Using the graphical method to find the feasible point-69705

6.

a.   max. z = 5x₁ + 6x₂

s.t.   -2x₁ + 3x₂ = 3   ………………i

x₁ + 2x₂ <= 5  ………………ii

6x₁ + 7x₂ <= 3 …………….iii

For eqⁿ   i  if x₁ = 0 ; x₂ = 1

x₂ = 0 ; x₁ = -3/2

ii  if x₁ = 0 ; x₂ = 5/2

x₂ = 0 ; x₁ = 5

iii if   x₁ = 0 ; x₂ = 3/7

x₂ = 0 ; x₁ = ½

Using the graphical method to find the feasible point

The solution will be feasible about the point P and the values of x₁ and x₂ about this point are 5, 5/2

So the feasible solution will be-

Max. z =  5x₁ + 6x₂

                    = 5*5 + 6*5/2

= 40

b.   min. z =    5x₁ + 6x₂

           s.t. 4x₁ + 5x₂ > = 10  ………………..i

4x₁ + 8x₂ > = 5  ………………….ii

For i if x₁ = 0 ; x₂ = 2

x₂ = 0 ; x₁ = 5/2

ii  x₁ = 0 ; x₂ = 5/8

x₂ = 0 ; x₁ = 5/4

Applying the graphical method to find the feasible point P

The feasible solution will be about point P and the values of x₁ and x₂ about this point are x₁ = 5/4  and x₂ = 5/8

So the feasible solution will be

min. z =  5x₁ + 6x₂

=  5*5/4 + 6*5/8

= 10

7.   Given that min. z = 5x₁ + 6x₂ + 3x₃

s.t. 5x₁ + 5x₂ + 3x₃ >= 50  …………….i

x₁ + x₂ – x₃ >= 20  ………………….ii

7x₁ + 6x₂ – 9x₃ >= 30  ……………..iii

5x₁ + 5x₂ + 5x₃ >= 35  …………….iv

2x₁ + 4x₂ – 15x₃ >= 10  …………….v

12x₁ + 10x₂ >= 90  …………………..vi

x₂ – 10x₃ >=  20  ……………………vii

x₁ , x₂ , x₃ >= 0

I will use the dual method to solve the given linear programming problem  as in the dual method we can easily find the dual point and will not be a lengthy procedure.

Min. D = 5x₁ + 6x₂ + 3x₃

The dual point is D(10, 10, 50/3)

So the optimal solution will be D = 5*10 + 6*10 + 3*50/3

= 160