Question

Find the minimum value of the linear programming problem \(Z=200x+500y\) under the following constraints: \(x+2y\ge10\), \(3x+4y\le24\), \(x\ge0,y\ge0\) by graphical method.

Hint:

Plot the constraint lines, find the feasible region's corners, and evaluate \(Z\) at each corner.

Answer 1

Step 1: Understanding the Concept:
Graph the two boundary lines, shade the feasible region satisfying all four constraints, identify its corner points, then evaluate \(Z\) at each corner — the minimum of a linear objective over a convex feasible region always occurs at a corner point.

Step 2: Finding where the constraint lines meet the axes:
\(x+2y=10\) meets the axes at \((10,0)\) and \((0,5)\). \(3x+4y=24\) meets the axes at \((8,0)\) and \((0,6)\).

Step 3: Finding the intersection of the two lines:
Solve \(x+2y=10\) and \(3x+4y=24\) together: from the first, \(x=10-2y\); substituting, \(3(10-2y)+4y=24\Rightarrow30-2y=24\Rightarrow y=3,\ x=4\). So they meet at \((4,3)\).

Step 4: Identifying the feasible corner points:
Checking which side of each line is feasible (\(x+2y\ge10\) is \"above\" its line, \(3x+4y\le24\) is \"below\" its line) shows the bounded feasible region is the triangle with vertices \((0,5)\), \((0,6)\), and \((4,3)\).

Step 5: Evaluating Z at each corner:
At \((0,5)\): \(Z=200(0)+500(5)=2500\). At \((0,6)\): \(Z=200(0)+500(6)=3000\). At \((4,3)\): \(Z=200(4)+500(3)=800+1500=2300\).

Final Answer:
The minimum value is \(\boxed{Z=2300}\), attained at \((x,y)=(4,3)\).