Minimize : Z = 6x + 4y Subject to the conditions: 3x + 2y ≥ 12, x + y ≥ 5, 0 ≤ x ≤ 4, 0 ≤ y ≤ 4 - Mathematics and Statistics

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Sum

Minimize : Z = 6x + 4y

Subject to the conditions:

3x + 2y ≥ 12,

x + y ≥ 5,

0 ≤ x ≤ 4,

0 ≤ y ≤ 4

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Solution

To draw feasible region, construct table as follows:

Inequality 3x+2y ≥12  x+y ≥ 5 x ≤ 4  y ≤ 4
Corresponding
equation (of line)
3x + 2y = 12 x+y=5 x=4 y=4
Intersection of line
with X-axis
(4, 0) (5, 0) (4, 0)          -
Intersection of line
with Y-axis
(0, 6) (0, 5)          - (0, 4)
Region Non-origin side Non-origin side Origin side Origin side

Shaded portion is the feasible region,

Where B(4, 4), A(4, 1), C`(4/3 , 4)`

For D : 3x + 2y = 12

x + y = 5 

Multiply equation (ii) by ‘2’ and subtract it from (i), 

3x + 2y = 12 

2x + 2y = 10 

`(-      -       -)/(x   = 2)`  

∴ y = 3 and hence D (2,3)

Z = 6x + 4y

Z at A (4,1) = 6(4) + 4(1) = 28

Z at B(4,4) = 6(4) + 4(4) = 40

Z at C `(4/3 , 4) = 6 (4/3) + 4(4) = 28`

Z at D (2,3) = 6(2) + 4(3) = 24

Thus, Z is minimized at D (2,3) and its minimum value is 24.

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2014-2015 (October)

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