Q.Solve the following linear programming problem graphically: Maximise $Z = 4x + y$ subject to the constraints: $x + y \le 50$, $3x + y \le 90$, $x \ge 0,\ y \ge 0$.

Main: min $Z=1550$ at $(0,5)$. OR: $Z=-50x+20y$ is unbounded below on the feasible region, so no minimum exists.

Main part. Constraints: $x+y\le 8$, $x\le 5$, $y\le 5$, $x+y\ge 4$, $x,y\ge 0$. This is a bounded region with corner points $(0,4),(0,5),(3,5),(5,3),(5,0),(4,0)$. Evaluate $Z=10(x-7y+190)$:

• $(0,4)$: $10(0-28+190)=1620$
• $(0,5)$: $10(0-35+190)=1550$
• $(3,5)$: $10(3-35+190)=1580$
• $(5,3)$: $10(5-21+190)=1740$
• $(5,0)$: $10(5-0+190)=1950$
• $(4,0)$: $10(4-0+190)=1940$

The minimum is $Z=1550$ at $(0,5)$.

OR part. Minimize $Z=-50x+20y$ subject to $2x-y\ge -5$, $3x+y\ge 3$, $2x-3y\le 12$, $x,y\ge 0$. Corner points are $(0,3),(0,5),(1,0),(6,0)$, with $Z=60,100,-50,-300$ respectively; the smallest corner value is $-300$ at $(6,0)$. But the feasible region is unbounded (it extends without bound as $x$ increases). Testing the open half-plane $-50x+20y<-300$: along the edge $2x-3y=12$, writing $x=6+\tfrac{3}{2}y$ gives $Z=-50\left(6+\tfrac32 y\right)+20y=-300-55y$, which decreases without limit as $y\to\infty$ while all constraints stay satisfied. Hence $-300$ is not a minimum, and $Z$ has no minimum value.

Main: plot the constraints, find corner points of the feasible region, evaluate $Z$ at each — max and min occur at corner points (corner-point theorem). OR: same method for a different feasible region and objective.

Main: Maximize/minimize $Z=3x+5y$ subject to $2x+3y\le36$, $x+y\le15$, $y\ge3$, $x\ge0$ (with $y\ge0$ implicit).

Find the corner points of the feasible region by intersecting boundary lines:

• $x=0$ and $y=3$: $(0,3)$
• $x=0$ and $2x+3y=36$: $y=12$, point $(0,12)$
• $y=3$ and $x+y=15$: $x=12$, point $(12,3)$ [note: $y=3$ meeting $2x+3y=36$ gives $x=13.5$, but that violates $x+y\le15$, so it is not a feasible vertex]
• $x+y=15$ and $2x+3y=36$: substituting $x=15-y$ gives $2(15-y)+3y=36\implies y=6,x=9$, point $(9,6)$

Feasible region vertices: $(0,3),(0,12),(9,6),(12,3)$.

Evaluate $Z=3x+5y$ at each:

• $(0,3)$: $Z=15$
• $(0,12)$: $Z=60$
• $(9,6)$: $Z=27+30=57$
• $(12,3)$: $Z=36+15=51$

By the corner-point theorem, the maximum and minimum of $Z$ over the feasible region occur at vertices: maximum $Z=60$ at $(0,12)$, minimum $Z=15$ at $(0,3)$.

OR: Minimize $Z=3x+9y$ subject to $x+3y\le60$, $x+y\ge10$, $x\le y$, $x\ge0,y\ge0$.

Corner points of the feasible region:

• $x+y=10$ and $x=y$: $x=y=5$, point $(5,5)$
• $x+y=10$ and $x=0$: point $(0,10)$
• $x+3y=60$ and $x=0$: point $(0,20)$
• $x+3y=60$ and $x=y$: $4x=60\implies x=y=15$, point $(15,15)$

Evaluate $Z=3x+9y$:

• $(0,10)$: $Z=90$
• $(5,5)$: $Z=15+45=60$
• $(15,15)$: $Z=45+135=180$
• $(0,20)$: $Z=180$

The minimum value is $Z=60$ at $(5,5)$, which is the vertex closest to the origin-side boundary $x+y=10$ meeting $x=y$; every other feasible vertex gives a strictly larger $Z$, and the region between vertices along each edge varies monotonically since $Z$ is linear, so the vertex minimum is the true minimum.

Graph the constraints, find the feasible-region corner points, and evaluate the objective at each; the OR part is a max-profit LPP solved the same way.

Main question: Constraints: $x+2y\ge100$, $2x-y\le0$ (i.e. $y\ge2x$), $2x+y\le200$, $x,y\ge0$.

Find the corner points of the feasible region by pairwise intersection of the boundary lines (keeping only points satisfying all constraints):

• $x+2y=100$ meets the $y$-axis ($x=0$) at $(0,50)$ — check: $y\ge2x$ (50≥0 ✓), $2x+y=50\le200$ ✓.
• $x+2y=100$ meets $y=2x$: substituting, $x+4x=100\Rightarrow x=20,\,y=40$, point $(20,40)$.
• $y=2x$ meets $2x+y=200$: $2x+2x=200\Rightarrow x=50,\,y=100$, point $(50,100)$.
• $2x+y=200$ meets the $y$-axis at $(0,200)$ — check: $x+2y=400\ge100$ ✓, $y\ge2x$ (200≥0 ✓).

