Question

Solve the following Linear Programming Problem graphically :

Minimize \(Z = 4x + y\)

subject to

\(x + y \geq 3\)

\(x + 2y \geq 4\)

\(x, y \geq 0\).

Hint:

Draw the two lines, shade the region above both, and find the corners \((0, 3)\), \((2, 1)\) and \((4, 0)\). Compare the values of \(Z\) there.

Answer 1

Step 1: Draw the boundary lines:
Line \(x + y = 3\) passes through \((3, 0)\) and \((0, 3)\).
Line \(x + 2y = 4\) passes through \((4, 0)\) and \((0, 2)\).
For both inequalities, the point \((0, 0)\) fails (\(0 \geq 3\) is false), so we shade the side away from the origin.

Step 2: Describe the feasible region:
With \(x \geq 0\) and \(y \geq 0\), the feasible region lies in the first quadrant, above both lines.
It is an unbounded region. Its boundary runs down the y-axis from far above to \((0, 3)\), then along \(x + y = 3\) to \((2, 1)\), then along \(x + 2y = 4\) to \((4, 0)\), then along the x-axis to the right.

Step 3: Find the corner points:
On the y-axis (\(x = 0\)): \(y \geq 3\) and \(y \geq 2\) give the corner \((0, 3)\).
The two lines meet where \(x + y = 3\) and \(x + 2y = 4\). Subtracting gives \(y = 1\), and then \(x = 2\). The corner is \((2, 1)\).
On the x-axis (\(y = 0\)): \(x \geq 3\) and \(x \geq 4\) give the corner \((4, 0)\).

Step 4: Evaluate Z at each corner:
\[ Z(0, 3) = 4(0) + 3 = 3 \]
\[ Z(2, 1) = 4(2) + 1 = 9 \]
\[ Z(4, 0) = 4(4) + 0 = 16 \]
The smallest corner value is \(3\) at \((0, 3)\).

Step 5: Check the unbounded region:
The region is unbounded, so a minimum exists only if the open half-plane \(4x + y < 3\) has no point in common with the region.
If \(4x + y < 3\), then \(y < 3 - 4x\). But the region needs \(y \geq 3 - x\). Both can hold only if \(3 - x < 3 - 4x\), that is \(x < 0\), which is not allowed.
So there is no feasible point with \(Z < 3\), and the minimum value is \(3\).

Final Answer:
The minimum value of \(Z\) is \(3\), at the point \((0, 3)\). \[ \boxed{Z_{\min} = 3 \text{ at } (x, y) = (0, 3)} \]