Question

Refer to question 14. How many sweaters of each type should the company make in a day to get a maximum profit? What is the maximum profit.

Answer 1

Referring to the solution of Question No.14,

We have Maximise Z = 200x + 120y

Subject to the constraints

x + y ≤ 300  ......(i)

3x + y ≤ 600  ......(ii)

x – y ≥ – 100  ......(iii)

x ≥ 0, y ≥ 0

On solving equation (i) and (iii)

We have x = 100, y = 200

On solving eq. (i) and (ii)

We get x = 150, y = 150

Let x + y = 300

x	0	300
y	300	0

Let 3x + y = 600

x	0	200
y	600	0

Let x + y = –100

x	0	–100
y	100	0

Here, the shaded region is the feasible region whose corner points are O(0, 0), A(200, 0), B(150, 150), C(100, 200), D(0, 100).

Let us evaluate the value of Z.

Corner points	Value of Z = 200x + 120y
O(0, 0)	Z = 200(0) + 120(0) = 0
A(200,0)	Z = 200(200) + 120(0) = 40000
B(150, 150)	Z = 200(150) + 120(150) = 48000	← Maximum
C(100, 200)	Z = 200(100) + 120(200) = 44000
D(0, 100)	Z = 200(0) + 120(100) = 12000

Hence, the maximum value of Z is 48000 at (150, 150)

i.e., 150 sweaters of each type.