Question

Minimize \(z=3x+2y\) under the following constraints by graphical method: \(x+y\ge8,\ x\ge0,\ 3x+5y\le15,\ y\ge0\).

Hint:

Check if the region x+y≥8 overlaps with 3x+5y≤15 in the first quadrant; it does not.

Answer 1

Step 1: Understanding the Concept:
Before minimizing anything, the feasible region (the set of points satisfying every constraint at once) must actually exist. Graph each constraint line and check where all shaded regions overlap.

Step 2: Plotting the boundary lines:
\(x+y=8\) passes through \((8,0)\) and \((0,8)\); the constraint \(x+y\ge8\) is the region ON or ABOVE this line. \(3x+5y=15\) passes through \((5,0)\) and \((0,3)\); the constraint \(3x+5y\le15\) is the region ON or BELOW this line.

Step 3: Checking for overlap:
On the line \(3x+5y=15\) with \(x,y\ge0\), the largest possible value of \(x+y\) occurs at a corner: at \((5,0)\), \(x+y=5\); at \((0,3)\), \(x+y=3\); anywhere in between is even smaller (since \(3x+5y\le15\) with \(y\ge0\) gives \(3(x+y)\le3x+5y\le15\), so \(x+y\le5\)).

Step 4: Comparing with the other constraint:
The first constraint needs \(x+y\ge8\), but the second constraint forces \(x+y\le5\) in the first quadrant. Since \(8>5\), no point can satisfy both together.

Final Answer:
The feasible region is empty, so this linear programming problem has no feasible solution, and \(z\) cannot be minimized. \[ \boxed{\text{No feasible solution exists}} \]