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 $3 \frac{10}{11}$.
A manufacturer produces two models of bikes-model $X$ and model $Y$. Model $X$ takes a 6 man-hours to make per unit, while model $Y$ takes 10 man hours per unit. There is a total of 450 man-hour available per week. Handling and marketing costs are ₹ 2000 and ₹ 1000 per unit for models $X$ and $Y$, respectively. The total funds available for these purposes are ₹ 80000 per week. Profits per unit for models $X$ and $Y$ are ₹ $1000$ and ₹ 500 , respectively. How many bikes of each model should the manufacturer produce, so as to yield a maximum profit? Find the maximum profit.
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.
In order to supplement daily diet, a person wishes to take some $X$ and some wishes $Y$ tablets. The contents of iron, calcium and vitamins in $X$ and $Y$ (in mg/tablet) are given as below
| Tablets | Iron | Calcium | Vitamin |
|---|---|---|---|
| X | 6 | 3 | 2 |
| Y | 2 | 3 | 4 |
The person needs atleast 18 mg of iron, 21 mg of calcium and 16 mg of vitamins. The price of each tablet of $X$ and $Y$ is ₹ $2$ and ₹1, respectively. How many tablets of each should the person take in order to satisfy the above requirement at the minimum cost?
Let the person takes $x$ units of tablet $X$ and $y$ units of tablet $Y$.
So, from the given information, we have
$$\begin{aligned} & 6 x+2 y \geq 18 \Rightarrow 3 x+y \geq 9 \quad\text{.... (i)}\\ & 3 x+3 y \geq 21 \Rightarrow x+y \geq 7 \quad\text{.... (ii)}\\ & 2 x+4 y \geq 16 \Rightarrow x+2 y \geq 8 \quad\text{.... (iii)} \end{aligned}$$
Also, we know that here, $x \geq 0, y \geq 0\quad\text{.... (iv)}$
The price of each tablet of $X$ and $Y$ is ₹ $2$ and ₹ $1$, respectively.
So, the corresponding LPP is minimise $Z=2 x+y$, subject to $3 x+y \geq 9, x+y \geq 7$, $x+2 y \geq 8, x \geq 0, y \geq 0$
From the shaded graph, we see that for the shown unbounded region, we have coordinates of corner points $A, B, C$ and $D$ as $(8,0),(6,1),(1,6)$, and $(0,9)$, respectively.
[on solving $x+2 y=8$ and $x+y=7$, we get $x=6, y=1$ and on solving $3 x+y=9$ and $x+y=7$, we get $x=1, y=6]$
| Corner points | Value of $Z=2x+y$ |
|---|---|
| (8, 0) | 16 |
| (6, 1) | 13 |
| (1, 6) | 8 $\leftarrow$ Minimum |
| (0, 9) | 9 |
Thus, we see that 8 is the minimum value of $Z$ at the corner point $(1,6)$. Here, we see that the feasible region is unbounded. Therefore, 8 may or may not be the minimum value of $Z$. To decide this issue, we graph the inequality
$$2 x+y<8\quad\text{.... (v)}$$
and check whether the resulting open half has points in common with feasible region or not. If it has common point, then 8 will not be the minimum value of $Z$, otherwise 8 will be the minimum value of $Z$.
Thus, from the graph it is clear that, it has no common point.
Therefore, $Z=2 x+y$ has 8 as minimum value subject to the given constraints.
Hence, the person should take 1 unit of $X$ tablet and 6 units of $Y$ tablets to satisfy the given requirements and at the minimum cost of ₹ $8$.
A company makes 3 model of calculators; $A, B$ and $C$ at factory $I$ and factory II. The company has orders for atleast 6400 calculators of model $A, 4000$ calculators of model $B$ and 4800 calculators of model $C$. At factory I, 50 calculators of model $A, 50$ of model $B$ and 30 of model $C$ are made everyday; at factory II, 40 calculators of model $A, 20$ of model $B$ and 40 of model $C$ are made everyday. It costs ₹ 12000 and ₹ 15000 each day to operate factory I and II, respectively. Find the number of days each factory should operate to minimise the operating costs and still meet the demand.
Let the factory I operate for $x$ days and the factory II operate for $y$ days.
At factory I, 50 calculators of model A and at factory II, 40 calculators of model $A$ are made everyday. Also, company has ordered for atleast 6400 calculators of model $A$.
$$\therefore \quad 50 x+40 y \geq 6400 \Rightarrow 5 x+4 y \geq 640\quad\text{.... (i)}$$
Also, at factory I, 50 calculators of model $B$ and at factory II, 20 calculators of modal $B$ are made everyday.
Since, the company has ordered atleast 4000 calculators of model $B$.
$\therefore \quad 50 x+20 y \geq 4000 \Rightarrow 5 x+2 y \geq 400\quad\text{.... (ii)}$
$$ \begin{array}{lc} \text { Similarly, for model C, } & 30 x+40 y \geq 4800 \\ \Rightarrow & 3 x+4 y \geq 480 \quad\text{.... (iii)}\\ \text { Also, } & x \geq 0, y \geq 0\quad\text{.... (iv)} \end{array}$$
[since, $x$ and $y$ are non-negative] It costs ₹ 12000 and ₹ $15000$ each day to operate factories I and II, respectively.
$\therefore$ Corresponding LPP is,
Minimise $Z=12000 x+15000 y$, subject to
$$\begin{gathered} 5 x+4 y \geq 640 \\ 5 x+2 y \geq 400 \\ 3 x+4 y \geq 480 \\ x \geq 0, y \geq 0 \end{gathered}$$
On solving $3 x+4 y=480$ and $5 x+4 y=640$, we get $x=80, y=60$.
On solving $5 x+4 y=640$ and $5 x+2 y=400$, we get $x=32, y=120$
Thus, from the graph, it is clear that feasible region is unbounded and the coordinates of corner points $A, B, C$ and $D$ are $(160,0),(80,60),(32,120)$ and $(0,200)$, respectively.
| Corner points | Value of $Z=12000x+15000y$ |
|---|---|
| (160, 0) | $160 \times 12000=1920000$ |
| (80, 60) | $(80 \times 12+60 \times 15) \times 1000=1860000 \leftarrow$ Minimum |
| (32, 120) | $(32 \times 12+120 \times 15) \times 1000=2184000$ |
| (0, 200) | $0+200 \times 15000=3000000$ |
From the above table, it is clear that for given unbounded region the minimum value of $Z$ may or may not be 1860000.
Now, for deciding this, we graph the inequality
$$\begin{gathered} 12000 x+15000 y<1860000 \\ 4 x+5 y<620 \end{gathered}$$
and check whether the resulting open half plane has points in common with feasible region or not.
Thus, as shown in the figure, it has no common points so, $Z=12000 x+15000 y$ has minimum value 1860000 .
So, number of days factory I should be operated is 80 and number of days factory II should be operated is 60 for the minimum cost and satisfying the given constraints.