Refer to question 12 . What will be the minimum cost?
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 $\left(\frac{30}{7}, \frac{30}{7}\right)$.
| Corner points | Corresponding value of $Z=400x+200y$ |
|---|---|
| (0, 15) | 3000 |
| (5, 5) | 3000 |
| $\left(\frac{30}{7}, \frac{30}{7}\right)$ | $\begin{gathered}400 \times \frac{30}{7}+200 \times \frac{30}{7}=\frac{18000}{7} \\ =2571.43 \leftarrow \text { Minimum }\end{gathered}$ |
Hence, the minimum cost is ₹2571.43.
Refer to question 13 . Solve the linear programming problem and determine the maximum profit to the manufacturer.
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.
Refer to question 14 . How many sweaters of each type should the company make in a day to get a maximum profit? What is the maximum profit?
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 = ₹ $48000$.
Refer to question 15. Determine the maximum distance that the man can travel.
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 $54 \frac{2}{7} \mathrm{~km}$.
Maximise $Z=x+y$ subject to $x+4 y \leq 8,2 x+3 y \leq 12,3 x+y \leq 9$, $x \geq 0$ and $y \geq 0$.
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 $3 \frac{10}{11}$.