To From | Transportation cost per 1000 bricks (in Rs.) | ||
P | Q | R | |
A | 40 | 20 | 20 |
B | 20 | 60 | 40 |
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, \[x\ge 0,y\ge 0\]
Correct Answer: C
Solution :
[c] The given information can be expressed as given in the diagram: |
In order to simply, 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 form 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: |
\[x+y\ge 15,x+y\le 30,\] |
\[x\le 15,y\le 20,x\ge 0,y\ge 0\] |
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