For the given Linear Programming Problem, shade the feasible region on the graph and write the coordinates of the corner points.
Minimize \(Z = 5x + 7y\)
subject to constraints
\[ 2x + y \geq 8 \]
\[ x + 2y \geq 10 \]
\[ x,\ y \geq 0. \]
For the given Linear Programming Problem, shade the feasible region on the graph and write the coordinates of the corner points.
Minimize \(Z = 5x + 7y\)
subject to constraints
\[ 2x + y \geq 8 \]
\[ x + 2y \geq 10 \]
\[ x,\ y \geq 0. \]
Show Hint
Solution and Explanation
Treat each inequality as an equation to get its line.
The line \(2x + y = 8\) passes through \((0, 8)\) and \((4, 0)\).
The line \(x + 2y = 10\) passes through \((0, 5)\) and \((10, 0)\).
Step 2: Decide which side to shade.
Test the origin in \(2x + y \geq 8\). We get \(0 \geq 8\), which is false. So shade the side away from the origin.
The origin also fails \(x + 2y \geq 10\), so shade away from the origin again.
The conditions \(x, y \geq 0\) keep the region in the first quadrant.
The feasible region is the unbounded region above both lines.
Step 3: Find the corner points.
On the y-axis, the higher point is \((0, 8)\), since \(2x + y = 8\) lies above \(x + 2y = 10\) there.
On the x-axis, the farther point is \((10, 0)\).
Solve the two lines together. From \(2x + y = 8\), \(y = 8 - 2x\). Put this in \(x + 2y = 10\).
\[ x + 16 - 4x = 10 \Rightarrow 3x = 6 \Rightarrow x = 2, \quad y = 4 \]
The corner points are \((0, 8)\), \((2, 4)\) and \((10, 0)\).
Step 4: Evaluate Z at each corner.
\[ Z(0,8) = 56, \qquad Z(2,4) = 10 + 28 = 38, \qquad Z(10,0) = 50 \]
The smallest value is 38 at \((2, 4)\).
The region is unbounded, so 38 is the minimum only if \(5x + 7y < 38\) has no point in the region. That open half plane has no common point with the feasible region, so the minimum exists.
Final Answer:
Corner points are \((0, 8)\), \((2, 4)\) and \((10, 0)\). The minimum value of Z is 38 at \((2, 4)\). \[ \boxed{Z_{\min} = 38 \text{ at } (2, 4)} \]
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