Maximise `Z=5x+3y`
Subject to `3x+5yle15, 5x+2yle10,xge0,yge0`.

Explanation

Maximise `Z=5x+3y`………………1
Subject to `3x+5yle15` …………..2
`5x+2yle10`………………..3
`xg30,yge0`………………….4
First we draw the graph of the line `3x+5y=15`

Put `(0,0)` in the inequation `3x+5yle15`
`3xx0+5xx0le15implies0le15` (True)
Therefore, half plane contains origin.
Since `x,yge0`
Therefore feasible region in first quadrant
Now draw the graph of the line `5x+2y=10`.

Put `(0,0)` in the inequation `5x+2yle10`
`5xx0+2xx0le10implies0le10`
Therefore half plane contains origin.
From the equation `3x+5y=15`
adn `5x+2y=10`,
We get `x=20/19` and `y=45/19`
Therefore, coordinates of point B are `(20/19, 45/19)`
Therefore, feasible region of OABCO.
Thus, the vertices of the feasible region ar `O(0,0), A(2,0),B(20/19, 45/19)` and `C(0,3)`. We find the value of `Z` at these vertices.

Therefore the maximum value of `Z` is `235/9` which is obtained at point `B(20/19, 45/19)`