A firm has to transport 1200 packages using large vans which can carry 200 packages each and small vans which can take 80 packages each. The cost for engaging each large van is ₹ 400 and each small van is ₹ 200. Not more than ₹ $3000$ is to be spent on the job and the number of large vans cannot exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimise cost.
Let the firm has $x$ number of large vans and $y$ number of small vans. From the given information, we have following corresponding constraint table.
| Large vans ($x$) |
Small vans ($y$) | Maximum / Minimum | |
|---|---|---|---|
| Packages | 200 | 80 | 1200 |
| Cost | 400 | 200 | 3000 |
Thus, we see that objective function for minimum cost is $Z=400 x+200 y$. Subject to constraints
$$\begin{aligned} & 200 x+80 y \geq 1200 \quad \text { [package constraint] }\\ \Rightarrow \quad & 5 x+2 y \geq 30 \quad\text{.... (i)}\\ \text { and } \quad & 400 x+200 y \leq 3000 \quad \text { [cost constraint] }\\ \Rightarrow \quad & 2 x+y \leq 15 \quad\text{.... (ii)}\\ \text { and } \quad & x \leq y \text { [van constraint] ... (iii)}\\ \text { and } \quad & x \geq 0, y \geq 0 \text { [non-negative constraints] .... (iv) } \end{aligned}$$
Thus, required LPP to minimise cost is minimise $Z=400 x+200 y$, subject to $5 x+2 y \geq 30$.
$$\begin{aligned} 2 x+y & \leq 15 \\ x & \leq y \\ x & \geq 0, y \geq 0 \end{aligned}$$
A company manufactures two types of screws $A$ and $B$. All the screws have to pass through a threading machine and a slotting machine. A box of type $A$ screws requires 2 min on the threading machine and 3 $\min$ on the slotting machine. A box of type $B$ screws requires 8 min on the threading machine and 2 min on the slotting machine. In a week, each machine is available for 60 h . On selling these screws, the company gets a profit of ₹ 100 per box on type $A$ screws and ₹ 170 per box on type $B$ screws. Formulate this problem as a LPP given that the objective is to maximise profit.
Let the company manufactures $x$ boxes of type $A$ screws and $y$ boxes of type $B$ screws. From the given information, we have following corresponding constraint table
| Type A ($x$) |
Type B ($y$) | Maximum time available on each machine in a week | |
|---|---|---|---|
| Time required for screws on threading machine | 2 | 8 | 60 $\times$ 60 (min) |
| Time required for screws on slotting machine | 3 | 2 | 60 $\times$ 60 (min) |
| Profit | ₹ 100 | ₹ 170 |
Thus, we see that objective function for maximum profit is $Z=100 x+170 y$.
Subject to constraints
$2 x+8 y \leq 60 \times 60$ [time constraint for threading machine]
$$\Rightarrow \quad x+4 y \leq 1800\quad\text{.... (i)}$$
$$\begin{array}{lr} \text { and } & 3 x+2 y \leq 60 \times 60 \quad \text{ [time constraint for slotting machine]}\\ \Rightarrow & 3 x+2 y \leq 3600 \quad\text{.... (ii)}\\ \text { Also, } & x \geq 0, y \geq 0\quad\text{[non-negative constraints] .... (iii)} \end{array}$$
$\therefore$ Required LPP is,
Maximise $\quad Z=100 x+170 y$
Subject to constraints $x+4 y \leq 1800,3 x+2 y \leq 3600, x \geq 0, y \geq 0$.
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.