A company manufactures two types of sweaters type $A$ and type $B$. It costs ₹ 360 to make a type $A$ sweater and ₹ $120$ to make a type $B$ sweater. The company can make atmost 300 sweaters and spend atmost ₹ $72000$ a day. The number of sweaters of type $B$ cannot exceed the number of sweaters of type $A$ by more than 100 . The company makes a profit of ₹ 200 for each sweater of type A and ₹ 120 for every sweater of type $B$. Formulate this problem as a LPP to maximise the profit to the company.
Let the company manufactures $x$ number of type $A$ sweaters and $y$ number of type $B$ sweaters.
From the given information we see that cost to make a type A sweater is ₹ $360$ and cost to make a type $B$ sweater is ₹ $120$. Also, the company spend atmost ₹ $72000$ a day.
$$\therefore \quad 360 x+120 y \leq 72000$$
$$\Rightarrow \quad 3 x+y \leq 600\quad\text{..... (i)}$$
Also, company can make atmost 300 sweaters.
$$\therefore x+y \leq 300\quad\text{.... (ii)}$$
Further, the number of sweaters of type $B$ cannot exceed the number of sweaters of type $A$ by more than 100 i.e.,
$$\begin{aligned} x+100 & \geq y \\ \Rightarrow x-y & \geq-100\quad\text{.... (iii)} \end{aligned}$$
Also, we have non-negative constraints for $x$ and yi.e., $x \geq 0, y \geq 0$ .... (iv)
Hence, the company makes a profit of ₹200 for each sweater of type $A$ and ₹ $120$ for each sweater of type B i.e.,
$$\operatorname{Profit}(Z)=200 x+120 y$$
Thus, the required LPP to maximise the profit is
Maximise $Z=200 x+120 y$ is subject to constraints.
$$\begin{aligned} 3 x+y & \leq 600 \\ x+y & \leq 300 \\ x-y & \geq-100 \\ x \geq 0, y & \geq 0 \end{aligned}$$
A man rides his motorcycle at the speed of $50 \mathrm{~km} / \mathrm{h}$. He has to spend ₹ 2 per km on petrol. If he rides it at a faster speed of $80 \mathrm{~km} / \mathrm{h}$, the petrol cost increases to ₹ 3 per km. He has atmost ₹ 120 to spend on petrol and one hour's time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.
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 $2 x+3 y \leq 120,8 x+5 y \leq 400, x \geq 0, y \geq 0$
Refer to question 11. How many of circuits of type $A$ and of type $B$, should be produced by the manufacturer, so as to maximise his profit? Determine the maximum profit.
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 $x=6$ ]
| Corner points | Corresponding value of $Z=50x+60y$ |
|---|---|
| (0, 0) | 0 |
| (10, 0) | 500 |
| $\left(\frac{28}{3}, \frac{4}{3}\right)$ | $\frac{1400}{3}+\frac{240}{3}=\frac{1640}{3}=546.66 \leftarrow$ Maximum |
| (6, 3) | 480 |
| (0, 5) | 300 |
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.
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.