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Minimize z=4x+5y subject to 2x+y>=7, 2x+3y<=15, x<=3,x>=0, y>=0 solve using graphical method.

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Question

Minimize `z=4x+5y ` subject to `2x+y>=7, 2x+3y<=15, x<=3,x>=0, y>=0` solve using graphical method.

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Solution

Consider equations obtained by converting all inequations representing the constraints.

`2x+y=7 " i.e " x/3.5+y/7=1`

`2x+3y=15 " i.e " x/7.5+y/5=1`

`x=3,x=0,y=0`

Plotting these lines on graph we get the feasible region.

From the graph we can see that ABC is the feasible region.

Take any one point on the feasible region say P (2, 3) .
Draw initial isocost line z passing through the point (2, 3) .

`therefore z_1=4(2)+5(3)=8+15=23`

`therefore " intial isocost line is " 4x+5y=23.`

Since the objective function is of minimization type, from the graph we can see that the line z3 contains only one point A(3, 1) of the feasible region ABC.

Minimum value of z = 4(3)+ 5(1)= 12 + 5 =17
z is minimum when x = 3 and y = 1.

 

 

 

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2015-2016 (March)

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