Referring to solution 13 , we have Maximise $Z=100 x+170 y$ subject to

$$3 x+2 y \leq 3600, x+4 y \leq 1800, x \geq 0, y \geq 0$$

From the shaded feasible region it is clear that the coordinates of corner points are $(0,0)$, $(1200,0),(1080,180)$ and $(0,450)$.

On solving $x+4 y=1800$ and $3 x+2 y=3600$, we get $x=1080$ and $y=180$

Corner points	Corresponding value of $Z=100x+170y$
(0, 0)	0
(1200, 0)	$1200 \times 100=120000$
(1080, 180)	$100 \times 1080+170 \times 180=138600 \leftarrow$ Maximum
(0, 450)	$0+170\times450=76500$

Hence, the maximum profit to the manufacturer is 138600.

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.