In the linear programming problem: Minimize \(z = x + y\), subject to the constraints: \(x + 3y \leq 60,\ x + y \geq 10,\ x \leq y,\ x \geq 0,\ y \geq 0\) Which of the following region in graph given below represents the feasible region of the above LPP?
Show Hint
Test one point of each region in all constraints. Feasible points lie above \(y=x\) and \(x+y=10\), below \(x+3y=60\).
Step 1: Understanding the Concept:
The feasible region is the set of points that satisfy all the constraints at once. We test one point from each labelled region against every constraint.
Step 2: Key Formula or Approach:
The constraints are \(x+3y\leq 60\), \(x+y\geq 10\), \(x\leq y\), \(x\geq0\), \(y\geq0\). The line \(y=x\) passes through the origin. The condition \(x\leq y\) keeps points on or above it.
Step 3: Test region A.
A lies near the x-axis, below the line \(y=x\). Take the point \((20,5)\). Here \(x\leq y\) reads \(20\leq 5\), which is false. So A is not feasible.
Step 4: Test region B.
B is the hatched region near the y-axis, above the line \(x+y=10\), below \(x+3y=60\) and above \(y=x\). Take \((2,12)\): \(2+36=38\leq 60\) is true, \(2+12=14\geq 10\) is true, and \(2\leq 12\) is true. So B satisfies every constraint.
Step 5: Test region C.
C lies above the line \(x+3y=60\). Take \((10,20)\): \(10+60=70\leq 60\) is false. So C is not feasible.
Step 6: Test region D.
D lies below the line \(y=x\), to the right. Take \((30,10)\): \(x\leq y\) reads \(30\leq10\), which is false. So D is not feasible.
Step 7: Corner points of B.
The corners are \((0,10)\), \((5,5)\), \((15,15)\) and \((0,20)\). These match the shape of region B in the graph.
Final Answer:
Only region B satisfies all the constraints, which is option 2.
\[ \boxed{\text{Region B (Option 2)}} \]
