Question

For the following Linear Programming Problem, shade the feasible region on the graph and find the coordinates of its corner points :

Minimize \(Z = 18x + 10y\)

subject to constraints

\[ \begin{aligned} 4x + y &\ge 20 \\ 2x + 3y &\ge 30 \\ x, y &\ge 0. \end{aligned} \]

Hint:

Draw both lines, shade away from the origin, and solve the two equations together to get the middle corner \((3, 8)\).

Answer 1

Step 1: Draw the boundary lines:
Line \(4x + y = 20\) cuts the axes at \((5, 0)\) and \((0, 20)\).
Line \(2x + 3y = 30\) cuts the axes at \((15, 0)\) and \((0, 10)\).
The conditions \(x, y \ge 0\) keep us in the first quadrant.

Step 2: Pick the side to shade:
Test the origin \((0, 0)\). It gives \(0 \ge 20\), which is false.
So for \(4x + y \ge 20\) we shade the side away from the origin.
The same test fails for \(2x + 3y \ge 30\), so again we shade away from the origin.

Step 3: Describe the feasible region:
The feasible region is the part of the first quadrant lying on or above both lines.
It is unbounded and it touches the y-axis from \((0, 20)\) upward and the x-axis from \((15, 0)\) to the right.

Step 4: Find the meeting point of the two lines:
From \(4x + y = 20\) we get \(y = 20 - 4x\).
Put it in \(2x + 3y = 30\): \(2x + 60 - 12x = 30\), so \(-10x = -30\) and \(x = 3\).
Then \(y = 20 - 12 = 8\). The point is \((3, 8)\).

Step 5: List the corner points:
The corner points are \((0, 20)\), \((3, 8)\) and \((15, 0)\).
The point \((5, 0)\) is not a corner point, because \(2(5) + 3(0) = 10 < 30\) lies outside the region.
The point \((0, 10)\) is also outside, because \(4(0) + 10 = 10 < 20\).

Final Answer:
The corner points of the feasible region are \((0, 20)\), \((3, 8)\) and \((15, 0)\). \[ \boxed{(0, 20),\ (3, 8),\ (15, 0)} \]