RD Sharma Class 12 Exercise 29.4 Linear Programming Solutions Maths

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RD Sharma Class 12 Solutions Chapter29Linear Programming - Other Exercise

Linear Programming Excercise:29.4

Linner Programming Exercise 29.4 Question 1

Answer:
Distance travelled at speed of $25 Km/hr=\frac{50}{3} Km$ and at a speed of $40 Km/hr=\frac{40}{3}Km$
Hint:
$Time\: taken=\frac{Distance}{Speed}$
Given:
Man spend $Rs\; 2/ Km \; on \; p\left | x_{0} \right |$ when speed is 23km/hr and Rs 5/km when speed Rs 50 Km/hr.
Solution:
Let he drives x Km at sped of 25 Km/hr. and y Km/hr. at a speed of 40 Km/hr.
Let Z be total distance travelled by him so,
$z=x+y$
Since he spend Rs 2 per Km on petrol when speed is 25 Km/hr. and Rs5 per Km on petrol when speed is 40 Km/hr, So, expense on x Km and y Km or Rs2x and Rs2y respectively. But he has only Rs100, 50.
2x+5y≤100 (1st constraint)
$Time\; taken\; to \; travel\; x\; Km=\frac{Distance}{Speed}$
$= \frac{x}{25}hr$
Time taken to travel $y\: Km=\frac{y}{40}hr$
Given he has 1 hr. to travel, so,
$\begin{aligned} &\frac{x}{25}+\frac{y}{40} \leq 1 \\ & \end{aligned}$
$\Rightarrow 40 x+25 y \leq 1000 \\$
$\Rightarrow 8 x+5 y \leq 200(\text { second constraint })$
Hence, mathematical formulation of LPP is find x and y which
Maximum $z=x+y$
Subject to constraint
$\begin{aligned} &2 x+5 y \leq 100 \\ & \end{aligned}$
$8 x+5 y \leq 200 \\$
$x, y \geq 0$ since distance cannot be less than zero
The corresponding line is
$\begin{aligned} &\Rightarrow 2 x+5 y \leq 100 \\ & \end{aligned}$
$\Rightarrow \frac{x}{50}+\frac{y}{20}=1$ ....(i)
$And\: 8 x+5 y=200$
$\Rightarrow \frac{x}{25}+\frac{y}{40}=1$ …(ii)
Point $P\left(\frac{50}{3}, \frac{40}{3}\right)$ is obtain by solving
$\begin{aligned} &8 x+5 y=200 \\ &2 x+5 y=100 \end{aligned}$

Points	$z=x+y$
$O\left ( 0,0 \right )$	$0$
$A _{2}\left ( 25,0 \right )$	$25$
$P\left(\frac{50}{3}, \frac{40}{3}\right)$	$30$
$B_{1}\left ( 0,20 \right )$	$20$

And $z=30\: at \: z=\frac{50}{3}, z=\frac{40}{3}$
Distance travelled at speed of $25 Km/hr=\frac{50}{3} Km$ and at a speed of $40 Km/hr=\frac{40}{3}Km$
Maximum distance $=30 Km$

Linear Programming Exercise 29.4 Question 2

Answer:
Maximum Profit = Rs 40 (A = 4, B = 4)
Hint:
Let $x_{1}$ and y are required quantity of x and y.
Given:
He makes a profit of ?6.00 on item A and ?4.00 on item B.
Solution:
So, profits of one X item =s of type A and Y item of type B are 6x and Rs 4y respectively.
Maximum $z=6x+4y$ (Where x and y are required quantity)
Given machine I works 1 hour and 2 hours on item A and B respectively., so x number of item A and y number of item B need x hour and 2y hours on machine I respectively, but machine I works at most 12 hours so,
$x+2y\leq 12$ ( first constraint)
Given, machine II works 2 hours and 1 hour on item A and B respectively, but machine II works maximum 12 hours, So
$2x+y\leq 12$ (second constraint)
Given, machine III works 1 hour and $\frac{5}{4}$ hour on one item A and B respectively, but machine III works maximum 5 hours, So

$x+54y\geq 5$
$\Rightarrow 4x+5y\geq 20$ (third constraint)
The required LPP
$z=6x+4y$
Subject to constraint
$\begin{aligned} &x+2 y \leq 12 \\ & \end{aligned}$
$2 x+y 1 \leq 2 \\$
$4 x+5 y \geq 20$
$x, y\geq 0$ since the number of item A and B not be less than zero
The corresponding line is
$\begin{aligned} &x+2 y=12 \Rightarrow \frac{x}{12}+\frac{y}{6}=1 \\ & \end{aligned}$
$2 x+y=12 \Rightarrow \frac{x}{6}+\frac{y}{12}=1 \\$
$4 x+5 y=20 \Rightarrow \frac{x}{5}+\frac{y}{4}=1$

Shaded region $A_{2}A_{3}PB_{3}B_{1}$ represent feasible region

Points	$z=6x+4y$
$A_{2}\left ( 6,0 \right )$	$36$
$A_{3}\left ( 5,0 \right )$	$\\30$
$B_{3}\left (0 ,4 \right )$	$16$
$B_{2}\left (0 ,6 \right )$	$\\24$
$P\left ( 4,4 \right )$	$40$

Hence z is maximum at x = 4 and y = 4
Required number of product A = 4, B = 4
Maximum profit=Rs 40

Linear Programming Exercise 29.4 Question 3

Answer:
A should work for 5 days and B should for 3 days.
Hint:
Let tailor A and B work for x and y days respectively.
Given:
Tailor A and B earn Rs15 and Rs20 respectively.
Solution:
Min $z =15x+20y$
Since tailor A and B stitch 6 and 10 shirts respectively in a day. But it is desired to produce 60 shirts at least. So,

