Q.A man rides his motorcycle at the speed of 50 km/hour. He has to spend Rs 2 per km on petrol. If he rides it at a faster speed of 80 km/hour, the petrol cost increases to Rs 3 per km. He has at most Rs 120 to spend on petrol and one hour's time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.

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 $(0,5)$.

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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 $(0,20)$ …

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) …

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): …

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. …

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. $x+y\le200,\ x\ge20,\ y\ge4x,\ x,y\ge0$

…

In each LPP, plot the constraint lines, find the vertices of the feasible region, and evaluate the objective at each vertex — the optimum occurs at a vertex (corner-point method).

Main question. Maximize $z=20x+15y$ subject to $2x+y\le200$, $x+y\le150$, $x,y\ge0$.

Find the corner points of the feasible region.

• Intersection of $2x+y=200$ and $x+y=150$: subtracting, $x=50$, so $y=100$. Point $(50,100)$.
• $y=0$: $2x+y=200\Rightarrow x=100$ (binding, since $x+y\le150$ allows $x$ up to $150$, so $200$-line is tighter) — vertex $(100,0)$.
• $x=0$: $x+y=150\Rightarrow y=150$ (binding, since $2x+y\le200$ allows $y$ up to $200$) — vertex $(0,150)$.
• Origin $(0,0)$.

Evaluate $z=20x+15y$:

Maximum is $z=2500$ at $(50,100)$.

OR question. Maximize and minimize $z=5x+2y$ subject to $x-2y\le2$, $3x+2y\le12$, $-3x+2y\le3$, $x,y\ge0$.

Find the vertices.

• $x=0$: constraints give $y\le6$ (from $3x+2y\le12$) and $y\le1.5$ (from $-3x+2y\le3$); tightest is $y\le1.5$ — vertex $(0,1.5)$; also $(0,0)$.
• $y=0$: constraints give $x\le2$ (from $x-2y\le2$) and $x\le4$ (from $3x+2y\le12$); tightest is $x\le2$ — vertex $(2,0)$.
• Intersection of $x-2y=2$ and $3x+2y=12$: adding, $4x=14\Rightarrow x=3.5$, $y=\dfrac{x-2}{2}=0.75$. Vertex $(3.5,\ 0.75)$ — check $-3(3.5)+2(0.75)=-9\le3$ ✓ feasible. …

The feasible region is a triangle with corners $(2,72),(15,20),(40,15)$; evaluating $z=6x+3y$ gives min $150$ and max $285$.

Maximize/minimize $z = 6x + 3y$ subject to $4x + y \ge 80$, $x + 5y \ge 115$, $3x + 2y \le 150$, $x,y \ge 0$.

Find the corner points by intersecting the boundary lines:

• $4x + y = 80$ and $x + 5y = 115$: solving gives $(15, 20)$.
• $4x + y = 80$ and $3x + 2y = 150$: solving gives $(2, 72)$.
• $x + 5y = 115$ and $3x + 2y = 150$: solving gives $(40, 15)$.

Each of these satisfies all constraints, so the feasible region is the triangle with vertices $(2,72),\ (15,20),\ (40,15)$.

Evaluate $z = 6x + 3y$ at each corner:

• $(2, 72): z = 12 + 216 = 228$
• $(15, 20): z = 90 + 60 = 150$
• $(40, 15): z = 240 + 45 = 285$

So the minimum is $150$ at $(15,20)$ and the maximum is $285$ at $(40,15)$.

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