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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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A man rides his motorcycle at the speed of 50km/hour.He has to speed Rs 2 per km on petrol.If he rides it at at a faster speed of 80km/hour,the petrol cost increases to Rs 3 per km.He has at most Rs120 to spend on petrol and one hour's time.He wishes to find the maximum distance that he can travel.Determine the maximum distance that the man can travel.

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Toolbox:
  • First formulate the objective function and identify the constraints from the problem statement, To solve a Linear Programming problem graphically, first plot the constraints for the problem. This is done by plotting the boundary lines of the constraints and identifying the points that will satisfy all the constraints.
  • 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:
It is given that the normal speed of the motorcycle is $50km/hr$
His increased speed is 80km/hr
he has at most of 1hour's time.
Let $x$ and $y$ be the time,which we takes,when the speed is normal
$\therefore 80x+50y\leq 4000$
$\Rightarrow 8x+5y\leq 400$
The cost for the petrol when goes in the normal speed is Rs.2 per kms.
The cost for the petrol when goes in the increased speed is Rs.3 per kms.
He has at most of Rs 120 to spend.
$\therefore 2x+3y\leq 120$
Step 2:
Let us draw the graph for the lines $AB :8x+5y=400$ and $CD=2x+3y=120$
Let us solve these two lines to find the point of intersection
$8x+5y=400$-----(1)
$2x+3y=120$------(2)
Multiply equ(2) by 4
$8x+5y=400$
$8x+12y=480$
_________________
$-7y=-80$
$y=\large\frac{80}{7}$$\Rightarrow x=\large\frac{300}{7}$
Step 3:
Consider the line $AB: 8x+5y=400$
Put $x=0,y=0$ then $0\leq 400$ which is true.
Hence the region $8x+5y\leq 400$ lies below the line AB
Consider the line $CD :2x+3y=120$
Put $x=0,y=0$,then $0\leq 120$,which is true.
Hence the region $2x+3y\leq 400$ lies below the line CD.
The region $O(0,0)$ is also included in the feasible region.
The feasible region is the shaded portion OCPB as shown in the fig.
The corner points of the feasible region are $O(0,0),C(0,40),P(\large\frac{80}{7},\frac{300}{7})$$,B(50,0)$
Step 4:
Now let us find the value of the objective function $Z=x+y$ subjected to constraints $8x+5y\leq 400,2x+3y\leq 120,x,y\leq 0$
At the points(x,y) the value of the objective function subjected to $Z=x+y$
At $O(0,0)$ the value of the objective function $Z=0$
At $C(0,40)$ the value of the objective function $Z=0+40=40$
At $P(\large\frac{300}{7},\frac{80}{7})$ the value of the objective function $Z=\large\frac{300}{7}+\frac{80}{7}=\frac{380}{7}=$$54\large\frac{2}{7}$
At $B(50,0)$ the value of the objective function $Z=50+0=50$
Hence the maximum distance is $54\large\frac{2}{7}$km
answered Aug 29, 2013 by sreemathi.v
 

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