The corner points of the feasible region of a LPP with the constraints $x + 2y \leq 40$, $3x + y \geq 30$, $4x + 3y \geq 60$, $x, y \geq 0$ are

We need to find all corner points of the region satisfying all constraints simultaneously:

$x + 2y \leq 40$

$3x + y \geq 30$

$4x + 3y \geq 60$

$x \geq 0,\ y \geq 0$

Corner points occur where the boundary lines intersect. We find all candidate intersections and check if they satisfy every constraint.

---

Intersection of $3x + y = 30$ and $4x + 3y = 60$

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

Substituting into the second:

$4x + 3(30 - 3x) = 60$

$4x + 90 - 9x = 60$

$-5x = -30$

$x = 6, \quad y = 30 - 18 = 12$

Candidate point: $(6, 12)$

Check: $6 + 2(12) = 30 \leq 40$ ✓ | $3(6) + 12 = 30 \geq 30$ ✓ | $4(6) + 3(12) = 60 \geq 60$ ✓

$(6, 12)$ is feasible. ✅

---

Intersection of $x + 2y = 40$ and $4x + 3y = 60$

From the first equation: $x = 40 - 2y$

Substituting into the second:

$4(40 - 2y) + 3y = 60$

$160 - 8y + 3y = 60$

$-5y = -100$

$x = 0, \quad y = 20$

Candidate point: $(0, 20)$

Check: $3(0) + 20 = 20 \geq 30$? ❌

$(0, 20)$ is not feasible.

---

Intersection of $x + 2y = 40$ and $3x + y = 30$

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

Substituting into the first:

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

$x + 60 - 6x = 40$

$-5x = -20$

$x = 4, \quad y = 30 - 12 = 18$

Candidate point: $(4, 18)$

Check: $4 + 2(18) = 40 \leq 40$ ✓ | $3(4) + 18 = 30 \geq 30$ ✓ | $4(4) + 3(18) = 70 \geq 60$ ✓

$(4, 18)$ is feasible. ✅

---

Intersection of $3x + y = 30$ with the $x$-axis ($y = 0$)

$3x = 30 \implies x = 10$

Candidate point: $(10, 0)$

Check: $4(10) + 3(0) = 40 \geq 60$? ❌

$(10, 0)$ is not feasible.

---

Intersection of $4x + 3y = 60$ with the $x$-axis ($y = 0$)

$4x = 60 \implies x = 15$

Candidate point: $(15, 0)$

Check: $15 + 0 = 15 \leq 40$ ✓ | $3(15) + 0 = 45 \geq 30$ ✓ | $4(15) + 0 = 60 \geq 60$ ✓

$(15, 0)$ is feasible. ✅

---

Intersection of $x + 2y = 40$ with the $x$-axis ($y = 0$)

$x = 40$

Candidate point: $(40, 0)$

Check: $40 \leq 40$ ✓ | $3(40) = 120 \geq 30$ ✓ | $4(40) = 160 \geq 60$ ✓

$(40, 0)$ is feasible. ✅

---

The corner points of the feasible region are:

$(4, 18), \quad (6, 12), \quad (15, 0), \quad (40, 0)$