Question

For the LPP Max Z=3x+4y, x+y≤40; x+2y≤ 60, x≥0, y≥0 the solution is:

• A. x= 20, y = 20, Max Z = 140
• B. x= 40, y = 0, Max Z = 120
• C. x = 0, y = 60, Max Z = 240
• D. x = 10, y = 30, Max Z = 130

Answer 1

The Correct Option is A

To solve the given Linear Programming Problem (LPP), identify the constraints and the objective function:

• Objective Function: Maximize \( Z = 3x + 4y \)
• Constraints:
  • \( x + y \leq 40 \)
  • \( x + 2y \leq 60 \)
  • \( x \geq 0 \)
  • \( y \geq 0 \)

Steps to solve:

1. Identify corner points of the feasible region by solving the system of linear inequalities.
2. Find intersections of boundary lines:
   • \( x + y = 40 \)
   • \( x + 2y = 60 \)

Find intersection by solving:

• Subtract the first equation from the second:
  \( (x + 2y) - (x + y) = 60 - 40 \)
  \( y = 20 \)
• Substitute \( y = 20 \) into \( x + y = 40 \):
  \( x + 20 = 40 \)
  \( x = 20 \)

Intersection point is \((20, 20)\).

Evaluate \( Z \) at each corner point:

1. For \((20, 20)\): \( Z = 3(20) + 4(20) = 60 + 80 = 140 \)
2. For \((40, 0)\): \( Z = 3(40) + 4(0) = 120 \)
3. For \((0, 30)\): Invalid, as it doesn't satisfy \( x + 2y \leq 60 \).
4. For \((0, 60)\): Invalid, as it doesn't satisfy \( x + 2y \leq 60 \).

The maximum value of \( Z \) is 140 at \( x = 20, y = 20 \).

Therefore, the solution is \( x = 20, y = 20, \text{Max } Z = 140 \).

Point	x	y	Z
(20, 20)	20	20	140
(40, 0)	40	0	120