Question

Solve the following LPP graphically : 

Maximize Z = 45x + 80y 

subject to 5x + 20y ≤ 400 

10x + 15y ≤ 450

and x ≥ 0, y ≥ 0.

Answer 1

Equation are

5x + 20y = 400

((x/80) + (y/20)) = 1 ...(i)

and 10x + 15y = 450 

((x/45) + (y/30)) = 1 ...(ii)

Its graph has been shown below x = 0, y = 0 are respectively y-axis and x-axis.

Again (0, 0) satisfies the equations So, feasible region is OAEDO (is shaded). 

The vertices of the feasible region are O = (0, 0), A = (45, 0), E = (24, 14) 

Solving (i) and (ii) D = (0,20) 

Now, at O(0, 0), Z = 45 x 0 + 80 x 0 = 0

at A(45, 0), Z = 45 x 45 + 80 x 14 = 2200

at D(0, 20), Z = 45 x 0 + 80 x 20 = 1600

So, maximum value of z is 2200 which obtained at (24, 14) 

So, Solution of LPP Maximum Z = 2200 where x = 24, y = 14