Find the common region of given inequation `3x + 4y le 60, x + 3y le 30, x ge 0, y ge 0`

Answer

N/a

Explanation

`x ge 0, y ge 0`
`rArr ` Solution lies I n first quadrant.
First inequation : `3x + 4y le 60`
Corresponding equation : `3x + 4y =60` ,brgt This line passes through the points A(20, 0) and B (0, 15) . Join AB. At point (0, 0), from the inequation, `0le 60`(True)
`:.` The solution of this inequation is that region of XY-place divided by line AB in which (0, 0) lies .
Second inequation : `x + 3y le 30`
Corresponding equation: `x + 3y 30`
This line passes through the points C(30, 0) and D (0, 10). Join CD.
At point (0, 0), from the inequation , `0 le 30` (True)

`:.` Solution of this inequatin is that region of XY-plane divided by line CD in which (0, 0) lies.
Therefore, the solution of given inequation is shown by the shaded region.