 to make a type B cardigan. The company can make at most 300 cardigans and spend at most ₹ 72,000 a day. The number of cardigans of type B cannot exceed the number of cardigans of type A by more than 200. The company makes a profit of ₹ 100 for each cardigan of type A and ₹ 50 for every cardigan of type B.

Formulate this problem as a linear programming problem to maximize the profit to the company. Solve it graphically and find maximum profit.

Let no. of cardigans of type A be x, and that of type B be x

and it is given that the company makes a profit 100 for type A and 50 for type B.

Hence to maximize the profit, the objective function is

max Z = 100x + 50y

Now, Subject to constraints,

It is given that it costs Rs 360 to make type A and Rs 120 to make type B.

It can spend at most of Rs 72000 per day.

∴ 360x + 120y ≤ 72000

⇒ 3x + y ≤ 600 …(i)

It is given that the company can sell product at most of 300 cardigans of both types per day.

∴ x + y ≤ 300 …(ii)

It is also given that type B no of cardigan cannot exceed no of type A cardigan by more than 200.

∴ y – x ≤ 200 …(iii)

Now let us draw the graph for the lines

AB: 3x + y = 600

CD: x + y = 300

EF: y – x = 200

Consider the line AB: 3x + y = 600

Put x = 0, y = 0 then 0 ≤ 600 is true.

Hence the region 3x + y ≤ 600 lies below the line AB.

Consider the line CD: x + y = 300

Put x=0, y=0 then 0 ≤ 300 is true.

Hence the region x + y ≤ 300 lies below the line CD.

Consider the line EF: y – x = 200

Put x=0, y=0 then 0 ≤ 200 is true.

Hence the region y – x ≤ 200 lies below the line EF.

The feasible region OXPQB is the shaded portion shown in the fig.

The coordinates of P is (50,250)

The coordinates of Q is (150,150)

The corner points are O(0,0), X(0,200), P(50,250), Q(150,150), B(200,0)

Let us obtain the values of the objective function Z = 100x + 50y

At the points (x,y)the value of the objective function subjected to

Z = 100x + 50y

At O (0,0)the value of the objective function

Z = 0

At X (0,200) the value of the objective function

Z = 100 × 0 + 50 × 200 = 10000

At P (50,250)the value of the objective function

Z = 100 × 50 + 50 × 250 = 5000 + 12500 = 17500

At Q (150,150) the value of the objective function

Z = 100 × 150 + 50 × 150 = 15000 + 7500 = 22500

At B (200,0)the value of the objective function

Z = 100 × 200 + 0 = 20000

The maximum profit is at Rs.22,500 at Q(150,150)

Hence, no. of cardigans of type A = 150 and type B = 150