Find graphically, the maximum value of z=2x+5y, subject to constraints given below: 2x+4y≤8, 3x+y≤6, x+y≤4, x≥0, y≥0.

Maximize z=2x+5y, subject to the constraints 2x+4y≤8 ⇒ x+2y≤4
3x+y≤6, x+y≤4, x≥0, y≥0.
Draw the lines x+2y=4 (passes through (4,0),(0,2)); 3x+y=6 (passes through (2,0),(0,6)) and x+y=4 (passes through (4,0),(0,4)). Shade the region satisfied by the given inequations. The shaded region in the figure gives the feasible region determined by the given inequations.
Solving 3x+y=6 and x+2y=4 simultaneously, we get x=8/5 and y=6/5
We observe that the feasible region OABC is a convex polygon and bounded and has corner points.
O(0,0), A(2,0), B(8/5, 6/5), C(0,2)
The optimal solution occurs at one of the corner points.
At O(0,0), z=2.0+5.0=0;
At A(2,0), z=2.2+5.0=4;
At B(8/5, 6/5), z=2.(8/5)+5.(6/5)=16/5+30/5=46/5;
At C(0,2), z=2.0+5.2=10;
Therefore, z maximum value at C and maximum value=10.