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Q.Solve the following Linear Programming Problem graphically: i. Maximize Z=3x+2yZ = 3x + 2y Subject to the constraints: −2x+y≤1-2x + y \leq 1 x≤2x \leq 2 x+y≤3x + y \leq 3, and x≥0,y≥0x \geq 0, y \geq 0 ii. Minimize Z=5x−2yZ = 5x - 2y Subject to the constraints: 2x+3y≥12x + 3y \geq 1, and x≥0,y≥0x \geq 0, y \geq 0 iii. Minimize Z=−x+2yZ = -x + 2y Subject to the constraints: −x+3y≤10-x + 3y \leq 10 x+y≤6x + y \leq 6 x−y≤2x - y \leq 2 and x≥0,y≥0x \geq 0, y \geq 0 iv. Maximize Z=−x+2yZ = -x + 2y Subject to the constraints: −0.5x+y≤2-0.5x + y \leq 2 x−y≤−1x - y \leq -1 and x≥0,y≥0x \geq 0, y \geq 0 v. Maximize Z=5x+4yZ = 5x + 4y Subject to the constraints: x−2y≤1x - 2y \leq 1 x+2y≤6x + 2y \leq 6 x−y≥3x - y \geq 3 and x≥0,y≥0x \geq 0, y \geq 0

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Each LPP is solved by plotting the constraints, identifying the feasible region, and evaluating the objective function at the corner points. The optimal value is found at a vertex of the feasible region.


i. Maximize Z=3x+2yZ = 3x + 2y

Constraints:

−2x+y≤1-2x + y \leq 1

x≤2x \leq 2

x+y≤3x + y \leq 3

x≥0,y≥0x \geq 0, y \geq 0

Concept:

We want the highest possible value of ZZ within the region where all constraints hold. Since ZZ is linear, the maximum occurs at a corner of the feasible polygon.

Steps:

  1. Plot each line as an equality:

    • −2x+y=1-2x + y = 1 → y=2x+1y = 2x + 1 (passes through (0,1)(0,1) and (1,3)(1,3))
    • x=2x = 2 (vertical line)
    • x+y=3x + y = 3 → y=3−xy = 3 - x (passes through (0,3)(0,3) and (3,0)(3,0))
    • x=0x=0 and y=0y=0 are the axes.
  2. Shade the feasible side for each inequality:

    • For −2x+y≤1-2x + y \leq 1: test (0,0)(0,0) → 0≤10 \leq 1 true, so shade below the line.
    • For x≤2x \leq 2: shade left of x=2x=2.
    • For x+y≤3x + y \leq 3: test (0,0)(0,0) → 0≤30 \leq 3 true, shade below the line.
    • x≥0,y≥0x \geq 0, y \geq 0: first quadrant.
  3. Find the feasible region:

    It is a polygon with vertices at intersections of boundary lines.

    • Intersection of x=0x=0 and y=0y=0: (0,0)(0,0)
    • Intersection of x=0x=0 and −2x+y=1-2x+y=1: (0,1)(0,1)
    • Intersection of −2x+y=1-2x+y=1 and x+y=3x+y=3: Solve y=2x+1y=2x+1 and y=3−xy=3-x → 2x+1=3−x2x+1 = 3-x → 3x=23x=2 → x=2/3x=2/3, y=7/3y=7/3
    • Intersection of x+y=3x+y=3 and x=2x=2: (2,1)(2,1)
    • Intersection of x=2x=2 and y=0y=0: (2,0)(2,0)

    Check which are feasible: all satisfy all constraints. So vertices: (0,0)(0,0), (0,1)(0,1), (2/3,7/3)(2/3, 7/3), (2,1)(2,1), (2,0)(2,0).

  4. Evaluate Z=3x+2yZ = 3x + 2y at each vertex:

    • (0,0)(0,0): Z=0Z = 0
    • (0,1)(0,1): Z=2Z = 2
    • (2/3,7/3)(2/3, 7/3): Z=3(2/3)+2(7/3)=2+14/3=20/3≈6.67Z = 3(2/3) + 2(7/3) = 2 + 14/3 = 20/3 \approx 6.67
    • (2,1)(2,1): Z=6+2=8Z = 6 + 2 = 8
    • (2,0)(2,0): Z=6Z = 6
  5. Maximum is 88 at (2,1)(2,1). …

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