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.
Show Hint
Plot the constraint lines, find the feasible region's corners, and evaluate \(Z\) at each corner.
Step 1: Reasoning about the direction to push Z down:
To shrink \(Z=200x+500y\), we want to move toward smaller \(x\) and especially smaller \(y\) (since \(y\) has the larger coefficient), while staying inside the feasible region.
Step 2: Identifying the binding constraints at the minimum:
The feasible region's boundary closest to the origin (where \(Z\) would be smallest without constraints) is the edge of \(x+2y\ge10\); moving along it towards smaller \(y\) is blocked by \(3x+4y\le24\), so the minimum occurs exactly where these two boundaries cross.
Step 3: Solving the two boundary equations together:
\(x+2y=10\) and \(3x+4y=24\) intersect at \((4,3)\) (solved as before).
Step 4: Evaluating Z there and confirming it beats the other corners:
\(Z(4,3)=800+1500=2300\), which is indeed lower than \(Z(0,5)=2500\) and \(Z(0,6)=3000\).
Final Answer:
Minimum \(Z=\boxed{2300}\) at \((4,3)\), matching the full corner-point method.