$6x+10y\geq 60$
$\Rightarrow 3x+5y\geq 30$ (First constraint)
Since tailor A and B stitch 4pants per day but it is desired to produce at least 32 pants so
$4x+4y\geq 32$
$\Rightarrow x+y\geq 8$ (Second constraint)
Hence the required LPP .
Min $z =15x+20y$
Subject to constraints
$3x+5y\geq 30$
$x+y\geq 8$
$x,y\geq 0$ Since x and y not be less than zero.
The corresponding equation is

Point P(5,3) obtain by solving (i) and (ii)
Unbounded shaded region $A_{1}PB_{2}$ represents feasible region with corner points $A_{1}(10,0), P(5,3), B_{2}(0,8)$

Points	$z=15x+20y$
$A_{1}(10,0)$	$150$
$P\left ( 5,3 \right )$	$135$
$B_{2}\left ( 0,8 \right )$	$160$

Min z=135 at x=5 and y=3
Tailor A should work for 5 days and B should for 3 days.

Linear Programming exercise 29.4 question 4

Answer:
The factory produces 30 and 20 packages of screw A and screw B to get the maximum profit.
Hint:
let factory manufacture x screws of type A and y screw of type B.
Given:
The information can be competed in table follow

	Screw A	Screw B	Availability
Automatic machine (min)	$4$	$6$	$4\times 60 = 120$
Hand operated machine (min)	$6$	$3$	$4\times 60 = 120$

Solution:
Let the factory manufacture x screws of type A and y screws of type B on each day.
$x\geq 0, y\geq 0$
The profit on a package of screws A is Rs 7 and on the package of screws B is Rs 10.
The constraints are

$4x+6y\leq 240$
$6x+3y\leq 240$
Total profit $z=7x+10y$
The required LPP
Max $z=0.7x+y$
Subject to constraints

$4x+6y\leq 240$ ....(i)
$6x+3y\leq 24$ ....(ii)
$x, y\geq 0$
The feasible region determined by the system of constraints is

Points (30,20) solution obtain by solving (i) and (ii). The corner points A(40,0), B(30,20) and C(0,40).

Points	$z=0.7x+y$
$A(40,0)$	$28$
$B\left ( 30,20 \right )$	$41$
$C\left (0,40 \right )$	$40$

The maximum value of z=41 at (30,20).
Thus, the factory should produce 30 packages of screw A and 20 packages of screw B to get the maximum profit of Rs41.

Linear Programming exercise 29.4 question 5

Answer:
Required number belt A is 200, while B is 600, maximum profit Rs1300.
Hint:
Let required number of belt A and B be x and y.
Given:
Profit on belt A and B be Rs2 and Rs1.50
Solution:
Let z be total profit
$z =2x+1.5y$
Where x and y be required number of belt A and belt B.
Since each belt of type A required twice as much time as belt B. let each belt B require to make, so A requires 2 hours but total time available is equal to production 1000 belt B that is 1000 hours, so
$2x+y\leq 1000$ (first constraint)
Given supply of lather only for 800 belts per day both A and B combined, so
$x+y\leq 800$ (second constraint)
Buckets available for A is only 400 and for B only 700, so
$x\leq 400$ (third constraint)
$y\leq 700$ (forth constraint)
Hence the required LPP
$z =2x+1.5y$
Subject to constraints

$2x+y\leq 1000\\$ ....(i)
$x+y\leq 800\\$ ....(ii)
$x\leq 400$ .....(iii)
$y\leq 700$ ...(iv)
$x,y\geq 0$ ,number of belt cannot less than zero
The feasible region determined by the system of constraints

Obtain point Q(200,600) by solving (i) and (ii) and P(400,200) by solving (i) and (iii) and R(100,700) by solving (ii) and (iv)
The shading region is $OA_{3}PQRB_{2}$

Corner Points	Value of $z=2x+1.5y$
$O\left ( 0,0 \right )$	0
$A_{3}\left (400,0 \right )$	$600$
$P(400,200)$	$1100$
$Q\left (200,600 \right )$	$1300$
$R\left (0,700 \right )$	$1050$

Maximum profit = 1300, required number belt A = 200, belt B = 600

Linear Programming exercise 29.4 question 6

Answer:
Maximum profit =350
Deluxe model = 10
Ordinary model = 20
Hint:
Let required number of deluxe and ordinary model be x and y.
Given:
Since, Profit on each model of deluxe and ordinary type of Rs15 and Rs10 respectively.
Solution:
Let z be total profit then
$z=15x+10y$
Where x and y are required deluxe and ordinary model
Since each deluxe and ordinary model required 2 and 1 hour of skilled men, but twice available skilled men is 5×8 = 4 hours so,
$2x+y\leq 40$ (first constraint)
Given each deluxe and ordinary model require 2 and 3 hour of semi-skilled men, but total ratable by semi -skilled men is 100×8 = 80 hours so
$2x+3y\leq 80$ (second constraint)
Hence the required LPP
$z =15x+10y$
Subject to constraints

$2x+y\leq 40$
$2x+3y\leq 80$
$x,y\geq 0$ ,since number of ordinary model cannot less than zero
The feasible region obtain by the system of constraint

Point (10,20) obtain by solving (i) and (ii).
The corner point of feasible region is $O(0,0), A_{1}(20,0), P(0,20), B_{2}(0,40)$

