Let there is `x` kg of food `X` and `y` kg of food `Y` in the mixture.
`:.` Minimize cost `z=50x+60y`
Subject to `xge0`
`yge0`
`2x+5yge8` brgt `5x+2yge11`
Draw the graph of equations `x=0, y=0, 2x+4y=8` and `5x+2y=11`.
Now the feasible region for the inequations `xge0,yge0,2x+5yge8,5x+2yge11`, and shade it. The vertices of the shaded region are `A(4,0),B(0,11/2)` and `C(13/7,6/7)`. We will find the value of `z` at these vertices.
Therefore the minimum cost is Rs. `144 2/7` and food `X` and `Y` are taken `13/7` kg and `6/7` respectively.