Question

The objective function is \( Z = 3x + 5y \). The difference between the maximum and minimum values of \( Z \) is \( 3a \), subject to constraints: \( x + y \ge 1 \), \( 5x + 10y \le 50 \), \( x, y \ge 0 \). Find the value of \( a \).

• A. \( 3 \)
• B. \( 6 \)
• C. \( 9 \)
• D. \( 27 \)

Hint:

In LPP questions, never test random points. Only the corner points of the feasible region are needed because the objective function always attains its optimum value at vertices.

Answer 1

The Correct Option is C

Concept: In Linear Programming Problems (LPP), the maximum and minimum values of the objective function occur at the corner points (vertices) of the feasible region. Thus the procedure is:
• Convert inequalities into equations.
• Find the feasible region.
• Determine all corner points.
• Evaluate the objective function at each vertex.

Step 1: Writing the constraint equations.
Given: \[ x+y\ge1 \] and \[ 5x+10y\le50 \] Divide second equation by \(5\): \[ x+2y\le10 \] Also: \[ x\ge0,\qquad y\ge0 \] Thus the boundary lines are: \[ x+y=1 \] and \[ x+2y=10 \]

Step 2: Finding the intercepts of the lines.
For: \[ x+y=1 \] If \(y=0\): \[ x=1 \] giving point: \[ (1,0) \] If \(x=0\): \[ y=1 \] giving point: \[ (0,1) \] Now for: \[ x+2y=10 \] If \(y=0\): \[ x=10 \] giving: \[ (10,0) \] If \(x=0\): \[ y=5 \] giving: \[ (0,5) \]

Step 3: Evaluating the objective function.
Objective function: \[ Z=3x+5y \] Evaluate at all corner points. At \((1,0)\): \[ Z=3(1)+5(0)=3 \] At \((0,1)\): \[ Z=3(0)+5(1)=5 \] At \((10,0)\): \[ Z=3(10)+5(0)=30 \] At \((0,5)\): \[ Z=3(0)+5(5)=25 \] Thus, \[ Z_{\min}=3 \] and \[ Z_{\max}=30 \]

Step 4: Finding the value of \(a\).
Difference: \[ Z_{\max}-Z_{\min}=30-3=27 \] Given: \[ 3a=27 \] Therefore, \[ a=9 \] Hence, \[ \boxed{9} \]