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Consider a L.P.P, Maximize Z=3x+5yZ = 3x + 5y subject to constraints 2x+6y≤62x + 6y ≤ 6, x−y≥0x - y ≥ 0, x≥0x ≥ 0, y≥0y ≥ 0. Then which of the following are true?

(A) The feasible region of L.P.P is bounded region

(B) The corner points of the feasible region are (0, 0), (2, 2), (0, 1)

(C) Maximum value of Z is 9

(D) Point (1, 3) lies in the feasible region

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 1

The objective function is Maximize Z=3x+5yZ = 3x + 5y with constraints:

2x+6y≤62x + 6y ≤ 6

x−y≥0x - y ≥ 0 (means x≥yx ≥ y)

x≥0,y≥0x ≥ 0, y ≥ 0


To find corner points, convert inequalities to equations:

  1. 2x+6y=62x + 6y = 6
  2. x−y=0x - y = 0 (i.e., x=yx = y)
  3. x=0x = 0 (y-axis)
  4. y=0y = 0 (x-axis)

Origin: (0,0)(0, 0) where x-axis meets y-axis


Line 2x+6y=62x + 6y = 6 meets y-axis (x=0x = 0):

6y=66y = 6

y=1y = 1

Point: (0,1)(0, 1)

Check constraint x−y≥0x - y ≥ 0:

0−1=−1≥00 - 1 = -1 ≥ 0 is false

This point is NOT in the feasible region.


Line 2x+6y=62x + 6y = 6 meets x-axis (y=0y = 0):

2x=62x = 6

x=3x = 3

Point: (3,0)(3, 0) is valid.


Lines 2x+6y=62x + 6y = 6 and x=yx = y intersect:

2x+6x=62x + 6x = 6

8x=68x = 6

x=34x = \dfrac{3}{4}

Therefore y=34y = \dfrac{3}{4}

Point: (34,34)(\dfrac{3}{4}, \dfrac{3}{4}) is valid.


The actual corner points are (0,0)(0, 0), (3,0)(3, 0), (34,34)(\dfrac{3}{4}, \dfrac{3}{4}).


Statement (A): The feasible region is bounded

The feasible region is enclosed by the constraint lines and doesn't extend to infinity.

Statement (A) is TRUE.


Statement (B): Corner points are (0,0)(0, 0), (2,2)(2, 2), (0,1)(0, 1)

Check (2,2)(2, 2): 2(2)+6(2)=16>62(2) + 6(2) = 16 > 6 fails the constraint

(0,1)(0, 1) fails x−y≥0x - y ≥ 0 as shown above

Statement (B) is FALSE.


Statement (C): Maximum value of Z is 9

Calculate ZZ at each corner point:

Z(0,0)=3(0)+5(0)Z(0, 0) = 3(0) + 5(0)

Z(0,0)=0Z(0, 0) = 0

Z(3,0)=3(3)+5(0)Z(3, 0) = 3(3) + 5(0)

Z(3,0)=9Z(3, 0) = 9

Z(34,34)=3(34)+5(34)Z(\dfrac{3}{4}, \dfrac{3}{4}) = 3(\dfrac{3}{4}) + 5(\dfrac{3}{4})

Z(34,34)=94+154Z(\dfrac{3}{4}, \dfrac{3}{4}) = \dfrac{9}{4} + \dfrac{15}{4}

Z(34,34)=244Z(\dfrac{3}{4}, \dfrac{3}{4}) = \dfrac{24}{4}

Z(34,34)=6Z(\dfrac{3}{4}, \dfrac{3}{4}) = 6

Maximum Z=9Z = 9

Statement (C) is TRUE.


Statement (D): Point (1,3)(1, 3) lies in the feasible region

Check constraints:

2(1)+6(3)=20≤62(1) + 6(3) = 20 ≤ 6 is false

Statement (D) is FALSE.


Only statements (A) and (C) are true.

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