Corner Points	Value of $z=15x+10y$
$O\left ( 0,0 \right )$	$0$
$A_{1}(20,0)$	$300$
$P(10,20)$	$350$
$B_{2}(0,40/3)$	$8003$

Maximum z = 350 at x = 10, y = 20
Required number of deluxe model = 10 and required number of ordinary model = 20
Maximum profit = 350

Linear Programming Exercise 29.4 Question 7

Answer:
15 tea cups of type A and 30 tea cups of type B.
Hint:
Let required number of tea cups of type A and B are x and y.
Given:
Profit on each tea cups of type A and B are 75 paisa and 50 paisa.
Solution:
Let the total profit of on tea cups be z.
$z=75x+50y$
Where x and y are the required number of tea cups.
Since each tea cup of type A and B require the work machine 1 for 12 and 6 min but total time available on machine I is 6×60 = 360min

$12x+6y\leq 360$
$\Rightarrow 2x+6y\leq 60$
Since each tea cup of type A and B require to work machine II for 6 and 0 min but total time available for machine II is 6×60=360min.

$18x+y\leq 360$
$x\leq 20$
Since each tea cup of type A and B require the work machine III for 6 and 9 min. but total time available for machine III is 6×60=360min.

$6x+9y\leq 360$
$\Rightarrow 2x+3y\leq 120$
The required LPP is
Max $z=75x+50y$
Subject to constraints

$2x+6y\leq 60$ ...(i)
$x\leq 20$ ....(ii)
$2x+3y\leq 120$ ...(iii)
$x,y\geq 0,$
The feasible region obtains by the system of constraint

Point (20,20) and Q(15,30) is obtain by solving (ii) and (iii) and (i) and (iii). $OA_{1}PQB_{3}.$

Corner Points	Value of $z=2x+1.5y$
$O\left ( 0,0 \right )$	$0$
$A_{1}(20,0)$	$1500$
$P(20,20)$	$2500$
$Q(15,30)$	$2624$
$B_{2}(0,40)$	$2000$

Hence z is maximum at x=15, y=30.
15 tea cups of type A and 30 tea cups of type B, we need to maximize the profit.

Linear Programming Exercise 29.4 Question 8

Answer:
Output is maximum when type A = 4, type B = 3 or type A = 6, type B = 0.
Hint:
Let required number of machine A and B are x and y.
Given:
Production of each machine A and B are 60 and 40 units daily.
Solution:
Let z donate total output daily, so
$z=60x+40y$
Since each machine of type A and type B require 100sq.m and 1200sq.m but total area available for machine is 76000sq.m

$1000x+1200y\leq 7600$
$5x+6y\leq 38$
Each machine of type A and B require 12 men and 8 men to work respectively. But total 72 men available for work so
$\begin{aligned} &12 x+8 y \leq 72 \\ & \end{aligned}$
$3 x+2 y \leq 18$
The required LPP is
Max $z=60x+40y$
Subject to constraints
$5x+6y\leq 38$
$3x+2y\leq 18$
$x,y\geq 0,$
The feasible region obtains by the system of constraint

P(4,3) is obtain by solving (i) and (ii).
$OA_{2}PB_{1}$ are the shaded region

Corner Points	Value of $z=60x+40y$
$O\left ( 0,0 \right )$	$0$
$A_{2}(6,0)$	$360$
$P(4,3)$	$360$
$B_{1}(0,193)$	$760/3$

Therefore, max z=360 at x=4, y=3 or x=6, y=20.
Output is maximum when 4 machine of type A and 3 machines of type B or 6 machine of type A and no machine of type B.

Linear Programming Exercise 29.4 Question 9

Answer:
Max profit =230 at type A=2 , type B=3.
Hint:
Let number of goods A and B are x and y respectively.
Given:
Profit of each A and B are Rs40 and Rs50
Solution:
Max $z=40x+50y$
Since each A and B require 3gm and 1gm of silver but total silver available are 9gm.
$3x+y\leq 9$
Since each A and B require 1gm and 2gm of gold but total gold available are 8gm.
$x+2y\leq 8$
The required LPP is
Max $z=40x+50y$
Subject to constraints
$\begin{aligned} &3 x+y \leq 9 \end{aligned}$
$x+2 y \leq 8 \\$
$x, y \geq 0,$
The feasible region obtains by the system of constraints

P(2,3) is obtain by solving (i) and (ii).
$OA_{1}PB_{2}$ are the shaded region

Corner Points	Value of $z=40x+50y$
$O\left (0,0 \right )$	$0$
$A_{1}(3,0)$	$120$
$P(2,3)$	$230$
$B_{2}(0,4)$	$200$

Therefore, max z=230 at x=2, y=3.
Hence maximum profit = 230, number of goods of type A =2, type B =3

Linear Programming exercise 29.4 question 10

Answer:
Maximum profit Rs22.2.
Hint:
Let daily production of chairs and table be x and y.
Given:
Profit on each chair and table are Rs3 and Rs5.
Solution:
Let 2 be total profit on table and chair
Max $z=3x+5y$ [when x and y are daily production on table and chair]
Since each chair and table require 2hrs and 4hrs on a machine A but maximum time available on machine A be 16 hrs.
$\begin{aligned} &2 x+4 y \leq 16 \\ & \end{aligned}$
$x+2 y \leq 8$
Since each chair and table require 6hrs and 2hrs on machine B, but maximum time available on machine B be 30 hrs.
$\begin{aligned} &6 x+2 y \leq 30 \\ & \end{aligned}$
$3 x+y \leq 15$
The required LPP is
Max $z=3x+5y$
Subject to constraints
$\begin{aligned} &x+2 y \leq 8 \ldots(i) \\ & \end{aligned}$
$3 x+y \leq 15 \ldots(i i)$
The feasible region obtains by the system of constraints

