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Given,
Objective function is: Z = 3x1 + 5x2 Constraints are:x1 + 3x2 ≥ 3
x1 + x2 ≥ 2
x1, x2 ≥ 0
First convert the given inequations into corresponding equations and plot them:
x1 + 3x2 ≥ 3 → x1 + 3x2 = 3 (corresponding equation)
Two coordinates required to plot the equation are obtained as:
Put, x1 = 0 ⇒ x2 = 1 (0,1) - - - - first coordinate.
Put, x2 = 0 ⇒ x1 = 3 (3,0) - - - - 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 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.
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 is the y-axis and x2 = 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 x1 + x2 = 2 and x1 + 3x2 = 3 gives (3/2,1/2)
Similarly solve other combinations by observing graph to get other coordinates.
From the figure we have obtained coordinates of corners as:
(3/2,1/2) ,(3,0) and (0,2)

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 = 3x1 + 5x2
∴ Z at (3,0) = 3× 3 + 5× 0 = 9
Z at (0,2) = 3× 0 + 5×2 = 10
Z at (3/2,1/2) = 3×(3/2) + 5×(1/2) = 7
Note: As the region is unbounded, so we need to check whether any value less than 7 is possible for Z or not. If it is unique, we will say that under given constraints we found the minimum Z
For this we define inequation as: 3x + 2y < 7

As (0,0) satisfies the inequation, so this is the region specified by above inequation. As our feasible region does not coincide with the region specified by 3x + 2y < 7.
∴ Z can be minimized
We can see that Z is minimum at (3/2,1/2) and min. value is 7
∴ Z is minimum at x = 3/2 and y = 1/2 ; and min value is 7