Solve the following Linear Programming Problem graphically :
Maximize \( Z = 20x + 10y \)
subject to constraints
\( x + 2y \leq 28 \)
\( 3x + y \leq 24 \)
\( x \geq 2 \)
\( x, y \geq 0 \)
Solve the following Linear Programming Problem graphically :
Maximize \( Z = 20x + 10y \)
subject to constraints
\( x + 2y \leq 28 \)
\( 3x + y \leq 24 \)
\( x \geq 2 \)
\( x, y \geq 0 \)
Show Hint
Solution and Explanation
In a linear programming problem, the best value of the objective function lies at a corner point of the feasible region, when that region is bounded.
Step 2: Draw each constraint:
The line \(x + 2y = 28\) passes through \((28,0)\) and \((0,14)\). The origin satisfies \(x+2y \le 28\), so we shade the side toward the origin.
The line \(3x + y = 24\) passes through \((8,0)\) and \((0,24)\). The origin satisfies the inequality, so we shade toward the origin.
The line \(x = 2\) is vertical. We keep the region to its right. Also \(y \ge 0\) keeps us above the x-axis.
Step 3: Find the corner points of the feasible region:
The region is bounded by \(x=2\), \(y=0\), \(3x+y=24\) and \(x+2y=28\).
\(x=2\) with \(y=0\) gives \(A(2,0)\).
\(y=0\) with \(3x+y=24\) gives \(B(8,0)\).
The two slanted lines meet where \(3x+y=24\) and \(x+2y=28\). Doubling the first gives \(6x+2y=48\). Subtracting the second gives \(5x = 20\), so \(x=4\) and \(y=12\). This is \(C(4,12)\).
\(x=2\) with \(x+2y=28\) gives \(y=13\), so \(D(2,13)\).
The region \(ABCD\) is a closed, bounded quadrilateral.
Step 4: Evaluate \(Z = 20x + 10y\) at each corner:
\[ Z(2,0) = 40 \]
\[ Z(8,0) = 160 \]
\[ Z(4,12) = 80 + 120 = 200 \]
\[ Z(2,13) = 40 + 130 = 170 \]
Step 5: Pick the maximum:
The largest value is 200, at \((4,12)\). The region is bounded, so this is the maximum.
Final Answer:
The maximum value of \(Z\) is 200, at \(x = 4\) and \(y = 12\). \[ \boxed{Z_{\max} = 200 \text{ at } (4,12)} \]
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