Let the man rides to his motorcycle to a distance $x \mathrm{~km}$ at the speed of $50 \mathrm{~km} / \mathrm{h}$ and to a distance $y \mathrm{~km}$ at the speed of $80 \mathrm{~km} / \mathrm{h}$.

Therefore, cost on petrol is $2 x+3 y$.

Since, he has to spend ₹ $120$ atmost on petrol.

$$\therefore \quad 2 x+3 y \leq 120\quad\text{.... (i)}$$

Also, he has atmost one hour's time.

$$\begin{array}{ll} \therefore & \frac{x}{50}+\frac{y}{80} \leq 1 \\ \Rightarrow & 8 x+5 y \leq 400 \quad\text{.... (ii)} \end{array}$$

Also, we have $x \geq 0, y \geq 0$ [non-negative constraints]

Thus, required LPP to travel maximum distance by him is Maximise $Z=x+y$, subject to $2 x+3 y \leq 120,8 x+5 y \leq 400, x \geq 0, y \geq 0$

Referring to solution 11 , we have

Maximise $Z=50 x+60 y$, subject to

$$2 x+y \leq 20, x+2 y \leq 12, x+3 y \leq 15, x \geq 0, y \geq 0$$

From the shaded region it is clear that the feasible region determined by the system of constraints is $O A B C D$ and is bounded and the coordinates of corner points are $(0,0)$, $(10,0),\left(\frac{28}{3}, \frac{4}{3}\right),(6,3)$ and $(0,5)$, respectively.

[since, $x+2 y=12$ and $2 x+y=20 \Rightarrow x=\frac{28}{3}, y=\frac{4}{3}$ and $x+3 y=15$ and $x+2 y=12 \Rightarrow y=3$ and $x=6$ ]

Since, the manufacturer is required to produce two types of circuits $A$ and $B$ and it is clear that parts of resistor, transistor and capacitor cannot be in fraction, so the required maximum profit is 480 where circuits of type $A$ is 6 and circuits of type $B$ is 3.

Referring to solution 12 , we have minimise $Z=400 x+200 y$, subject to $5 x+2 y \geq 30$,

$$2 x+y \leq 15, x \leq y, x \geq 0, y \geq 0$$

On solving $x-y=0$ and $5 x+2 y=30$, we get

$$y=\frac{30}{7}, x=\frac{30}{7}$$

On solving $x-y=0$ and $2 x+y=15$, we get $x=5, y=5$

So, from the shaded feasible region it is clear that coordinates of corner points are $(0,15)$, $(5,5)$ and $\left(\frac{30}{7}, \frac{30}{7}\right)$.

Hence, the minimum cost is ₹2571.43.

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 $y=180$

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 $(0,100)$.

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