(The point $(100,0)$, where $x+2y=100$ meets the $x$-axis, is rejected since it violates $y\ge2x$.)

So the feasible region is the quadrilateral with vertices $(0,50),\,(20,40),\,(50,100),\,(0,200)$.

Evaluate $Z=x+2y$:

Since $(0,50)$ and $(20,40)$ both lie on the line $x+2y=100$ and give the same $Z=100$, the minimum $Z=100$ is attained at every point of the segment joining them (multiple optimal solutions). The maximum $Z=400$ occurs uniquely at $(0,200)$.

OR: Let $x$ = number of desktops, $y$ = number of portables.

Budget: $25000x+40000y\le70{,}00{,}000\Rightarrow5x+8y\le1400$ (dividing by 5000).

Demand: $x+y\le250$. Also $x,y\ge0$.

Maximize profit $P=4500x+5000y$.

Corner points:

• $(0,0)$
• $(250,0)$ (demand line meets $x$-axis; budget $5(250)=1250\le1400$, feasible)
• Intersection of $5x+8y=1400$ and $x+y=250$: substitute $x=250-y$: $5(250-y)+8y=1400\Rightarrow1250+3y=1400\Rightarrow y=50,\,x=200$, giving $(200,50)$
• $(0,175)$ (budget line meets $y$-axis: $8y=1400\Rightarrow y=175$; demand check $175\le250$ ✓)

Evaluate $P=4500x+5000y$:

Maximum profit $=$ Rs. $11{,}50{,}000$, attained at $(200,50)$ — i.e. $200$ desktop models and $50$ portable models.

Plot the corner points of the unbounded feasible region and test, via the half-plane method, whether $Z$ is bounded in either direction — here it is unbounded both ways.

Constraints: $x\ge2$, $x+y\ge5$, $x+2y\ge6$, $y\ge0$. Since all inequalities are "$\ge$" (plus $y\ge0$), the feasible region lies above/right of these boundary lines and is unbounded.

Corner points (intersections of the boundary lines, checked for feasibility):

• $x=2$ and $x+y=5$: gives $(2,3)$. (Check $x+2y=2+6=8\ge6$ ✓.)
• $x+y=5$ and $x+2y=6$: subtracting gives $y=1,\ x=4$, i.e. $(4,1)$. (Check $x\ge2$ ✓.)
• $x+2y=6$ and $y=0$: gives $(6,0)$. (Check $x+y=6\ge5$ ✓, $x\ge2$ ✓.)

The region is bounded by these three segments but extends unboundedly: upward along $x=2$ (for $y\ge3$) and rightward along $y=0$ (for $x\ge6$).

Evaluate $Z=-x+2y$ at the corners:

$Z(2,3) = -2+6=4,\qquad Z(4,1)=-4+2=-2,\qquad Z(6,0)=-6+0=-6.$

Testing for a maximum: Consider the open half-plane $-x+2y>4$. The point $(2,100)$ lies in the feasible region (satisfies $x\ge2$, $x+y=102\ge5$, $x+2y=202\ge6$, $y\ge0$) and gives $Z=-2+200=198>4$. Since this half-plane intersects the feasible region, $Z$ can be made arbitrarily large along $x=2$ as $y\to\infty$ — so $Z$ has no maximum value.

Testing for a minimum: Consider the open half-plane $-x+2y<-6$. The point $(100,0)$ is feasible (satisfies all constraints) and gives $Z=-100<-6$. Since this half-plane also intersects the feasible region, $Z$ can be made arbitrarily small (large negative) along $y=0$ as $x\to\infty$ — so $Z$ has no minimum value either.

Hence, over this unbounded feasible region, $Z=-x+2y$ is unbounded in both directions: neither a maximum nor a minimum exists.

OR: Toys $A,B$; machine-minute constraints (converting 6 hours $=360$ minutes each):

$\text{I: } 12x+6y\le360 \Rightarrow 2x+y\le60,\quad \text{II: } 18x\le360\Rightarrow x\le20,\quad \text{III: } 6x+9y\le360\Rightarrow 2x+3y\le120,\quad x,y\ge0.$

Maximize $Z=7.5x+5y$.

Corner points of the feasible region: $(0,0)$, $(20,0)$, $(20,20)$ [intersection of $x=20$ and $2x+y=60$], $(15,30)$ [intersection of $2x+y=60$ and $2x+3y=120$], $(0,40)$ [intersection of $2x+3y=120$ and $x=0$].

$Z(0,0)=0,\quad Z(20,0)=150,\quad Z(20,20)=7.5(20)+5(20)=250,\quad Z(15,30)=7.5(15)+5(30)=112.5+150=262.5,\quad Z(0,40)=200.$

The maximum is $Z=262.5$, attained at $(15,30)$ — confirming that 15 toys of type $A$ and 30 toys of type $B$ give the maximum profit of Rs. 262.50 per day.

