Which one of the following inequalities is redundant for the shaded feasible region (ABCDA) shown below?

A redundant constraint is an unnecessary rule that doesn't form any boundary of the feasible region. Even if removed, the shaded region stays exactly the same.

---

The four constraints are:

1. $x + 2y \leq 40$
2. $3x + y \geq 30$
3. $4x + 3y \leq 60$
4. $x \geq 0, y \geq 0$

The feasible region is ABCDA with vertices A, B, C, D.

---

The boundary lines are:

Line 1: $x + 2y = 40$

Line 2: $3x + y = 30$

Line 3: $4x + 3y = 60$

Lines 4: $x = 0$ and $y = 0$

---

Point A (Line 2 meets y-axis):

Put $x = 0$ in $3x + y = 30$

$y = 30$

Point A $= (0, 30)$

---

Point B (Line 1 meets Line 2):

$x + 2y = 40$

$3x + y = 30$

From the second equation: $y = 30 - 3x$

Substituting into the first equation:

$x + 2(30 - 3x) = 40$

$x + 60 - 6x = 40$

$-5x = -20$

$x = 4$

$y = 30 - 12 = 18$

Point B $= (4, 18)$

---

Point C (Line 1 meets x-axis):

Put $y = 0$ in $x + 2y = 40$

$x = 40$

Point C $= (40, 0)$

---

Point D (Line 2 meets x-axis):

Put $y = 0$ in $3x + y = 30$

$x = 10$

Point D $= (10, 0)$

---

Checking if $4x + 3y = 60$ passes through any vertex:

At A $(0, 30)$: $4(0) + 3(30) = 90 \neq 60$

At B $(4, 18)$: $4(4) + 3(18) = 70 \neq 60$

At C $(40, 0)$: $4(40) + 3(0) = 160 \neq 60$

At D $(10, 0)$: $4(10) + 3(0) = 40 \neq 60$

Constraint 3 doesn't form any edge of the feasible region.

---

The feasible region ABCDA is completely defined by constraints 1, 2, and 4. Constraint 3 plays no role in forming the boundaries.

Therefore, the inequality $4x + 3y \leq 60$ is redundant.

The answer is Option 3.