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

Maximize Z = 50x + 30y

Subject to :

2x + y ≤ 18

3x + 2y ≤ 34

x, y ≥ 0

Given,

Objective function is: Z = 50x + 30y

Constraints are:

2x + y ≤ 18

3x + 2y ≤ 34

x, y ≥ 0

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

2x + y ≤ 18 → 2x + y = 18 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 18 (0,18) - - - - first coordinate.

Put, y = 0 ⇒ x = 9 (9,0) - - - - second coordinate

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 the 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.

3x + 2y ≤ 34 → 3x + 2y = 34 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 17 (0, 17) - - - - first coordinate.

Put, y = 0 ⇒ x = 34/3 (34/3, 0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x – axis

Hence we obtain a plot as shown in figure:

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 3x + 2y = 34 and 2x + y = 18 gives (2,14)

Similarly solving 3x + 2y = 34 and x = 0 gives (0, 17)

And solving 2x + y = 18 and y = 0 gives (9,0)

And solving x = 0 and y = 0 gives (0,0)

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 = 50x + 30y

∴ Z at (2,14) = 50 × 2 + 30 × 14 = 520

Z at (0,17) = 50 × 0 + 30 × 17 = 510

Z at (9,0) = 50 × (9) + 30 × 0 = 450

Z at (0,0) = 0

We can see that Z is maximum at (2, 14) and max. value is 520

∴ Z is maximum at x = 2 and y = 14 ; and max value is 520.