Consider the objective function \( Z=3x+4y \) subject to the following constraints: \( x-6y\le 6,x+y-1\ge 0,x-y+1\ge 0,x+2y\le 6,\ x\ge 0,y\ge 0 \) . Which of the following inequalities does not contribute to the feasible region?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( x-6y\le 6 \)
A constraint contributes to the feasible region only if its line forms one of the region’s edges. The corners of the region are:
\( \displaystyle x+y=1\ \text{and}\ y=0:\ (1,0) \)
\( \displaystyle x+2y=6\ \text{and}\ y=0:\ (6,0) \)
\( \displaystyle x-y+1=0\ \text{and}\ x+2y=6:\ 3x+2=6\ \Rightarrow\ \left(\frac{4}{3},\frac{7}{3}\right) \)
\( \displaystyle x+y=1\ \text{and}\ x-y+1=0:\ (0,1) \)
So the feasible region is the quadrilateral with corners (1, 0), (6, 0), \( \left(\frac{4}{3},\frac{7}{3}\right) \) and (0, 1). Its edges lie on y = 0, x + 2y = 6, x − y + 1 = 0 and x + y = 1, so the constraints in Options (1), (3) and (4) each cut off part of the plane and shape the region.
Now x − 6y ≤ 6 holds at every corner:
\( \displaystyle (1,0):\ 1\le 6 \)
\( \displaystyle (6,0):\ 6\le 6 \)
\( \displaystyle \left(\frac{4}{3},\frac{7}{3}\right):\ \frac{4}{3}-14\le 6 \)
\( \displaystyle (0,1):\ -6\le 6 \)
All corners satisfy it, so the whole region satisfies it. In fact, for y ≥ 0, \( x+2y\le 6 \) already gives \( x-6y\le 6-8y\le 6 \). The line x − 6y = 6 only touches the region at the corner (6, 0), so removing this constraint does not change the feasible region.