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Given,
Z = 20x + 10y
Constraints:
x + 2y ≤ 28
3x + y ≤ 24
x ≥ 2
First convert the given inequations into corresponding equations and plot them:
x + 2y ≤ 28 → x + 2y = 28 (corresponding equation)
Two coordinates required to plot the equation are obtained as:
Put, x = 0 ⇒ y = 14 (0,14) - - - - first coordinate.
Put, y = 0 ⇒ x = 28 (28,0) - - - - 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.
3x + y ≤ 24 → 3x + y = 24 (corresponding equation)
Two coordinates required to plot the equation are obtained as:
Put, x = 0 ⇒ y = 24 (0,24) - - - - first coordinate.
Put, y = 0 ⇒ x = 8 (8,0) - - - - second coordinate
x = 2 is the line parallel to y-axis passing through (2,0)

Hence, we have the following plot:

The shaded region in the above figure represents the region of a feasible solution.
Now to minimize 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 x + 2y = 28 and 3x + y = 24 gives (4,12)
Similarly solve other combinations by observing graph to get other coordinates.
From the figure we have obtained coordinates of corners as:
(2,0),(8,0),(2,13) and (4,12)

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 = 20x + 10y ∴ Z at (2,0) = 20× 2 + 10× 0 = 40
Z at (8,0) = 20×8 + 10×0 = 160
Z at (2,13) = 20×2 + 10×13 = 170
Z at (4,12) = 20×(4) + 10×(12) = 200
Clearly Z is maximum at x = 4 and y = 12 and maximum value is 200