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Solve the following Linear Programming problem graphically: Maximize: Z = 3x + 3.5y Subject to constraints: x + 2y ≥ 240, 3x + 1.5y ≥ 270, 1.5x + 2y ≤ 310, x ≥ 0, y ≥ 0. - Mathematics

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Question

Solve the following Linear Programming problem graphically:

Maximize: Z = 3x + 3.5y

Subject to constraints:

x + 2y ≥ 240,

3x + 1.5y ≥ 270,

1.5x + 2y ≤ 310,

x ≥ 0, y ≥ 0.

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Solution

Changing inequations to equations

x + 2y = 240

3x + 1.5y = 270

1.5x + 2y = 310

For equation (i)

x 0 240
y 120 0

For equation (ii)

x 0 90
y 180 0

For equation (iii)

x 0 206.7
y 155 0


For A:

(3x + 1.5y = 270) × 2

(1.5x + 2y = 310) × 3/2

`\implies`       6x + 3y = 540
      2.25x + 3y = 465
         –     –        –     
              3.75x = 75

x = `75/3.75` = 20

y = `(310 - (1.5 xx 20))/2`

= `(310 - 30)/2`

= 140

A = (20, 140)

For B:

3x + 1.5y = 270

(x + 2y = 240) × 3

`\implies` 3x + 1.5y = 270
        3x + 6y = 720
        –     –       –        
           – 4.5y = – 450

`\implies` y = 100

x = 40

B = (40, 100)

For C:

     x + 2y = 240
1.5x + 2y = 310
–     –          –     
     – 0.5x = – 70

x = 140

y = 50

C = (140, 50)

Corner points Max. Z = 3x + 3.5y
A(20, 140) (3 × 20) + (3.5 × 140) = 550
B(40, 100) (3 × 40) + (3.5 × 100) = 470
C(140, 50) (3 × 140) + (50 × 3.5) = 595 `rightarrow` Max.

Maximum value is 595 at (140, 50).

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