Question

The difference between the maximum and minimum values of the objective function \(Z = 3x+5y\), subject to the constraints \(x+3y\leq 60\), \(x+y\geq 10\), \(x-y\leq 0\), \(x,y\geq 0\) is

• A. \(50\)
• B. \(60\)
• C. \(70\)
• D. \(80\)

Hint:

Find the corner points of the feasible region and evaluate Z at each.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The optimum of a linear objective over a bounded feasible region occurs at a corner point. The constraints are \(x + 3y \leq 60\), \(x + y \geq 10\), \(x \leq y\), \(x, y \geq 0\).

Step 2: Find corner points
Intersections of the boundary lines:
1. \(x = y\) with \(x + y = 10\): \((5, 5)\).
2. \(x = y\) with \(x + 3y = 60\): \((15, 15)\).
3. \(x = 0\) with \(x + y = 10\): \((0, 10)\).
4. \(x = 0\) with \(x + 3y = 60\): \((0, 20)\).
The feasible region is the quadrilateral with these four vertices (it lies above the line \(y = x\) and above \(x + y = 10\)).

Step 3: Evaluate Z
\(Z(5, 5) = 15 + 25 = 40\), \(Z(15, 15) = 45 + 75 = 120\), \(Z(0, 10) = 50\), \(Z(0, 20) = 100\).
Maximum is 120 and minimum is 40. The difference is \(120 - 40 = 80\), option (D).

Final Answer:
The difference is 80, option (D). \[ \boxed{80} \]