Solution:

The given inequalities are \[3x+4y\le 60\], \[x+3y\le 30\], \[x\ge 0\], \[y\ge 0\]

For \[3x+4y\le 60\],

Let us put value of \[x=0\] and \[y=0\] in equation one by one, we get

\[y=15\]and \[x=20\]

We get the required points as \[(0,15)\]and \[(20,0)\]

To check if the origin is included in the line`s graph \[(0,0)\]

\[0\le 60\], which is true, hence the origin would lie in the solution area of the line`s graph.

Therefore, required solution area would be on the left of the line`s graph.

For \[x+3y\le 30\],

Let us put value of \[x=0\] and \[y=0\] in equation one by one, we get

\[y=10\]and \[x=30\]

We get the required points as \[(0,10)\] and \[(30,0)\]

To check if the origin is included in the line`s graph \[(0,0)\]

\[0\le 30\], which is true, hence the origin lies in the solution area which is given by the left side of the line`s graph.

Let us Consider \[x\ge 0\], \[y\ge 0\],

We know that the given inequalities imply the solution lies in the first quadrant only.

In the below graph the shaded area in the graph is the required solution of the given inequalities.