Q.The feasible region of a linear programming problem is a bounded quadrilateral with corner points $P\left(\dfrac{3}{13}, \dfrac{24}{13}\right)$, $Q\left(\dfrac{3}{2}, \dfrac{15}{4}\right)$, $R\left(\dfrac{7}{2}, \dfrac{3}{4}\right)$ and $S\left(\dfrac{18}{7}, \dfrac{2}{7}\right)$. Determine the maximum and minimum values of $Z = x + 2y$ over this region.

The key idea is that a corner point (also called an extreme point or vertex) of a feasible region is a point that lies on the boundary of the region and cannot be expressed as a convex combination of two other distinct points in the region. For a system of linear inequalities, corner points occur at intersections of constraint boundaries (lines) that also satisfy all inequalities.

We must check each candidate point against all constraints. If a point fails even one constraint, it is not in the feasible region at all — and therefore cannot be a corner point.

1. List the constraints clearly

The feasible region is the set of points satisfying all four inequalities simultaneously.

2. Test each candidate

   • (A) $(0,2)$

     Check (1): $0 + 2(2) = 4 \le 4$ ✓

     Check (2): $0 + 2 = 2 \ge 2$ ✓

     Check (3): $x=0, y=2$ both nonnegative ✓

     So $(0,2)$ is in the feasible region. It lies at the intersection of $x=0$ and $x+2y=4$, and also satisfies $x+y=2$ (it lies on that line too). It is a corner.

   • (B) $(4,0)$

     Check (1): $4 + 0 = 4 \le 4$ ✓

     Check (2): $4 + 0 = 4 \ge 2$ ✓

     Check (3): both nonnegative ✓

     So $(4,0)$ is in the feasible region. It lies at the intersection of $y=0$ and $x+2y=4$. It is a corner.

   • (C) $(0,0)$

     Check (1): $0 + 0 = 0 \le 4$ ✓

     Check (2): $0 + 0 = 0 \ge 2$ ✗ Fails

     Since it violates $x+y \ge 2$, $(0,0)$ is not in the feasible region. Therefore it cannot be a corner point. …

Concept & Intuition

In linear programming, the optimal solution (if it exists) lies at a corner point (vertex) of the feasible region. The feasible region is the set of all points satisfying every constraint. To identify which of the given points is a corner point, we must first find all intersections of the boundary lines, then verify they satisfy all inequalities. The trick: not every intersection is a corner — it must lie inside (or on) every constraint.

Step-by-step reasoning

1. List the constraints as equations (boundary lines):

   • $4x + 3y = 60$
   • $x = 3$
   • $y = 2x$
   • $y = 0$

   Also note: $x \geq 3$ and $y \geq 0$ are half-planes, so the feasible region is to the right of $x=3$ and above $y=0$.

2. Find all intersection points of these lines (potential corner points):

   • Intersection of $x=3$ and $y=2x$: $x=3 \Rightarrow y = 2(3) = 6$ → point $(3,6)$.
   • Intersection of $x=3$ and $4x+3y=60$: $4(3)+3y=60 \Rightarrow 12+3y=60 \Rightarrow 3y=48 \Rightarrow y=16$ → point $(3,16)$.
   • Intersection of $y=2x$ and $4x+3y=60$: Substitute $y=2x$: $4x+3(2x)=60 \Rightarrow 4x+6x=60 \Rightarrow 10x=60 \Rightarrow x=6$, then $y=12$ → point $(6,12)$.
   • Intersection of $y=0$ and $x=3$: $(3,0)$.
   • Intersection of $y=0$ and $4x+3y=60$: $4x=60 \Rightarrow x=15$ → point $(15,0)$.
   • Intersection of $y=0$ and $y=2x$: $0=2x \Rightarrow x=0$ → point $(0,0)$, but note $x=0$ violates $x \geq 3$, so this is not in the feasible region.

