Q. 185.0( 1 Vote )

Solve each of the following linear programming problems by graphical method.

Maximize Z = - x1 + 2x2

Subject to :

- x1 + 3x2 ≤ 10

x1 + x2 ≤ 6

x1 - x2 ≤ 2

x1, x2 ≥ 0

Answer :

Given,


Objective function is: Z = - x1 + 2x2


Constraints are:


- x1 + 3x2 ≤ 10


x1 + x2 ≤ 6


x1 - x2 ≤ 2


x1, x2 ≥ 0


First convert the given inequations into corresponding equations and plot them:


- x1 + 3x2 ≤ 10 - x1 + 3x2 = 10 (corresponding equation)


Two coordinates required to plot the equation are obtained as:


Put, x1 = 0 x2 = 10/3 (0,10/3) - - - - first coordinate.


Put, x2 = 0 x1 = - 10 ( - 10,0) - - - - second coordinate


x1 + x2 ≤ 6 x1 + x2 = 6 (corresponding equation)


Two coordinates required to plot the equation are obtained as:


Put, x1 = 0 x2 = 6 (0,6) - - - - first coordinate.


Put, x2 = 0 x1 = 6 (6,0) - - - - second coordinate


x1 - x2 ≤ 2 x1 - x2 = 2 (corresponding equation)


Two coordinates required to plot the equation are obtained as:


Put, x1 = 0 x2 = - 2 (0, - 2) - - - - first coordinate.


Put, x2 = 0 x1 = 2 (2,0) - - - - second coordinate


x1 ≥ 0 and x2 ≥ 0


Join them to get the line.


As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,


If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.


Similarly, we repeat the steps for other inequation also and find the common region.


Hence, we obtain the following plot:



The shaded region in the above figure represents the region of a feasible solution.


Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.


We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.


Solving x1 - x2 = 2 and x1 + x2 = 6 gives (6,6)


Similarly solve other combinations by observing graph to get other coordinates.


From figure we have obtained coordinates of corners as:


(0,0),(2,0),(4,2),(2,4) and (0,10/3)


Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.


Z = - x1 + 2x2


Z at (0,0) = 0


Z at (2,0) = - 2 + 2×0 = - 2


Z at (4,2) = - 4 + 2× 2 = 0


Z at (2,4) = - 2 + 2×4 = 6


Z at (0,10/3) = - 0 + 2× (10/3) = 20/3


We can see that Z is maximum at (0,10/3) and max. value is 20/3


Z is maximum at x = 0 and y = 10/3 ; and max value is 20/3


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