Q. Minimize$z = 3x + y$, subject to constraints $2x + 3y \leq 6,\ x + y \geq 1,\ x \geq 0,\ y \geq 0$. Then
A) $x\text{=}1,\ y = 1$
B) $x = 0,\ y = 1$
C) $x = 1,\ y = 0$
D) $x = - 1,\ y = - 1$
Explanation
(b)
Feasible region is $ABDCA$ and $z = 3x + y$
[image:12_image22.jpeg]
Now, at $A(1,\ 0),\ z = 3$
At $B(3,\ 0),\ z = 9$
At $C(0,\ 1),\ z = 1$
And at $D(0,\ 2),\ z = 2$
It is clear that minimum value of $z$ is 1 at (0, 1)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.