Q. $z = 30x + 20y,\ x + y \leq 8,\ x + 2y \geq 4,\ 6x + 4y \geq 12,\ x \geq 0,\ y \geq 0$ has
A) Unique solution
B) Infinity many solution
C) Minimum at (4, 0)
D) Minimum 60 at point (0, 3)
Explanation
(d)
Feasible region is $ABCDFA$ and $z = 30x + 20y$
[image:12_image11.jpeg]
Now, at $A(4,\ 0),\ z = 30 \times 4 + 0 = 120$
$B(8,\ 0),\ z = 30 \times 8 + 0 = 240$
$C(0,\ 8),\ z = 0 + 20 \times 8 = 160$
$D(0,\ 3),\ z = 0 + 20 \times 3 = 60$
And $F\left( 1,\frac{3}{2} \right),\ \ z = 30 \times 1 + 20 \times \frac{3}{2} = 60$
It is clear that minimum value of $z$ is 60 at points $D(0,\ 3)$ and $F\left( 1,\frac{3}{2} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.