Question

Minimize: Z = 6x + 4y

Subject to the conditions:

3x + 2y ≥ 12,

x + y ≥ 5,

0 ≤ x ≤ 4,

0 ≤ y ≤ 4

Answer 1

Graph the Feasible Region

(i) 3x + 2y ≥ 12

3x + 2y = 12 ⇒ intercepts: `{:{(x=0=> y=6),(y=0=>x=4):}`

(ii) x + y ≥ 5

x+y=5⇒intercepts: `{:{(x=0=>y=5),(y=0=>y=5):}`

(iii) 0 ≤ x ≤ 4, 0 ≤ y ≤ 4

This defines a square bounded by (0,0), (4,0), (4,4), and (0,4).

Find Corner Points of the Feasible Region

• x + y ≥ 5

• 3x + 2y ≥ 12

1) 3x + 2y = 12 and x + y = 5

From (2): y = 5 − x

3x + 2(5 − x) = 12 ⇒ 3x + 10 − 2x = 12 ⇒ x = 2, y = 3

Point A = (2, 3)

2) 3x + 2y = 12 and y = 4

3x + 2(4) = 12 ⇒ 3x = 4 ⇒ x = `4/3, y =4`

Point B = `(4/3, 4)`

3) x + y = 5 and y = 4

x + 4 = 5 ⇒ x = 1, y = 4

Point C = (1, 4)

Points where boundary lines meet square limits:

(4, 0): Check constraints

3(4) + 2(0) = 12

4 + 0 = 4 → Doesn't satisfy x + y ≥ 5

(4, 1):

3(4) + 2(1) = 14

4 + 1 = 5

Point D = (4, 1)

Evaluate Z = 6x + 4y at all feasible corner points

Point	Z = 6x + 4y
A = (2, 3)	6(2) + 4(3) = 12 + 12 = 24
B = (4/3, 4)	6(4/3) + 4(4) = 8 + 16 = 24
C = (1, 4)	6(1) + 4(4) = 6 + 16 = 22
D = (4, 1)	6(4) + 4(1) = 24 + 4 = 28

The minimum value of Z is 22, which occurs at point C = (1, 4).