If $x ≥ 03y-2x≥-12, 2x+5y ≤20$

The area of the triangle formed in the xy plane by the system of inequalities above is:

30

given, x≥0

3y−2x≥−12 and 2x+5y≤20

first, draw the graph for equations x=0

3y−2x=−12 and 2x+5y=20

x=0 is Y-axis. Hence x≥0 includes the above region of the line.

for 3y−2x=−12

substitute y=0 we get, −2x=−12⟹x=6

substitute x=0 we get, 3y=−12⟹y=−4

therefore, 3y−2x=−12 line passes through (6,0) and (0,-4) as shown in fig.

Hence, 3y−2x≥−12includes the region above the line.

similarly for 2x+5y=20

substitute y=0 we get, 2x=20⟹x=10

substitute x=0 we get, 5y=20⟹y=4

therefore, 2x+5y=20 line passes through (10,0) and (0,4) as shown in fig.

Hence, 2x+5y≤20includes the region below the line

the shaded region as shown in figure is intersection region

distance between (0,4) and (0,-4) is 8 units

adding 3y−2x=−12 and 2x+5y=20⟹y=1

substituting y=1 in any one of equation gives x=7.5

therefore, shaded triangle base =8 units and

height = 7.5 unit

Area = $\frac{1}{2}$ × base × height

$=\frac{1}{2}×8×7.5=30$