Math, asked by sauravvashishtha, 1 month ago

Solve by graphical method max z=5x+8y, subject to constraints 3x+2y<=36,3x+4y>=24,x,y>=0

Answers

Answered by pulakmath007
3

SOLUTION

TO DETERMINE

Solve by graphical method max z = 5x + 8y

subject to constraints

3x + 2y ≤ 36

3x + 4y ≥ 24

x , y ≥ 0

EVALUATION

We have to maximize z = 5x + 8y

subject to constraints

3x + 2y ≤ 36

3x + 4y ≥ 24

x , y ≥ 0

Here the given inequalities are

3x + 2y ≤ 36 , 3x + 4y ≥ 24

Corresponding equations are

3x + 2y = 36 , 3x + 4y = 24

We Draw the lines in the graph

ABCDA is the bounded region

The corner points are A(8,0) , B(0,6) , C(0,18) , D(12,0)

At A(8,0) we have z = 40

At B(0,6) we have z = 48

At C(0,18) we have z = 144

At D(12,0) we have z = 60

So max z = 144 at the point C(0,18)

Hence the required solution is

max z = 144 when x = 0 , y = 18

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