$P(\frac{22}{5},\frac{9}{5})$ obtain by solving equation (i) and (ii).
$OA_{2}PB_{1}$ are the shaded region

Corner Points	Value of $z=2x+1.5y$
$O\left ( 0,0\right )$	$0$
$A_{2}(5,0)$	$15$
$P(\frac{22}{5},\frac{9}{5})$	$22.2$
$B\left ( 10,4 \right )$	$20$

Hence maximum profit $=Rs 22.2 \left ( \frac{22}{5} ,\frac{9}{5}\right )$

Linear Programming exercise 29.4 question 11

Answer:
Maximum profit = Rs. 2025 could be obtained if 45 units of chairs and no units of table are produce.
Hint:
Use graph and simultaneous equation
Given:
Resources available 400square feet of teak wood and 450 man hours.
Solution:
Let required production of chairs and tables be z=x and y respectively.
Since, profits of each chair and table is Rs45 and Rs80 respectively
So, profit on x number of type A and y number of type B are 45x and 80y respectively.
Let z denotes total output daily so,
$z=45x+80y$
Since each chair and table require 5 sq. and 80sq.ft of wood respectively. So, x number of chair and y number of table require 5x and 80y sq. of wood respectively. But 400sq.ft of wood available
So,
.
$5x+80y\leq 400$
$x+4y\leq 80$ (first constraints)
Since, each chair ad table requires 10 and 25 man hours respectively, so, x number of chair and y number of tables are require 10x and 25y men hours respectively. But, only 450 hours are available.
So,
$10x+25y\leq 450$
$2x+5y\leq 90$ (second constraints)
Hence mathematical formulation of the given LPP is
Max $z=45x+80y$
Subject to constraints
$\begin{aligned} &x+4 y \leq 80 \ldots(i) \\ & \end{aligned}$
$2 x+5 y \leq 90 \ldots(i i)$
$\\x, y \geq 0$ [since production of chair and table can not be less than 0]
Region $x+4y\leq 80$ : line $x+4y=80$ meets the axes at A980,0), B(1,20) respectively.
Region containing the origin represents $x+4y\leq 80$ as origin satisfies $x+4y\leq 80$
Region $2x+5y\leq 90$ : line $2x+5y= 90$ meets the axes at C(45,0), D(0,20) respectively.
Region containing the origin represents $2x+5y\leq 90$ as origin satisfies $2x+5y\leq 90/$
Regionx, $y\geq 0$ : it represents the first quadrant.

The corner points are 0(0,0), D(1,18),C(45,0).
The value of z at these corner points are as follows.

Corner Points	Value of $z=2x+1.5y$
$(0,0)$	$0$
$(45,0)$	$2025$
$(0,18)$	$1440$

The maximum value of z is 2025 which is attained at C(45,0)
Thus, maximum profit of Rs2025 is obtained when 45 units of chairs and no units of tables are produced.

Linear Programming exercise 29.4 question 12

Answer:
$Max \: Profit \Rightarrow Rs. 500$ could be obtained when no units of product A and 125 units of product B are manufactured.
Hint:
Form Equation and solve graphically.
Given:
A firm manufactures two products A and B. Each product is processed on two machines. M1 and M2 . Product A requires 4 minutes of processing time on M1 and 8 min on M2 ; Product B requires 4 min on M1 and 4 min on M2
Solution:
Let required production of product A and B be x and y respectively.
Since profit on each product A and B are Rs. 3 and Rs. 4 respectively. So, profits on x number of type A and y number of type B are 3x and 4y respectively.
Let Z denotes total output daily, so,

$z=3x+4y$
Since, each A and B requires 4 minutes each on machine M1 . So, x of type A and y of type B require 4x and 4y minutes respectively, But total number available on machine M1 is 8 hours 20 minutes is equal to 500 minutes.
So,
$\begin{aligned} &4 x+4 y \leq 500 \\ &x+y \leq 125 \end{aligned}$ { first constraint}
Since each A and B requires 8 minutes and 4 minutes on machine M2 So , 2x of type A and y of type B require 8x and 4y minutes respectively. But
Total time available on Machine M1 is 10 hours = 600 minutes
So,
$\begin{aligned} &8 x+4 y \leq 600 \\ &2 x+y \leq 150 \end{aligned}$ {Second constraint}
Hence, mathematical formulation of the given L.P.P is,
Max $z = 3x + 4y$
Subject to constraints,
$\begin{aligned} &x+y \leq 125 \\ &2 x+y \leq 150 \end{aligned}$
$x,y\geq 0$ [Since production of chairs and tables cannot be less than 0]
Region: $x+4y\leq 125$ : Line $x+y=125$ meets the axes at $A(125,0) , B(0,125)$ respectively.
Region containing the origin represents $x+4y\leq 125$ as origin satisfies $x+4y\leq 125$
Region $2x+5y\leq 150$ : Line $2x+y=150$ meets the axes at $C(75,0), D(0,150)$ respectively.

Region containing the origin represents $2x+5y\leq 150$ as origin satisfies $2x+5y\leq 150$
Region $x,y\geq 0$ : it represents the first quadrant.

The corner points are 0(0,0),B(0,125), ?(25,100),C(75,0).
The value of z at these corner points are as follows.

Corner Points	$z=3x+4y$
$O$	$0$
$B$	$500$
$\epsilon$	$475$
$C$	$225$

The maximum value of z is 500 which is attained at B(0,125)
Thus, maximum profit is Rs500 obtained when no units of product A and 125 units of product B are manufactured.

