Consider the LPP, Max z = 2x + 3y, subject to the conditions, x + y ≤…

$x \geq 0, y \geq 0$

$x \leq 2$

$x+y \leq 2$

Function to be maximised

Z = 2x + 3y

x ≥ 0, y ≥ 0

solution is in first quadrant

plotting for x + y = 2

x	0	2
y	2	0

now checking for (0, 0) in x + y ≤ 2

0 ≤ 2

solution lies in part containing (0, 0) below line

corner points obtained 

→ A (0, 0)

B (0, 2)

C (2, 0)

Z (x, y) = 2x + 3y

Z(0, 0) = 2(0) + 3(0) = 0

Z (0, 2) = 2(0) + 3(2)

= 0 + 6 = 6

Z (2, 0) = 2(2) + 3(0)

= 4 + 0 = 4

for (B (0, 2))

Z is maximum

Z_max = 6