Evaluating $Z=3x+5y$ at the feasible region's corner points gives the minimum $7$ at $(3/2,1/2)$. (OR: the classic two-constraint LPP has minimum $300$ at $(60,0)$ and maximum $600$ along the whole edge from $(120,0)$ to $(60,30)$.)

Minimize $Z=3x+5y$ subject to $x+3y\ge3,\ x+y\ge2,\ x,y\ge0$

Find the corner points of the feasible region (intersection of the boundary lines with each other and the axes, keeping only feasible points):

• On $y=0$: need $x\ge3$ (from $x+3y\ge3$) and $x\ge2$ (from $x+y\ge2$) — the binding one is $x=3$, giving corner $(3,0)$.
• On $x=0$: need $y\ge1$ and $y\ge2$ — binding is $y=2$, giving corner $(0,2)$.
• Intersection of $x+3y=3$ and $x+y=2$: subtracting, $2y=1\Rightarrow y=\tfrac12$, then $x=\tfrac32$ — corner $\left(\tfrac32,\tfrac12\right)$.

The feasible region is unbounded, with corners $(3,0)$, $\left(\tfrac32,\tfrac12\right)$, $(0,2)$ (and extending outward).

Evaluate $Z=3x+5y$:

$Z(3,0)=9, \qquad Z\left(\tfrac32,\tfrac12\right)=\tfrac92+\tfrac52=7, \qquad Z(0,2)=10.$

Since both coefficients of $Z$ are positive and the region extends only outward (away from the origin), $Z$ cannot go below the smallest corner value; the open half-plane $3x+5y<7$ has no point in common with the feasible region. So the minimum is $Z=7$ at $\left(\tfrac32,\tfrac12\right)$.

OR: Minimise and Maximise $Z=5x+10y$ subject to $x+2y\le120,\ x+y\ge60,\ x-2y\ge0,\ x,y\ge0$

Finding all feasible corner points (checking each pairwise intersection against all constraints):

• $x+2y=120$ and $x=2y$: $y=30,x=60$ — $(60,30)$, feasible.
• $x+2y=120$ and $y=0$: $(120,0)$, feasible.
• $x+y=60$ and $x=2y$: $y=20,x=40$ — $(40,20)$, feasible.
• $x+y=60$ and $y=0$: $(60,0)$, feasible.

(Points like $(0,60)$ or $(0,0)$ fail the $x\ge2y$ or $x+y\ge60$ constraint and are not corners of the feasible region.)

Evaluate $Z=5x+10y$:

$Z(60,0)=300,\quad Z(120,0)=600,\quad Z(60,30)=300+300=600,\quad Z(40,20)=200+200=400.$

Minimum $Z=300$ at $(60,0)$.

Maximum $Z=600$, attained at both $(120,0)$ and $(60,30)$ — since these both lie on the line $x+2y=120$ and $Z=5x+10y=5(x+2y)$ is constant ($=600$) all along that edge, the maximum is attained at every point of the line segment joining $(120,0)$ and $(60,30)$ (multiple optimal solutions).

Plot the feasible region from the constraints, find its corner points, then evaluate the objective function at each corner.

Max/Min $Z=x+2y$ s.t. $x+2y\ge100,\ 2x-y\le0,\ 2x+y\le200,\ x,y\ge0$

Rewrite $2x-y\le0$ as $y\ge2x$.

Corner points (solving pairs of boundary lines and checking they satisfy all constraints):

• $x+2y=100$ and $y=2x$: $x+4x=100\Rightarrow x=20,y=40$ → $(20,40)$
• $y=2x$ and $2x+y=200$: $2x+2x=200\Rightarrow x=50,y=100$ → $(50,100)$
• $x=0$ with $x+2y=100$: $y=50$ → $(0,50)$ (since at $x=0$, need $y\ge50$ from constraint 1)
• $x=0$ with $2x+y=200$: $y=200$ → $(0,200)$ (upper bound at $x=0$)

These four points $(0,50),(20,40),(50,100),(0,200)$ form the (bounded) feasible region.

Evaluate $Z=x+2y$:

• $(0,50)$: $Z=100$
• $(20,40)$: $Z=100$
• $(50,100)$: $Z=250$
• $(0,200)$: $Z=400$

Minimum $Z=100$ (attained all along the edge joining $(0,50)$ and $(20,40)$, since that edge lies exactly on $x+2y=100$). Maximum $Z=400$ at $(0,200)$.

OR: Maximize $Z=1000x+600y$ s.t. $x+y\le200,\ x\ge20,\ y\ge4x,\ x,y\ge0$

At $x=20$: $y\ge80$ and $y\le180$, giving corners $(20,80)$ and $(20,180)$.

$y=4x$ meets $x+y=200$: $x+4x=200\Rightarrow x=40,y=160$ → $(40,160)$.

Feasible region is the triangle with vertices $(20,80),(20,180),(40,160)$.

Evaluate $Z=1000x+600y$:

• $(20,80)$: $Z=20000+48000=68000$
• $(20,180)$: $Z=20000+108000=128000$
• $(40,160)$: $Z=40000+96000=136000$

Maximum $Z=136000$ at $x=40,y=160$.