Linear Programming Exercise 29.4 Question 13

Answer:
$Max\: Profit \Rightarrow Rs.2150$ when 10 units of item A and 65 units of item B are manufactured.
Hint:
Form Linear Equations and solve graphically.
Given:
A firm manufacturing two types of electric items A and B can make a profit of Rs.20 per unit of A and E 30 per unit of B. Each unit of A requires 3 motors and 4 transformers and each unit of B requires 2 motors and 4 transformers. The total supply of these per month is restricted to 210 motors and 300 transformers Type B is an export model requiring a voltage stabilizer which has a supply restricted to 65 units per month.
Solution:
Let x units of item A and y units of item B are manufactured.
Numbers of items cannot be negative.
Therefore, $x, y \geq 0$
The given information can be tabulated as follows:

Product	Motors	Transformers
$A(x)$	$3$	$4$
$B(y)$	$2$	$4$
Availability	$210$	$300$

Further it is given that type B is an export model, whose supply is restricted to 65 per month.
Therefore, the constraints are
$\begin{aligned} &3 x+2 y \leq 210 \\ &4 x+4 y \leq 300 \\ &y \leq 65 \end{aligned}$
A and B make profit of Rs20 and Rs30 per unit respectively.
Therefore, Profit gained from x units of item A and y units od Item B is Rs. 20 x and 30uy respectively.
$Max \: Profit \Rightarrow z=20x+30y$ which according to question is to be maximized
Thus, mathematical formulation of the given L.P.P is,
Max $z = 10x + 3y$
Subject to constraints,
$\begin{aligned} &3 x+2 y \leq 210 \\ &4 x+4 y \leq 300 \\ &Y \leq 65 \end{aligned}$

$x,y\geq 0$ ; Region represented by $3x+2y\leq \210$ .

The line $3x+2y\leq \210$ meets the axes at A(70,0) , B(0,105) respectively
Region containing the origin represents $3x+2y\leq \210$ as origin satisfies $3x+2y\leq \210$
Region $4x+4y< 300$ : The line $4x+4y< 300$ meets the axes at C75,0

, D0,75

respectively.
Region containing the origin represents $4x+4y< 300$ as origin satisfies $4x+4y< 300$
Y=65 is the line passing through the point ?(0,65) and is parallel to x axis
Region $x,y\geq 0:$ it represents the first quadrant.
Scale: On x-axis, 1 big division =20 units
On y-axis, 1 big division =20 units

The corner points are 0(0,0),?(0,65),G(10,65),F(60,15) and A(70,0).
The value of z at these corner points are as follows.

Corner Points	$z=20x+30y$
$O$	$0$
$\epsilon$	$1950$
$G$	$2150$
$F$	$1650$
$A$	$1400$

The maximum value of z is 2150 which is attained at G(10,65)
Thus, maximum profit is Rs 2150 obtained when 10 units of item A and 65 units of item B are manufactured.

Linear Programming Exercise 29.4 Question 14

Answer:
$Max\: Profit \Rightarrow Rs. 16$ when two units of first product and 4 units of second product were manufactured.
Hint:
Form Linear Equation and solve graphically.
Given:
A factory uses three different resources for the manufacture of two different products , 20 units of the resources A , 12 units of B and 16 units of C. 1 unit of the first Product requires 2, 2 and 4 units of the respective resources and 1 unit of the second product requires 4,2 and 0 units of respective resources. It is known that the first product gives a profit of 2 monetary units per unit and the second 3.
Solution:
Let number of product I and product II are x and y respectively.
Since profit on each product I and II requires 2 an 4 units of resources A:50 , x units of product I and y units of product II requires 2x and 4y minutes respectively. But maximum available quantity of resources A is 20 units.
So,

$\begin{aligned} &2 x+4 y \leq 20 \\ &x+2 y \leq 10 \end{aligned}$ { first constraint}
Since each I and II requires 2 and 2 units of resources B . So, x units of product I and y units of product II requires 2x and 2y minutes respectively. But maximum available quantity of resources A is 12 units
So ,
$\begin{aligned} &2 x+2 y \leq 12 \\ &x+y \leq 6 \end{aligned}$ {Second constraint}
Since each units of product I requires 4 units of resources C . It is not required product II. So x units of product I require 4x units of resource C . But maximum available quantity of resource C is 16 units.
So ,
$\begin{aligned} &4 x \leq 16 \\ &x \leq 4 \end{aligned}$ {Third constraint}
Hence mathematical formulation of the given L.P.P is,
Max $z = 2x + 3y$
Subject to constraints,
$\begin{aligned} &x+2 y \leq 10 \\ &x+y \leq 6 \\ &x \leq 4 \end{aligned}$

$x,y\geq 0$ [Since production of I AND II cannot be less than 0]

Region represented by $x+2y\leq 10$ The line $x+2y=10$ meets the axes at A(10,0) , B(0,5) respectively.
Region containing the origin represents $x+2y\leq 10$ as origin satisfies $x+2y\leq 10$
Region represented by $x+y\leq 6 :$ Line $x+y=6$ meets the axes at C(6,0), D(0,6) respectively.

Region containing the origin represents $x+y\leq 6$ as origin satisfies $x+y\leq 6$
Region $x,y\geq 0:$ it represents the first quadrant.

The corner points are 0(0,0),B(0,5), G(2,4),?(4,0).
The value of z at these corner points are as follows.

