Question

Maximize \(Z=105x+90y\) under the constraints \(2x+y\le80,\ x+y\le50,\ x\ge0,\ y\ge0\) by graphical method.

Hint:

Plot the constraints, find corner points of the feasible region, evaluate Z at each.

Answer 1

Step 1: Identifying the feasible region's corners:
The constraint lines are \(2x+y=80\) and \(x+y=50\), together with the axes. Their intersection: subtract \(x+y=50\) from \(2x+y=80\) to get \(x=30\), then \(y=50-30=20\).

Step 2: Listing all corner points:
The feasible region (a convex polygon) has corners \((0,0)\), \((40,0)\) [where \(2x+y=80\) meets \(y=0\)], \((30,20)\) [intersection], and \((0,50)\) [where \(x+y=50\) meets \(x=0\)].

Step 3: Evaluating Z at each corner:
\(Z(0,0)=0\); \(Z(40,0)=105(40)=4200\); \(Z(30,20)=105(30)+90(20)=3150+1800=4950\); \(Z(0,50)=90(50)=4500\).

Step 4: Picking the maximum:
The largest value among \(0,4200,4950,4500\) is at \((30,20)\).

Final Answer:
\[ \boxed{Z_{\max}=4950 \text{ at } x=30,\ y=20} \]