Question

$z=30x+20y,x+y\le 8,x+2y\ge 4,6x+4y\ge 12,x\ge 0,y\ge 0$ has

• A. Unique solution
• B. Infinitely many solution
• C. Minimum at $(4,0)$
• D. Minimum 60 at point $(0,3)$

Answer 1

The correct option is D Minimum 60 at point $(0,3)$

Since, $x+y=8$ ....(i)
This line meets axes at $(8,0)$ and $(0,8)$ respectively.
$x+2y=4$ .....(ii)
$\Rightarrow \frac{x}{4}+\frac{y}{2}=1$
This line meet axes at $(4,0)$ and $(0,2)$.
And $6x+4y=12$ ......(iii)
$\Rightarrow \frac{x}{2}+\frac{y}{3}=1$
This line meets axes at $(2,0)$ and $(0,3)$
The point of intersection of equations (ii) and (iii) is $F(1,\frac{3}{2})$
Now, at $A(4,0),z=30\times 4=120$
$B(8,0),z=30\times 8=240$
$C(0,8),z=20\times 8=160$
$D(0,3),z=20\times 3=60$
and $F(1,\frac{3}{2}),z=30\times 1+20\times \frac{3}{2}=60$
It is clear that $z$ is minimum 60 at points $D(0,3)$ and $F(1,\frac{3}{2})$
Hence, option (D) is correct.