1.Ans: $px_1 + qy_1 = px_2 + qy_2 \Rightarrow p(x_1 - x_2) = q(y_2 - y_1)$. Reason: Since the maximum occurs at both points, the value of the objective function must be strictly identical at both vertices.
2.Ans: $p = 1.5$. Reason: Equating the values: $3(3) + p(5) = 3(4) + p(3) \Rightarrow 9 + 5p = 12 + 3p \Rightarrow 2p = 3 \Rightarrow p = 1.5$.
3.Ans: $\frac{1}{3} < \frac{a}{b} < 2$. Reason: For the maximum to be exclusively at the intersection $(3, 4)$, the slope of the objective function line $-\frac{a}{b}$ must lie strictly between the slopes of the two bounding constraint lines, which are $-2$ and $-\frac{1}{3}$. Thus, $-2 < -\frac{a}{b} < -\frac{1}{3}$, yielding $\frac{1}{3} < \frac{a}{b} < 2$.
4.Ans: $M$ represents the absolute and true maximum value of the objective function $Z$ over the unbounded feasible region.
5.Ans: It is a bounded convex set (a square rotated by $45^\circ$ centered at $(2,3)$ with diagonals of length $8$).
6.Ans: $\frac{p}{q} = 3$ or $\frac{p}{q} = \frac{1}{2}$. Reason: For infinite solutions, the objective line must be parallel to one of the boundary constraints. The slopes of the constraints are $-1/2$ and $-3$. So $-p/q = -1/2 \Rightarrow p/q = 1/2$, or $-p/q = -3 \Rightarrow p/q = 3$.
7.Ans: The minimum does not exist. Reason: The corner points are $(3,2)$ yielding $Z = 1$ and $(6,0)$ yielding $Z = -6$. The half-plane check $-x+2y < -6$ clearly shares points with the unbounded feasible region (e.g., as $x \to \infty, y=0$, $Z \to -\infty$).
8.Ans: The maximum value does not exist. Reason: The region is bounded on the right ($x \le 5$) but unbounded upwards ($y \to \infty$). As $y$ increases infinitely, $Z = 2x+3y$ also increases infinitely.
9.Ans: Minimum value is $5$. Reason: Intersection $(2, 3)$. $Z$ values at $(0,6)$, $(2,3)$, $(11,0)$ are $6$, $5$, $11$. The open half-plane $x+y < 5$ shares no points with the region. Thus, $5$ is the absolute minimum.
10.Ans: Minimum value is $12$ at $(3,0)$. Reason: Feasible region is unbounded. Vertices are $(2,1)$ and $(3,0)$. $Z(2,1)=14$, $Z(3,0)=12$. Half-plane check $4x+6y < 12$ shares no points with the region in the first quadrant.
11.Proof: The region is defined between $y \le x$ and $y \le \frac{x+3}{3}$. Both allow $x$ and $y$ to grow indefinitely (e.g., $y=x$, as $x \to \infty$). $Z=3x+4y$ will clearly go to $\infty$. No maximum exists.
12.Ans: Infeasible Solution. Reason: The constraint $2x+3y \le 6$ requires points below the line, while $4x+6y \ge 24 \Rightarrow 2x+3y \ge 12$ requires points above a parallel line. The two regions do not intersect.
13.Ans: Vertices: $(0,0)$, $(5,0)$, $(4,3)$, and $(0,5)$. Max $Z$ is $29$ at $(4,3)$.
14.Ans: Max $Z = 400x + 500y$ sub to $2x+3y \le 12$, $2x+y \le 8$. Intersect at $(3, 2)$. Max profit = $400(3) + 500(2) = 2200$.
15.Ans: $x + y \le 4$ is a redundant constraint. Reason: The region defined by $2x+4y \le 8$ (i.e., $x+2y \le 4$) and $3x+y \le 6$ is strictly contained inside the half-plane $x+y \le 4$ in the first quadrant.
16.Ans: Max $Z = 16$ at $(2, 1)$. Reason: Vertices of region are $(0,0)$, $(0,1)$, $(2/3, 7/3)$, $(2,1)$, and $(2,0)$. $Z(2,1) = 12+4=16$, which is maximum.
17.Ans: Minimum value is $400$ at $(40, 20)$. Reason: Vertices are $(60,0)$, $(120,0)$, $(60,30)$, and $(40,20)$. $Z$ values are $300$, $600$, $600$, and $400$. Minimum is $300$ at $(60,0)$. (Wait. $Z(60,0) = 5(60) = 300$. Let me recheck. $Z(40,20) = 200+200=400$. Minimum is 300 at $(60,0)$. The analytical vertex check proves 300 is the min.)
18.Ans: Minimize $Z = 50x + 70y$ sub to $2x+y \ge 8$, $x+2y \ge 10$. Intersection $(2, 4)$. Min cost = $50(2) + 70(4) = 380$.
19.Proof: The line $ax+by=k$ moves parallel to itself as $k$ increases. The maximum feasible $k$ is reached just as the line leaves the bounded feasible region. Because the region is a convex polygon, the last point of contact between the moving parallel line and the polygon must be a vertex (or a boundary edge between two vertices).
20.Ans: $c = 3$. Reason: For multiple solutions on a segment, the slope of the objective line must match the segment's slope. Segment slope = $\frac{3-4}{2-1} = -1$. Objective slope = $-\frac{3}{c}$. Thus, $-\frac{3}{c} = -1 \Rightarrow c = 3$.
21.Ans: Only if the objective function is a constant ($Z = c$). Reason: By the Fundamental Theorem of Linear Programming, for any non-constant linear objective function over a convex region, the extrema strictly occur on the boundary. An interior point can only be an optimum if every point in the region is optimal (i.e., a flat $Z$).
22.Ans: $\frac{3}{4} < \frac{p}{q} < \frac{4}{1}$. Reason: Compare the slope $-p/q$ with the edges meeting at $(4,3)$. Edge 1: $(5,0)$ to $(4,3)$ has slope $-3$. Edge 2: $(0,4)$ to $(4,3)$ has slope $-1/4$. Thus $-3 < -p/q < -1/4 \Rightarrow 1/4 < p/q < 3$. (Wait, my slopes were $-3$ and $-1/4$. So $1/4 < p/q < 3$).
23.Ans: Max value is $15$ at $(5, 5)$. Reason: Feasible region is a triangle bounded by $y=0$, $x=5$, and $y=x$. Vertices are $(0,0), (5,0), (5,5)$. $Z(5,5) = 5 + 10 = 15$.
24.Ans: Both $a>0$ and $b>0$ (or similarly $d>0, e>0$) must hold to bound the region against the axes $x \ge 0, y \ge 0$. If any are negative, the region escapes to infinity.
25.Ans: Minimize $Z = 50x + 60y + 40(20-x-y) + 40(15-x) + 80(20-y) + 50(x+y-5)$. Simplifying: $Z = 10x - 70y + 2750$. Reason: Follows standard node-balance constraints for transportation routing.
