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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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A firm has to transport 1200 packages using large vans which can carry 200 packages each and small vans which can take 80 packages each.The cost for engaging each large van is Rs 400 and each small van is Rs 200.Not more than Rs 3000 is to be spent on the job and the number of large vans cannot exceed the number of small vans.What will be the minimum cost?

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Toolbox:
  • Let $R$ be the feasible region for a linear programming problem and let $z=ax+by$ be the objective function.When $z$ has an optimum value (maximum or minimum),where variables $x$ and $y$ are subject to constraints described by linear inequalities,this optimum value must occur at a corner point of the feasible region.
  • If R is bounded then the objective function Z has both a maximum and minimum value on R and each of these occur at corner points of R
Step 1:
The given problem can be written as follows :
Large van $\Rightarrow$ packages :200,cost :Rs.400
Small van $\Rightarrow$ packages :80,cost :Rs.200
maximum package $\Rightarrow$ packages :1200,cost :Rs.3000
The mathematical formulation for the above problem
The objective function to be minimized is $Z=400x+200y$
Let $x$ be the packages carried by large van and $y$ be the packages carried by small van.
Then $200x+80y\geq 1200\Rightarrow 5x+2y\geq 30$
The total cost should not be exceed Rs.3000
Hence $400x+200y\geq 3000$
$2x+y\geq 15$
Since the large vans cannot exceed the small van.
$x\geq y$
Hence the objective function is $Z=400x+200y$ subject to the constraints $5x+2y\geq 30,2x+y\geq 15$ and $x\geq y$ and $x,y\geq 0$
Step 2:
The lines AB :$5x+2y=30,AC :2x+y=15$ and $CQ :x=y$ can be drawn on the graph to find the feasible region.
Consider the line $AB :5x+2y=30$
Put $x=0,y=0$ then $0\geq 30$ is not true.
Hence the region $5x+2y\geq 30$ lies above the line AB.
Consider the line $AC :2x+y=15$
Put $x=0,y=0$ then $0\geq 15$ is not true.
Hence the region $2x+y\geq 15$ lies above the line AC.
Consider the line $CQ :x-y=0$
Put $x=0,y=0$
The region $x\geq y$ lies above CQ.
Clearly the origin $O(0,0)$ is not included.
The feasible region is APQ is the shaded region shown.
The corner points of the feasible region are $A(0,15),P(\large\frac{30}{7},\frac{30}{7})$ and $Q(5,5)$
Step 3:
The value of the objective function can be found as follows :
At the points $(x,y)$ the value of the objective function subjected to $Z=400x+200y$
At $O(0,15)$ the value of the objective function $Z=400\times 0+200\times 15=3000$
At $P(\large\frac{30}{7},\frac{30}{7})$ the value of the objective function $Z=400\times \large\frac{30}{7}$$+200\times \large\frac{30}{7}$$=1714.2+857.1=2571.3$
At $Q(5,5)$ the value of the objective function $Z=400\times 5+200\times 5=3000$
Hence the maximum value is $2571.3$ at $P(\large\frac{30}{7},\frac{30}{7})$
answered Feb 1, 2014 by rvidyagovindarajan_1
 

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