The area of the feasible region for the following constraints 3y + x ≥ 3, x ≥ 0, y ≥ 0 will be
a.Bounded
b.Unbounded
c.Concave
d.Convex
Answers
Answer:
Correct answer is the A
Because the feasible region is
of the bound contrats
Answer
The feasible region of the constraints is unbounded. Option b is correct.
Concept
A straight line ax + by = c divides the plane into three regions: ax + by < c, ax + by = c, ax + by > c.
The two halfplanes correspond to the inequalities: ax + by < c, and ax + by > c.
To determine which of the two half regions corresponds to which inequality pick a point in one of the half-planes. Whichever inequality that point satisfies the half-plane containing that point corresponds to that inequality.
Given
The three inequality:
- x ≥ 0,
- y ≥ 0,
- x + 3y ≥ 3.
Find
Which one of the options is true ut the feasible region of the inequality constraints.
Solution
Feasible region of the constraint x ≥ 0 ∧ y ≥ 0
x ≥ 0 constitutes I and III quadrants.
y ≥ 0 constitutes I and II quadrants.
So, x ≥ 0 ∧ y ≥ 0 is the intersection of them, which is the I quadrant.
So, the feasible region of the constraint x ≥ 0 ∧ y ≥ 0 is the I quadrant as shown in the figure as region A.
Feasible region of the inequality x + 3y ≥ 3
Plot the equality on the graph, as plotted in the figure.
Pick a point in the I quadrant, say (1,1).
Put it in the inequality
1 + 3 × 1 = 4 > 3.
The point (1,1) satisfies the inequality.
So, the feasible region includes is the region which contains the point (1,1).
The upward region of the line x + 3y = 3 is, thus, the feasible region of the inequality x + 3y ≥ 3 as shaded in the figure as region B.
Feasible region of all the three constraints
The Feasible region of the three constraints is the intersection of regions A and B. The intersection of the two regions is shaded in the figure as region C.
The region C as can be seen is not bounded on the positive sides of the x and y axes.
So, the feasible region of the constraints is unbounded.
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