3. Check which of these points satisfy ALL constraints (including $x \geq 3$ and $y \geq 0$):

   • $(3,6)$: $x=3$ ok, $y=6 \geq 0$, $4(3)+3(6)=12+18=30 \leq 60$, $y=6 \leq 2(3)=6$ → feasible.
   • $(3,16)$: $4(3)+3(16)=12+48=60 \leq 60$ ok, $y=16 \leq 2(3)=6$? No, $16 \leq 6$ is false → not feasible.
   • $(6,12)$: $4(6)+3(12)=24+36=60 \leq 60$ ok, $y=12 \leq 2(6)=12$ ok, $x=6 \geq 3$ ok → feasible.
   • $(3,0)$: $x=3$ ok, $y=0$ ok, $4(3)+0=12 \leq 60$, $y=0 \leq 2(3)=6$ ok → feasible. …

Concept & Intuition

We want to maximize $P = 500x + 400y$ under linear inequalities. This is a classic linear programming problem: the maximum (if it exists) occurs at a vertex (corner point) of the feasible region. The constraints are:

1. $x + y \leq 200$
2. $x \geq 20$
3. $y \geq 4x$
4. $y \geq 0$

The tricky part is that $y \geq 4x$ is a lower bound on $y$, not an upper bound. Combined with $x + y \leq 200$, this forces $x$ to be small enough that $4x$ doesn’t exceed the line $y = 200 - x$. Let’s find where they intersect.

Step-by-step solution

1. Find the intersection of $y = 4x$ and $x + y = 200$ Substitute $y = 4x$ into $x + y = 200$:

Then $y = 4(40) = 160$. So the point $(40, 160)$ is where the lower bound meets the upper bound.

2. Identify the feasible region

   • $x \geq 20$ is a vertical line to the right of $x=20$.
   • $y \geq 0$ is the x-axis.
   • $y \geq 4x$ is the region above the line through origin with slope 4.
   • $x + y \leq 200$ is the region below the line from $(0,200)$ to $(200,0)$.

   The feasible region is a polygon with vertices at intersections of these boundaries. The leftmost boundary is $x=20$. On $x=20$, the lower bound $y \geq 4x$ gives $y \geq 80$. The upper bound $x+y \leq 200$ gives $y \leq 180$. So on the line $x=20$, $y$ runs from 80 to 180. But also $y \geq 0$ is automatically satisfied.

3. List all corner points

   • Intersection of $x=20$ and $y=4x$: $(20, 80)$
   • Intersection of $x=20$ and $x+y=200$: $(20, 180)$
   • Intersection of $y=4x$ and $x+y=200$: $(40, 160)$
   • Also check where $y=0$ meets $x=20$? But $y=0$ violates $y \geq 4x$ because $0 \geq 80$ is false. So no point on x-axis is feasible. …

We are minimizing a linear objective subject to linear inequalities. The feasible region is unbounded (since constraints are "greater than or equal to"), but because coefficients are positive, the minimum will occur at a corner point where two constraints intersect. The key idea: the minimum of a linear function over a convex polygon (or unbounded region with positive coefficients) occurs at a vertex.

1. Find the corner points of the feasible region.

   The constraints are:

   • $3x + 5y \geq 30$
   • $x + y \geq 8$
   • $x \geq 0$, $y \geq 0$

   Since both inequalities are "≥", the feasible region is above both lines and in the first quadrant. The vertices are intersections of:

   • $3x + 5y = 30$ and $x + y = 8$
   • $3x + 5y = 30$ and $x = 0$
   • $x + y = 8$ and $y = 0$
   • Also check $x = 0$ with $y = 0$? But $(0,0)$ does not satisfy either inequality, so it's not feasible.

2. Solve each intersection:

   • Intersection of $3x + 5y = 30$ and $x + y = 8$:

     From $x + y = 8$, we have $y = 8 - x$. Substitute:

     $3x + 5(8 - x) = 30 \implies 3x + 40 - 5x = 30 \implies -2x = -10 \implies x = 5$, then $y = 3$.

     So point $(5, 3)$.

   • Intersection of $3x + 5y = 30$ and $x = 0$:

     $3(0) + 5y = 30 \implies y = 6$. So point $(0, 6)$.

