Solve the following linear programming problem graphically :
Maximize \(Z = 10500x + 9000y\)
Subject to constraints
\(x + y \le 50\)
\(2x + y \le 80\)
\(x, y \ge 0\)
Solve the following linear programming problem graphically :
Maximize \(Z = 10500x + 9000y\)
Subject to constraints
\(x + y \le 50\)
\(2x + y \le 80\)
\(x, y \ge 0\)
Show Hint
Solution and Explanation
Treat each inequality as an equation and find two points on it.
The line \(x+y=50\) passes through \((50,0)\) and \((0,50)\).
The line \(2x+y=80\) passes through \((40,0)\) and \((0,80)\).
The conditions \(x \ge 0\) and \(y \ge 0\) keep us in the first quadrant.
Step 2: Pick the side of each line:
Test the origin \((0,0)\). It gives \(0 \le 50\) and \(0 \le 80\), which are both true.
So the feasible region lies on the origin side of both lines.
Step 3: Find the point where the two lines meet:
Subtract \(x+y=50\) from \(2x+y=80\). This gives \(x=30\).
Then \(y = 50-30 = 20\).
So the lines meet at \(B(30,20)\).
Step 4: Describe the feasible region and its corners:
The feasible region is the bounded quadrilateral \(OABC\).
The line \(2x+y=80\) cuts the x-axis first (at 40), and the line \(x+y=50\) cuts the y-axis first (at 50).
So the corner points are \(O(0,0)\), \(A(40,0)\), \(B(30,20)\) and \(C(0,50)\).
Step 5: Evaluate Z at every corner:
At \(O(0,0)\): \(Z = 0\).
At \(A(40,0)\): \(Z = 10500(40) = 4,20,000\).
At \(B(30,20)\): \(Z = 10500(30) + 9000(20) = 3,15,000 + 1,80,000 = 4,95,000\).
At \(C(0,50)\): \(Z = 9000(50) = 4,50,000\).
Step 6: Pick the maximum:
The region is bounded, so the maximum of Z occurs at a corner point.
The largest value is 4,95,000 and it occurs at \(B(30,20)\).
Final Answer:
Z is maximum at \(x=30\), \(y=20\), and the maximum value is 4,95,000.\[ \boxed{Z_{\max} = 495000 \text{ at } (30,20)} \]
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