Referring to solution 14 , we have maximise $Z=200 x+120 y$ subject to $x+y \leq 300,3 x+y \leq 600, x-y \geq-100, x \geq 0, y \geq 0$.

On solving $x+y=300$ and $3 x+y=600$, we get

$$x=150, y=150$$

On solving $x-y=-100$ and $x+y=300$, we get

$$x=100, y=200$$

From the shaded feasible region it is clear that coordinates of corner points are $(0,0)$, $(200,0),(150,150),(100,200)$ and $(0,100)$.

Corner points	Corresponding value of $Z=100x+120y$
(0, 0)	0
(200, 0)	40000
(150, 150)	$150 \times 200+120 \times 150=48000 \leftarrow$ Maximum
(100, 200)	$100\times200+120\times200=44000$
(0, 100)	$120\times100=12000$

Hence, 150 sweaters of each type made by company and maximum profit = ₹ $48000$.

Referring to solution 15 , we have

Maximise $Z=x+y$, subject to

$$2 x+3 y \leq 120,8 x+5 y \leq 400, x \geq 0, y \geq 0$$

On solving, we get

$$\begin{aligned} 8 x+5 y & =400 \text { and } 2 x+3 y=120, \text { we get } \\ x & =\frac{300}{7}, y=\frac{80}{7} \end{aligned}$$

From the shaded feasible region, it is clear that coordinates of corner points are $(0,0)$, $(50,0),\left(\frac{300}{7}, \frac{80}{7}\right)$ and $(0,40)$.

Corner points	Corresponding value of $Z=x+y$
(0, 0)	0
(50, 0)	50
$\frac{300}{7}, \frac{80}{7}$	$\frac{380}{7}=54 \frac{2}{7} \mathrm{~km} \leftarrow$ Maximum
(0, 40)	40

Hence, the maximum distance that the man can travel is $54 \frac{2}{7} \mathrm{~km}$.

Here, the given LPP is,

Maximise $Z=x+y$ subject to,

$$x+4 y \leq 8,2 x+3 y \leq 12,3 x+y \leq 9, x \geq 0, y \geq 0$$

On solving $x+4 y=8$ and $3 x+y=9$, we get

$$x=\frac{28}{11}, y=\frac{15}{11}$$

From the feasible region, it is clear that coordinates of corner points are $(0,0),(3,0)$, $\left(\frac{28}{11}, \frac{15}{11}\right)$ and $(0,2)$.

Corner points	Corresponding value of $Z=x+y$
(0, 0)	0
(3, 0)	3
$\left(\frac{28}{11}, \frac{15}{11}\right)$	$\frac{43}{11}=3 \frac{10}{11} \mathrm{~km} \leftarrow$ Maximum
(0, 2)	2

Hence, the maximum value is $3 \frac{10}{11}$.

Let the manufacturer produces $x$ number of models $X$ and $y$ number of model $Y$ bikes. Model $X$ takes a 6 man-hours to make per unit and model $Y$ takes a 10 man-hours to make per unit.

There is total of 450 man-hour available per week.

$$\begin{gathered} \therefore 6 x+10 y \leq 450 \\ \Rightarrow\quad 3 x+5 y \leq 225\quad\text{.... (i)} \end{gathered}$$

For models $X$ and $Y$, handling and marketing costs are ₹ $2000$ and ₹ $1000$, respectively, total funds available for these purposes are ₹ 80000 per week.

$$\begin{array}{lc} \therefore & 2000 x+1000 y \leq 80000 \\ \Rightarrow & 2 x+y \leq 80 \quad\text{.... (ii)}\\ \text { Also, } & x \geq 0, y \geq 0 \end{array}$$

Hence, the profits per unit for models $X$ and $Y$ are ₹ $1000$ and ₹ $500$, respectively.

$\therefore$ Required LPP is

Maximise $Z=1000 x+500 y$

Subject to, $3 x+5 y \leq 225,2 x+y \leq 80, x \geq 0, y \geq 0$

From the shaded feasible region, it is clear that coordinates of corner points are $( 0,0 ), (40,0),(25,30)$ and $(0,45)$.

On solving $3 x+5 y=225$ and $2 x+y=80$, we get

$$x=25, y=30$$

Corner points	Value of $Z=1000x+500y$
(0, 0)	0
(40, 0)	40000 $\leftarrow$ Maximum
(25, 30)	25000 + 15000 = 40000 $\leftarrow$ Maximum
(0, 45)	22500

So, the manufacturer should produce 25 bikes of model $X$ and 30 bikes of model $Y$ to get a maximum profit of ₹ $40000$.

Since, in question it is asked that each model bikes should be produced.

Let the person takes $x$ units of tablet $X$ and $y$ units of tablet $Y$.

So, from the given information, we have

$$\begin{aligned} & 6 x+2 y \geq 18 \Rightarrow 3 x+y \geq 9 \quad\text{.... (i)}\\ & 3 x+3 y \geq 21 \Rightarrow x+y \geq 7 \quad\text{.... (ii)}\\ & 2 x+4 y \geq 16 \Rightarrow x+2 y \geq 8 \quad\text{.... (iii)} \end{aligned}$$

Also, we know that here, $x \geq 0, y \geq 0\quad\text{.... (iv)}$

The price of each tablet of $X$ and $Y$ is ₹ $2$ and ₹ $1$, respectively.

So, the corresponding LPP is minimise $Z=2 x+y$, subject to $3 x+y \geq 9, x+y \geq 7$, $x+2 y \geq 8, x \geq 0, y \geq 0$

From the shaded graph, we see that for the shown unbounded region, we have coordinates of corner points $A, B, C$ and $D$ as $(8,0),(6,1),(1,6)$, and $(0,9)$, respectively.

[on solving $x+2 y=8$ and $x+y=7$, we get $x=6, y=1$ and on solving $3 x+y=9$ and $x+y=7$, we get $x=1, y=6]$

Corner points	Value of $Z=2x+y$
(8, 0)	16
(6, 1)	13
(1, 6)	8 $\leftarrow$ Minimum
(0, 9)	9

Thus, we see that 8 is the minimum value of $Z$ at the corner point $(1,6)$. Here, we see that the feasible region is unbounded. Therefore, 8 may or may not be the minimum value of $Z$. To decide this issue, we graph the inequality

$$2 x+y<8\quad\text{.... (v)}$$

and check whether the resulting open half has points in common with feasible region or not. If it has common point, then 8 will not be the minimum value of $Z$, otherwise 8 will be the minimum value of $Z$.

Thus, from the graph it is clear that, it has no common point.

Therefore, $Z=2 x+y$ has 8 as minimum value subject to the given constraints.

Hence, the person should take 1 unit of $X$ tablet and 6 units of $Y$ tablets to satisfy the given requirements and at the minimum cost of ₹ $8$.