The inequalities are
3x + 4y ≤ 60 ... (1)
x + 3y ≤ 30 ... (2)
x ≥ 0 ... (3)
y ≥ 0 ... (4)

We first draw the graphs of lines
l1 : 3x + 4y = 60, l2 : x + 3y = 30, x = 0 and y = 0.
(i) The line 3x + 4y = 60 passes through (20, 0) and (0, 15) which is represented by AB. Consider the inequality 3x + 4y ≤ 60, putting x = 0, y = 0 in 3x + 4y ≤ 60, we get 0 + 0 ≤ 60, which is true.
∴ 3x + 4y ≤ 60 represents the region below AB and all the points on AB.
(ii) Further, x + 3y = 30 passes through (0, 10) and (30, 0), CD represents this line.
Consider the inequality x + 3y ≤ 30
Putting x = 0, y = 0 in x + 3y ≤ 30, we get 0 < 30 is true.
∴ Origin lies in the region x + 3y ≤ 30. This inequality represents the region below it and the line itself.
Thus, we note that inequalities (1) and (2) represent the two regions below the respective lines (including the lines).
Inequality (3) represents the region on the right of y-axis and the y-axis itself.
Inequality (4) represents the region above x-axis and the x-axis itself.
∴ Shaded area in the figure is the solution area.