Linear Programming — NCERT Solutions

We minimise $Z=x+2y$ subject to

• $2x+y\ge 3$
• $x+2y\ge 6$
• $x\ge 0,\ y\ge 0$

The line $x+2y=6$ itself gives $Z=6$. We check whether there are feasible points on this line.

If $x+2y=6$, then $y=(6-x)/2$.
Substitute into $2x+y\ge 3$:

$2x+\frac{6-x}{2}\ge 3$

Multiply by 2:

$4x+6-x\ge 6$

$3x\ge 0$

This is true for all $x\ge 0$.

So every feasible point on the line segment satisfying $x+2y=6$ gives $Z=6$.
For example, the points $(0,3)$ and $(6,0)$ are feasible, and both give $Z=6$.

Hence the minimum value is $6$, attained at more than two points.

Constraints:

• $x+2y\le 120$
• $x+y\ge 60$
• $x-2y\ge 0$
• $x\ge 0,\ y\ge 0$

Rewrite $x-2y\ge 0$ as $x\ge 2y$.

Corner points of the feasible region:

1. Intersection of $x+y=60$ and $x-2y=0$:

$x=2y$, so $2y+y=60\Rightarrow 3y=60\Rightarrow y=20$, $x=40$.
Point: $(40,20)$.

2. Intersection of $x+y=60$ and $x+2y=120$:

Subtract: $y=60$, then $x=0$.
Point: $(0,60)$.

3. Intersection of $x-2y=0$ and $x+2y=120$:

$x=2y$, so $2y+2y=120\Rightarrow y=30$, $x=60$.
Point: $(60,30)$.

Now evaluate $Z=5x+10y$:

• $(40,20)$: $Z=5(40)+10(20)=200+200=400$
• $(0,60)$: $Z=600$
• $(60,30)$: $Z=300+300=600$

The minimum value is $400$ at $(40,20)$, and the maximum value is $600$ at $(0,60)$ and $(60,30)$.

However, note that $Z=5x+10y=5(x+2y)$. Since $x+2y\le 120$, the largest possible value is $5\cdot120=600$, and the smallest feasible value occurs at the least value of $x+2y$ within the region, which is $80$ at $(40,20)$, giving $5\cdot80=400$.

We need minimise and maximise $Z=x+2y$ subject to

• $x+2y\ge 100$
• $2x-y\le 0$ i.e. $y\ge 2x$
• $2x+y\le 200$
• $x\ge 0,\ y\ge 0$

Since $Z=x+2y$, and the first constraint is exactly $Z\ge 100$, the minimum possible value is at least 100.

Check whether $Z=100$ is feasible. On the line $x+2y=100$, we need:

• $y\ge 2x$
• $2x+y\le 200$

Take the intersection of $x+2y=100$ and $y=2x$:

$x+2(2x)=100\Rightarrow 5x=100\Rightarrow x=20,\ y=40$.
This is feasible.
So the minimum value is $100$.

For the maximum, use $2x+y\le 200$ together with $y\ge 2x$ and non-negativity. Corner points of the feasible region are:

• $(20,40)$ from $x+2y=100$ and $y=2x$
• $(50,100)$ from $x+2y=100$ and $2x+y=200$
• $(0,200)$ from $y=200$ and $x=0$

Evaluate $Z=x+2y$:

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

So the maximum value is $400$ at $(0,200)$.

Thus, minimum and maximum are $100$ and $400$ respectively.

We maximise $Z=-x+2y$ subject to

• $x\ge 3$
• $x+y\ge 5$
• $x+2y\ge 6$
• $y\ge 0$

To maximise $-x+2y$, we want $y$ as large as possible relative to $x$.
But the constraints only give lower bounds on $y$; there is no upper bound on $y$.

For example, take $x=3$. Then the constraints become:

• $3+y\ge 5 \Rightarrow y\ge 2$
• $3+2y\ge 6 \Rightarrow y\ge 1.5$
• $y\ge 0$

So any $y\ge 2$ is feasible. Then

$Z=-3+2y$

which increases without bound as $y$ increases.

Therefore, the objective function has no maximum value. The feasible region is unbounded upward, so $Z$ can be made arbitrarily large.

Constraints:

• $x-y\le -1$ means $y\ge x+1$
• $-x+y\le 0$ means $y\le x$
• $x\ge 0,\ y\ge 0$

But $y\ge x+1$ and $y\le x$ cannot both hold for any real $x$, because $x+1>x$ always.
So there is no feasible region.

Hence there is no feasible solution, and therefore no maximum value of $Z=x+y$.

If the book intended the constraints exactly as written, the problem is infeasible.