Consider $z=x+3 y$ subject to the constraints $2 x+3 y ≥ 6, 2 x+y ≤ 8, x ≥ 0, y ≥ 0$. The minimum value of $z$ is :

3

$Z=x+3 y$ → function to minimise constraints 

$x \geq 0, y \geq 0$ → solution is in first quadrant

$2 x+3 y \geq 6, 2 x+y \leq 8$

first plotting

2x + 3y = 6

2x + y = 8

now checking inequality with points (0, 0)

for $2 x+3 y ≥ 6$

0 ≥ 6 → not correct

→ (solution lies in half not containing (0, 0))

for inequality

2x + y ≤ 8

with (0, 0) ⇒ 0 ≤ 8 → (solution lies in part containing (0, 0) origin)

Corner points are

A (0, 8)

B (0, 2)

C (4, 0)

D (3, 0)

Checking function for each point

$z(x, y)=x+3 y$

$z(0,8)=0+3 \times 8=24$

$z(0, z)=0+3 \times 2=6$

$z(4,0)=4+0=4$

$z(3,0)=3+0=3$

So at (3, 0) Minimum value of function is 3