Corner Points	$z=3x+4y$
O	0
B	15
G	16
F	14
?	8

The maximum value of z is 16 which is attained at G(12,4)
Thus, maximum profit is 16 monetary units obtained when 2 units of the first product and 4 units of the second product were manufactured.

Linear Programming Exercise 29.4 Question 15

Answer:
$Max \: Profit \Rightarrow Rs. 3360$ which is obtained by selling 360 copies of hardcover edition and 600 copies paperback edition.
Hint:
Form Linear Equation and solve graphically.
Given:
A publisher sells a hard cover edition of a text 600K for Rs.72.00 and a paperback edition of the same text for Rs.240.00. Costs of the publisher are Rs.56.00 and Rs.28.00 per book respectively in addition to weekly costs of Rs.9600.00. Both types require 5 minutes of printing time, although hard cover requires 10 minutes binding time and the paperback requires only 2 minutes. Bothe the printing back and binding operations have 4800 min available each week.
Solution:
Let the sale of hard cover edition be h and that of paperback editions be t
S.P. of a hard cover edition of the textbook $= Rs. 72.$
S.P. of a paperback edition of the textbook $= Rs. 40.$
Cost to the publisher for a paperback edition $= Rs.28$
weakly cost to the publisher $=Rs.9600.$
Profit to bae maximized, $z=(72-56)h+(40-28)t-9600$

The corner points are 0(0,0),B(0,960), ?(360,600),F(480,0).
The value of z at these corner points are as follows.

Corner Points	$z$
O	-9600
B	1920
?	3360
F	-1920

The maximum value of z is 3360 which is attained at ? (360,600)
Thus, maximum profit is 3360 which is obtained by selling 360 copies of hardware edition and 600 copies paperback edition.

Linear Programmig Excercise 29.4 Question 16

Answer:
Codeine Quantity = 24(approx.) and least quantity of pill A=5.86 and B =0.27
Hint:
Form Linear Equation and solve graphically.
Given:
A firm manufactures headache pills in two sizes A and B. Size A contains 2 grains of aspirin, 5 grains of bicarbonate and 1 grain of codeine, Size B contains 1 grain of aspirin and 18 grains of bicarbonate and 66 grains od codeine. It has been found by users that it requires at least 12 grains of aspirin. 7.4 grains of bicarbonate and 24 grains of codeine for providing immediate effect.
Solution:
The above LPP can be presented in a table below,

	$Pill\: size\: A$	$Pill\: size\: B$
	$x$	$y$
Aspirin	$2x$	$1y$	$\geq 12$
Bicarbonate	$5x$	$8y$	$\geq 7.4$
Codeine	$1x$	$66y$	$\geq 24$
Relief	$x$	$y$	minimize

Hence mathematical formulation of the given L.P.P is,
Max z = x + y
Subject to constraints,
$\begin{aligned} &2 x+y \geq 12 \\ &5 x+3 y \geq 7.4 \\ &x+66 y \geq 24 \end{aligned}$

$x,y\geq 0$ [Since production cannot be less than 0]

The corner points are B(0,12), P(5.86,0.27),?(24,0).
The value of z at these corner points are as follows.

Corner Points	$z=x+y$
(0,12)	12
(24,0)	24
(5.86,0.27)	6.13

The maximum value of z is 6.13 but the region is unbounded so check whether $x+y<6.13$
Clearly it can be seen that it does not has any common region
So, $x=5.86,y=0.27$
This is the least quantity of Pill A and B
Codeine quantity $= x+66y=5.86+(660.27)=24(approx)$

Linear Programmig Excercise 29.4 Question 17

Answer:
$Min\: cost = Rs 254$ when 14 units of compound A and 33 unit compound B are produced.
Hint:
Form Linear Equation and solve graphically.
Given:
A chemical company produces two compounds A and B . The following table gives the units of ingredients C and D per kg of Compounds A and B as well as minimum requirement of C and D and costs per kg of A and B

	compound		Minimum
	A	B	requirement
Ingredient C	1	2	80
Ingredient D	3	1	75
Cost(inE) perkg	4	6

Solution:
Let required quantity of compound A and B are x and y kg. Since, cost of 1 kg of compound A and B are Rs.4 and Rs.6 per kg. So, Cost of x kg compound A and y kg of compound B are Rs.4x and Rs.6 respectively.
Let z be the total cost of compounds.
So, $z= 4x + 6y$
Since compound A and B contain 1 and 2 units of ingredient C per kg respectively , So x kg of Compound A and y kg of Compound B contain x and 2y units of ingredient C respectively but minimum requirement of ingredient C is 80 units.
So ,
$x+2y\geq 80$ {first constraint}
Since, compound A and B contain 3 and 1 units of ingredient D per kg respectively.
So x kg of compound A and y kg of compound B contain 3x and y units of ingredient D respectively but minimum requirement of ingredient C is 75 units
So ,
$3x+y\geq 75$ {Second constraint}
Hence mathematical formulation of the given L.P.P is,
Min $z = 4x + 6 y$
Subject to constraints,
$\begin{aligned} &x+2 y \geq 80 \\ &5 x+y \geq 75 \\ & \end{aligned}$
$x, y \geq 0$ [Since production cannot be less than 0]
Region represented by $x+2y\geq 80$ : The line $x+2y=80$ meets the axes at A(80,0) , B(0,40) respectively.
Region not containing represents $x+2y=80$ as (0,0) does not satisfy satisfies $x+2y=80$
Region $3x-1y\geq 75$ : line $3x+y=75$ meets axes at $C(25,0),D(0,7)$ respectively.
Region not containing origin represents $3x-1y\geq 75$ as (0,0)

does not satisfy $3x-1y\geq 75$ .
Region $x,y\geq 0:$ it represents first quadrant.