   • Intersection of $x + y = 8$ and $y = 0$:

     $x + 0 = 8 \implies x = 8$. So point $(8, 0)$.

   • Intersection of $x = 0$ and $y = 0$ is not feasible (fails both inequalities).

   Also check intersection of $x + y = 8$ with $x = 0$ gives $(0, 8)$, but does it satisfy $3x + 5y \geq 30$? $3(0) + 5(8) = 40 \geq 30$, yes. So $(0, 8)$ is also a corner (where $x=0$ meets $x+y=8$). Similarly, intersection of $3x+5y=30$ with $y=0$ gives $(10,0)$, which also satisfies $x+y \geq 8$? $10+0=10 \geq 8$, yes. So $(10,0)$ is another corner.

   So the vertices are: $(0, 6)$, $(0, 8)$, $(5, 3)$, $(8, 0)$, $(10, 0)$. But note: $(0,6)$ and $(8,0)$ are not actually vertices of the feasible region because they lie on only one constraint and the other constraint is not satisfied at that point? Let's check carefully:

   • $(0,6)$: $3(0)+5(6)=30$ (ok), but $x+y=6 < 8$ → fails second constraint. So not feasible. …

Constraints: $x+y\ge2$, $x+3y\ge3$, $x,y\ge0$. The feasible corners:

• $(3,0)$: on $x+3y=3$; $Z=9$.
• $(0,2)$: on $x+y=2$; $Z=10$. …

Constraints: $x,y\ge0,\ x+y\le12,\ 2x+y\le20$.

Corner points:

• $(0,0)$
• $(10,0)$ from $2x+y=20$ on the $x$-axis
• $(0,12)$ from $x+y=12$ on the $y$-axis
• Intersection $x+y=12,\ 2x+y=20\Rightarrow x=8,\ y=4$: $(8,4)$ …

Constraints: $x,y\ge0,\ x+y\le5,\ 3x+y\le9$.

Corner points:

• $(0,0)$
• $(3,0)$ from $3x+y=9$ on the $x$-axis
• $(0,5)$ from $x+y=5$ on the $y$-axis
• Intersection $x+y=5,\ 3x+y=9\Rightarrow x=2,\ y=3$: $(2,3)$ …

Concept: Linear programming - the optimum of a linear objective over a polygon occurs at a corner point.

Maximise Z = 4x + 2y subject to 4x + 2y >= 46, x + 3y <= 24, x >= 0, y >= 0.

Find the corner points of the feasible region:

1. 4x + 2y = 46 meets y = 0 at x = 11.5 -> (11.5, 0)
2. x + 3y = 24 meets y = 0 at x = 24 -> (24, 0)
3. Intersection of 4x + 2y = 46 and x + 3y = 24: from the first, y = 23 - 2x; substitute: x + 3(23 - 2x) = 24 -> x + 69 - 6x = 24 -> -5x = -45 -> x = 9, y = 5 -> (9, 5) …

Concept: Linear programming, corner-point method.

Maximise z = 12x + 3y subject to x >= 0, y >= 0, 3x + y <= 9, and the fifth constraint.

As printed, the constraint reads x + y >= 5. Working that region: it is the triangle with vertices (0,5), (0,9) and (2,3) (the point where x + y = 5 meets 3x + y = 9). Evaluating z there gives 15, 27 and 33 respectively, so the maximum would be 33 - which is NOT one of the four options, so the constraint as printed cannot be what the paper intended.

The standard problem (and the one consistent with the given options) has x + y <= 5:

Constraints: x >= 0, y >= 0, x + y <= 5, 3x + y <= 9.

Corner points:

O(0, 0) …

Concept: Corner-point method for an LPP.

Maximize Z = x + y subject to 3x + 4y <= 12, x >= 0, y >= 0.

Feasible region: triangle with vertices

(0, 0), (4, 0) [3x = 12 -> x = 4], and (0, 3) [4y = 12 -> y = 3].

Evaluate Z = x + y:

• (0, 0): Z = 0
• (4, 0): Z = 4 <-- maximum
• (0, 3): Z = 3 …