Minimize : Z = x+2y;
x+2y ≥50.
2x - y ≤0
2x + y ≤ 100
x≥0,y≥0
Answers
(a)
Z=x+2y, subject to the constraints
x+2y≥100
2x−y≤0
2x+y≤200
x−y≥0 by graphical method.
On solving equations 2x−y=0 and x+2y=100 we get point B(20,40)
On solving 2x−y=0 and 2x+y=200 we get C(50,100)
∴ Feasible region is shown by ABCDA
The corner points of the feasible region are A(0,50),B(20,40),C(50,100),D(0,200)
Let us evaluate the objective function Z at each corner points as shown below
At A(0,50), Z=0+100=100
At B(20,40), Z=20+80=100
At C(50,100), Z=50+200=250
At D(0.200), Z=0+400=400
Hence, Maximum value of Z is 400 at D(0,200) and minimum value of Z is 100 at A and B.
(b)
Consider,
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b+c
b
c
a
c+a
c
a
b
a+b
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=(b+c)
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c+a
c
b
a+b
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−a
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b
c
b
a+b
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+a
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b
c
c+a
c
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=(b+c)[(c+a)(a+b)−bc]−a[ab+b
2
−bc]+a[bc−c
2
−ac]
=(b+c)a(b+c+a)−a[b(b−c+a)]+a[−c(−b+c+a)]
=4abc
Hence,
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b+c
b
c
a
c+a
c
a
b
a+b
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=4abc
Answer:
Step-by-step explanation: