Word problems based on formulation of linear programming problems.
2. A manufacturer produces nuts and bolts for industrial machinery. It takes 1 hour of work on machine 1 and 3 hours of work on machine 2 to produce a package of nuts. It takes 3 hours on machine 1 and 1 hour on machine 2 to produce a package of bolts. He earns a profit of Rs. 2.50 per package on nuts and Rs. 1 per package on bolts. Form a LPP to maximize his profit, if he operates each machine for almost 12 hours.
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solution
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Let x packages of nuts and y packages of bolts be produced. The objective of the manufacturer to maximize the profit is
Total Time required on machine 1 to produce x packages of nuts and y packages of bolts is equal to
Total Time required on the machine 2 to produce x packages of nuts and y packages of bolts is equal to
s
According to restrictions,
For machine 1
For machine 2
Maximize z is equal to
Subject to constraints
To solve this graphically, let us take
The lines are drawn using suitable points on the graph.
The lines intersect at P(3,3)
Now shade the region of intersection of the lines.
The feasible region is OAPB
For the corner point O(0,0), z=
For the corner point A(4,0)
For the corner point P(3,3,)
For the corner point B(0,4)
Clearly z is maximum at x=3 , y=3 and the maximum value is 10.50
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2.50x + 1y
1x + 3y
3x + 1y
X + 3y ≤ 12
3x + y ≤ 12 and x,y≥0
2.50x + y
X + 3y ≤ 12
3x + y ≤ 12
X ≥ 0, y ≥ 0
X + 3y =12
3x + y = 12, x=0, y=0
O(0,0),A(4,0),P(3,3,),B(0,4)
2.5(0) +1(0)=0
2.5(4)+1(0)=10
2.5(3)+1(3)=10.5
2.5(0)+1(4)=4
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3. A shopkeeper deals in 2 items → wall hangings and artificial plants. He had a space to store 80 pieces and Rs. 15000 to invest. A wall hangings cost him Rs. 300 and artificial plant Rs. 150. He can sell a wall hanging at a profit of Rs. 50 and artificial plant at a profit of Rs. 18. Assuming that he can sell all the items that he buys, formulate a LPP in order to maximize his profit.\
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solution
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Let x be the number of wall hangings and y be the number of artificial plants.
Profit of dealer is equal to
Objective function z is equal to
Since dealer invest atmost Rs. 15000
Therefore
Or
Also a dealer has space to store atmost 80 pieces.
Therefore,
Maximize z is equal to
Subject to constraints
To solve this graphically, we need to draw the graph
Let us the draw the lines 2x + y ≤100
X + y ≤ 80
X ≥ 0, y ≥ 0
On the graph by using suitable points.
The points of intersection are
Then shade the region of intersection of these two lines
The feasible points are OABC
Now to obtain the maximum value;
For The corner point O(0,0), z=
For the corner point A(50,0)
For the corner point B(20,60)
For the corner point C(0,100)
Clearly we get the maximum value of 2500 at A.(50,0)
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50x + 18y
50x + 18y
300x + 150y ≤ 15000
2x + y ≤ 100
X + y ≤ 80
X ≥ 0, y ≥ 0
Z = 50x + 18y
2x + y ≤100
X + y ≤ 80
X ≥ 0, y ≥ 0
(20,60)
O(0,0),A(50,0),B(20,60),(C(0,100)
50(0)+18(0)=0
50(50)+18(0)=2500
50(20)+18(60)=2080
50(0)+18(100)=1800
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