Show that the following system of linear inequalities has no solutin ` x+2y le 3, 3x +4y ge 12, x ge 0, y ge 1`.

Explanation

First we draw the graph of `x +2 y le 3 `
Consider , the consider the line `x+2y =3`
Clearly , the points A(1, 1) and B(-1, 2) lie on this line.
Then, line AB represents x + 2y = 3.
Now, (0, 0) satisfies `x + 2y le 3 `
So, the line AB and part of the plane containing O(0,0) represent the solution set of `x + 2y le 3`
(ii) Next, we draw the graph of ` 3x + 4y ge 12 `
Consider the line ` 3x + 4y ge 12 `
Now `3x+4y = 12 rArr x/4+ y/3= 1`
So, this line meets the axes at C(4, 0) and D(0, 3).
Thus, CD represents the line `3x +4y ge 12 `
Clearly, (0, 0) does not satisfy `3x+ 4y ge 12`
Therefore, the line CD and part of the plane not containing 0(0, 0) represent the solution set of ` 3x + 4y ge 12 `
(iii) Clearly, `x ge 0 ` is represented by the y-axis and the plane on its right.
`y ge 1 ` represents a line parallel to the x-axis at a distance of 1 unit above the x-axis and the plane above this line.

Clearly all these planes have no common intersection
Hence , the given system of inequations has no solution.