7.CSIR NET MATHS 2020-JUNE PART B

 

CSIR NET MATHEMATICS 2020-JUNE

HELD ON 26TH-NOVEMBER 

PART-B

 

7.The maximum and minimum values of $5x+7y$,when $\vert x\vert +\vert y \vert \leq 1$ 

 

(A)5 and -5

(B) 5 and -7 

(C) 7 and -5 

(D) 7 and -7 

 

Solution; Given $Z=5x+7y$
 

constraint  $\vert x\vert +\vert y \vert \leq 1$ \
 

Here 

 

  $ \vert x \vert $ =$  \begin{cases}
        x \enspace if     x \geq 0 \\
        -x \enspace if      x <0
    \end{cases}$

$ \vert y \vert $  = $ \begin{cases}
        y \enspace if        y \geq 0 \\
        -y \enspace if      y <0     \end{cases}$ 

 

Now constraint take as

x+y $ \leq $ 1 

-x+y $ \leq $ 1 

x-y $ \leq $ 1

-x-y $ \leq $  1 

 

Let take inequality constraint as equality

x+y = 1 

-x+y = 1

x-y = 1

-x-y = 1

 

From equation (1) becomes  $x+y = 1 $

x=0  $ \implies  $ 0+y=1 $ \implies $  y=1 $  \implies  $ (0,1) 

y =0  $ \implies $  x+0=1  $ \implies $  x=1  $ \implies $  (1,0) 

 The first constraint point (0,1) and  (1,0) 

From equation (2) becomes  $-x+y = 1 $

x =0 $  \implies $ 0+y=1 $ \implies$  y=1 $ \implies  $(0,1) 

y=0 $ \implies  $ -x+0=1 $  \implies  $ -x=1 $ \implies  $ x=-1 $ \implies $   (-1,0) 

The second constraint point (0,1) and (-1,0)

From equation (3) becomes $x-y = 1 $

x =0  $ \implies $  0-y=1 $  \implies  $ -y=1 $  \implies $   y=-1 $ \implies $  (0,-1) 

y=0  $ \implies $  x-0=1 $  \implies $  x=1 $  \implies  $ (1,0) 

The third constraint point (0,-1) and  (1,0)

From equation (4) becomes $-x-y = 1 $

x=0  $ \implies  $ 0-y=1 $  \implies $  -y=1 $  \implies $   y=-1 $ \implies  $ (0,-1) 

y=0  $ \implies $  -x-0=1 $  \implies $  -x=1  $  \implies $  x=-1  $ \implies $   (-1,0) 

        

 

 

 The fourth constraint point (0,-1) and  (-1,0)

The closed polygon $A-B-C-D$ is the feasible region

The feasible region is also known as solution  space of LPP.

The extreme points are $A(1,0),B(0,1),C(-1,0),D(0,-1)$

The extreme points of the solution space are

 Extreme point    Vertices        Value of Z=5x+7y

 

       A                     (1,0)              z=5(1)+7(0)=5

 

       B                      (0,1)              z=5(0)+7(1)=7

       C                      (-1,0)             z=5(-1)+7(0)=-1

       D                      (0,-1)              z=5(0)+7(-1)=-7

 

The maximum value of Z is 7 and minimum value of Z is -7

 

Hence correct option (D) 7 and -7