Question

A brick manufacturer has two depots $A$ and $B$, with stocks of $30000$ and $20000$ bricks respectively. He receives orders from three builders $P,Q$ and $R$ for $15000,20000$ and $15000$ bricks respectively. The cost (in $₹$) of transporting $1000$ bricks to the builders
from the depots are as given in the table :

The manufacturer wishes to find how to fulfil the order so that transportation cost is minimum.
Formulate the $LPP$.

• A. Minimize $Z=40x-20y$
  Subject to, $x+y\ge 15,x+y\le 30,x\ge 15,y\le 20,x\ge 0,y\ge 0$
• B. Minimize $Z=40x-20y$
  Subject to, $x+y\ge 15,x+y\le 30,x\le 15,y\ge 20,x\ge 0,y\ge 0$
• C. Minimize $Z=40x-20y$
  Subject to, $x+y\ge 15,x+y\le 30,x\le 15,y\le 20,x\ge 0,y\ge 0$
• D. Minimize $Z=40x-20y$
  Subject to, $x+y\ge 15,x+y\le 30,x\ge 15,y\ge 20,x\ge 0,y\ge 0$

Answer 1

Answer: (C)

The given information can be expressed as given in the diagram:
In order to simplify, we assume that $1$ unit $=1000$ bricks

Suppose that depot $A$ supplies $x$ units to $P$ and $y$ units to $Q$, so that depot $A$ supplies $(30-x-y)$ bricks to builder $R$.
Now as $P$ requires a total of $15000$ bricks, it requires $(15-x)$ units from depot $B$.
Similarly $Q$ requires $(20-y)$ units from $B$ and $R$ requires
$15-(30-x-y)=x+y-15$ units from $B$.
Using the transportation cost given in table, total transportation cost,
$Z=40x+20y+20(30-x-y)+20(15-x)+60(20-y)+40(x+y-15)$
$=40x-20y+1500$
Obviously the constraints are that all quantities of bricks supplied from $A$ and $B$ to $P,Q,R$ are non-negative.
$\therefore x\ge 0$, $y\ge 0$, $30-x-y\ge 0$, $15-x\ge 0$, $20-y\ge 0$,
$x+y-15\ge 0$
Since, $1500$ is a constant, hence instead of minimizing
$Z=40x-20y+1500$, we can minimize $Z=40x-20y$.
Hence, mathematical formulation of the given $LPP$ is
Minimize $Z=40x-20y$,
subject to the constraints :
$x+y\ge 15,x+y\le 30,x\le 15,y\le 20,x\ge 0,y\ge 0$