Step - 1

Step - 2

Step - 3

Step - 4

Given,
Z = 3x + 5y
Constraints:
- 2x + y ≤ 4
x + y ≥ 3
x – 2y ≤ 2
x, y ≤ 0
First convert the given inequations into corresponding equations and plot them:
- 2x + y ≤ 4 → - 2x + y = 4 (corresponding equation)
Two coordinates required to plot the equation are obtained as:
Put, x = 0 ⇒ y = 4 (0,4) - - - - first coordinate.
Put, y = 0 ⇒ x = - 2 ( - 2,0) - - - - 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.
x + y ≥ 3 → x + y = 3 (corresponding equation)
Two coordinates required to plot the equation are obtained as:
Put, x = 0 ⇒ y = 3 (0,3) - - - - first coordinate.
Put, y = 0 ⇒ x = 3 (3,0) - - - - second coordinate
x – 2y ≤ 2 → x – 2y = 2 (corresponding equation)
Two coordinates required to plot the equation are obtained as:
Put, x = 0 ⇒ y = - 1 (0, - 1) - - - - first coordinate.
Put, y = 0 ⇒ x = 2 ( 2,0) - - - - second coordinate
x = 0 is the y - axis and y = 0 is the x - axis.
Hence, we have the following plot:

There is no shaded region in the above figure represents that there is no region of a feasible solution.
x + y ≥ 3 will not be bounded by x, y ≤ 0. Thus, no feasible region is there.
∴ There is no possible minimum value Z.