Math, asked by rakesh329, 4 months ago

Minimize : Z = x+2y;
x+2y ≥50.
2x - y ≤0
2x + y ≤ 100
x≥0,y≥0​

Answers

Answered by jiyashah2901
2

(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,

b+c

b

c

a

c+a

c

a

b

a+b

=(b+c)

c+a

c

b

a+b

−a

b

c

b

a+b

+a

b

c

c+a

c

=(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,

b+c

b

c

a

c+a

c

a

b

a+b

=4abc

Answered by paramwadkar
2

Answer:

Step-by-step explanation:

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