The corner points are $D(0,75),\epsilon (14,33),A(80,0)$ .
The value of z at these corner points are as follows.

Corner Points	$z=4x+6y$
D	450
ε	254
A	320

The minimum value of z is 254 which is attained at $\epsilon (14,33)$
Thus, the minimum cost is Rs. 254 obtained when 14 units of compound A and 33 units compound B are produced.

Linear Programmig Excercise 29.4 Question 18

Answer:
Max Profit = Rs.16 when 8 souvenirs of Type A and 20 Souvenirs of Type B is produced.
Hint:
Form Linear Equation and solve graphically.
Given:
A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 min each for cutting and 10 min each for assembling. Souvenirs of type B require 8 min each for cutting and 8 min each for assembling. There are 3 hours 20 min available for cutting and 4 hours available for assembling. The profit is 50 paise each of type A and 60 paise each of type B souvenirs.
Solution:
Let the company manufacture x souvenirs of Type A and y souvenirs of Type B
Therefore,
$x\geq 0,y\geq 0$
The given information can be compiled in a table as follows:

	Type A	Type B	Availability
Cutting(min)	5	8	3×60+20=200
Assembling (min)	10	8	4×60=240

The profit on type A souvenirs is 50 paise and on Type B souvenirs is 60 paise. Therefore, profit gained on x souvenirs of type A and y souvenirs of type B is Rs.0.50 x and Rs. 0.60 y respectively
Total Profit, $z= 50x+60y$
The mathematical formulation of the given problem is,
max: $z= 50x+60y$ , Subject x constraint,
$\begin{aligned} &5 x+8 y \leq 200 \\ &10 x+8 y \leq 240 \\ &x \geq 0, y \geq 0 \end{aligned}$

Region $5x+8y \leq 200$ : The line $5x+8y=200$ meets the axes at A(40,0) , B(0,25) respectively.

Region containing origin represents the solution of the in equation $5x+8y \leq 200$ as (0,0) satisfy satisfies $5x+8y \leq 200$
Region $10x+8y\leq 240$ : line $10x+8y=240$ meets axes at $C(24,0),D(0,30)$ respectively.
Region containing origin represents the solution of in equation $10x+8y\leq 240$ as (0,0) satisfies $10x+8y\leq 240$
Region $x,y\geq 0$ : it represents first quadrant.

The corner points are $O(0,0),B(0,25),\epsilon (8,20),C(24,0)$
The value of z at these corner points are as follows.

Corner Points	$z=50x+60y$
O	0
B	1500
$\epsilon$	1600
C	1200

Thus, 8 souvenirs of Type A and 20 souvenirs of Type B should be produced each day to get the maximum profit of Rs16

Linear Programming Exercise 29.4 Question 19

Answer:
Max Profit = Rs1,440 when 48 units of product A and 16 units of product B are manufactured
Hint:
Form Linear Equation and solve graphically.
Given:
A manufacturer makes two products A and B. Product A sells at Rs.200 each and takes 2hrs to make. Product B sells at rs.300 each and rakes 1 hr. to make. There is a permanent order for 14 of product A and 16 of product B. A working week consists of 40 hrs. of production and weekly turn over must not be less that Rs. 10000. If the profit on product A is 20Rs and on Product B is Rs.300
Solution:
Let x be units of product A and y be units of product B are manufactured.
Number of units cannot be negative
Therefore, $\text { x, } y \geq 0$
According to question, the given information can be tabulated as:

	Selling Price (Rs)	Manufacturing time (hrs.)
Product A(x)	200	0.5
Product B(y)	300	1

Also, the availability of time is 40 hrs and the revenue should be at least Rs.10000
Further, it is given that there is a permanent order for 14 units of Product A and 16 units of Product B
Therefore, the constraints are,
$\begin{aligned} &200 x+300 y \geq 10000 \\ &0.5 x+y \leq 40 \\ &x \geq 14 \& y \geq 16 \end{aligned}$
If the profit on each of product A is Rs,20 and on product B is Rs,30. Therefore, profit gained on x units of product A and y units of Product B is Rs. 20x and Rs.30y resepectively.
Total Profit=20x+30y which is to be maximized.
Thus, the mathematical formulation of the given LPP is,
Max: $z = 20x +30y$
Subject to constraints,
$\begin{aligned} &200 x+300 y \geq 10000 \\ &0.5 x+y \leq 40 \\ &x \geq 14 \& y \geq 16 \\ &x, y \geq 0 \end{aligned}$
Region $200 x+300 y \geq 10000: \text { Line } 200 x+300 y=10000$ meets the areas at A(50,0) , $\mathrm{A}(50,0), \mathrm{B}\left(0, \frac{100}{3}\right)$ respectively.
Region not containing origin represents $200 x+300 y \geq 10000$ as (0,0) does not satisfy $200 x+300 y \geq 10000$
Region $0.5 x+y \leq 40: \text { line } 0.5 x+y=40$ meets the area at C(80,0) D(0,40) respectively.
Region containing origin represents $0.5 x+y \leq 40 \text { as }(0,0) \text { satisfies } 0.5 x+y \leq 40$
Region represented by $x\geq 14$
x=14 is the line passes through (14,0) and is parallel to the X-axis.
The region to the right of the line y=14 will satisfy the in equation
Region $x, y \geq 0$ it represents first quadrant.

The corner points of the feasible region are E(26,16), F(48,16), G(14,33) and H(14,24)
The value of Z at the corner points are as follows:

Corner Points	$z=20x+30y$
E	1000
F	1440
G	1270
H	1000

The maximum of Z is 1440 which is attained at F(48,16)
Thus, the maximum profit is Rs.1440 obtained when 48 units of Product A and 16 units of Product B are manufactured.

Linear Programming Exercise 29.4 Question 20

Answer:
Max Profit = Rs.165 when 3 units of each type of trunk is manufactured.
Hint:
Form Linear Equation and solve graphically.
Given:
A manufacturer produces two types of steel trunks. He has two machines A and B. For completing the first type of the trunk requires 3 hours on machine A and 3 hours on Machine B. whereas the second type of the trunk requires 3 hours on Machine A and 2 hours on Machine B. Machine A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of Rs.30 and Rs.25 per tank of the first type and the second type respectively.
Solution:
Let x be trunks of first type and y trunks of second type were manufactured. Number of trunks cannot be negative.
Therefore, $x, y \geq 0$
According to the question, the given information can be tabulated as

	Machine A (hours)	Machine B (hours)
First type (x)	3	3
Second type (y)	3	2
Availability	18	15

Therefore, the constraints are,
$\begin{aligned} &3 x+3 y \leq 18 \\ &3 x+2 y \leq 15 \end{aligned}$
He earns a profit of Rs.30 and Rs.25 per trunk of the first type and second type respectively. Therefore, profit gained by him from x trunks of first type and y trunks of second type is Rs.30x and Rs.25y respectively.
Total Profit: $Z=30x+25y$
Subject to
$\begin{aligned} &3 x+3 y \leq 18 \\ &3 x+2 y \leq 15 \\ &x, y \geq 0 \end{aligned}$
Region $3 x+3 y \leq 18$ : line $3x+3y=18$ meets areas at A(6,0), B(0,6) respectively. Region containing origin represents the solution of the in equation $3 x+3 y \leq 18$ as (0,0) satisfies
Region $3 x+2 y \leq 15$ : line $3 x+2 y=15$ meets areas at $C(5,0), \mathrm{D}\left(0, \frac{15}{2}\right)$ respectively. Region containing origin represents the solution of the in equation $3 x+2 y \leq 15$ as (0,0) satisfies $3 x+2 y \leq 15$
Region $x, y \geq 0$ : it represents first quadrant.
The corner points are O(0,0), B(0,6), E(3,3) and C(5,0)’

The value of Z at the corner points are as follows.

Corner Points	$z=30x+25y$
O	0
B	150
E	165
C	150

The maximum value of Z is 165 which is attained at E(3,3)
Thus, the maximum profit is of Rs.165 obtained when 3 units of each type of trunk is manufactured.

Linear Programming Exercise 29.4 Question 21

Answer:
Max Profit = Rs.3, 25,500 when 10500 bottles of A and 34500 bottles of B are manufactured.
Hint:
Form Linear Equation and solve graphically.
Given:
A manufacturer of patent medicines is preparing a production plan on medicines A and B. There are sufficient raw materials available to make 20000 bottles of A and 40000 bottles of B. But there are only 45000 bottles into which either of the medicines can be put. Further, it takes 3 hours to prepare enough material to fill 1000 bottles of B and there are 66 hours available for this operation.
Solution:
Let production of each bottle of A and B are x and y respectively.
Since profits on each bottle of A and B are Rs.8 and Rs.7 per bottle respectively. So, profit on x bottles of A and y bottles of B are 8x and 7y respectively. Let Z be total profit on bottles so,
Z = 8x + 7y
Since, it takes 3 hours and 1 hour to prepare enough material to fill 1000 bottles of Type A and Type B respectively, so x bottles of A and y bottles of B are preparing is $\frac{3 x}{1000}$ hours and $\frac{y}{1000}$ hours respectively, about only 66 hours are available, so,
$\begin{aligned} &\frac{3 x}{1000}+\frac{y}{1000} \leq 66 \\ &3 x+y \leq 66000 \end{aligned}$
Since raw materials available to make 2000 bottles of A and 4000 bottles of B but there are 45000 bottles in which either of these medicines can be put so,
$\begin{aligned} &x \leq 20000 \\ &y \leq 40000 \\ &x+y \leq 45000 \\ &x, y \geq 0 \end{aligned}$
[Since production of bottles cannot be negative]
Hence mathematical formulation of the given LPP is,
Max Z = 8x + 7y
Subject to constraints,
$\begin{aligned} &3 x+y \leq 66000 \\ &x \leq 20000 \\ &y \leq 40000 \\ &x+y \leq 45000 \\ &x, y \geq 0 \end{aligned}$
Region $3 x+y \leq 66000$ : line meets the axes at A (22000,0), B(0,66000) respectively.
Region containing origin represents $3 x+y \leq 66000$ as (0,0) satisfy $3 x+y \leq 66000$
Region $x+y \leq 45000$ : line $x+y=45000$ meets the axes at C (45000,0) and D(0,45000) respectively.
Region towards the origin will satisfy the in equation as (0,0) satisfies the in equation.
Region represented by $x \leq 20000$
$x= 200000$ is the line passes through (20000,0) and is parallel to the y-axis. The region towards the origin will satisfy the in equation.
Region represented by $y \leq 40000$ ,
Y=40000 is the line passes through (0, 40000) and is parallel to the x- axis. The region towards the origin will satisfy the in equation.
Region $x, y \geq 0$ it represents first quadrant.

Scale: On y-axis, 1 Big division=20000 units

On x-axis, 1 Big division=10000 units

The corner points are O(0,0), B(0,40000), G(10500,34500), H(20000,6000), A(20000,0)
The value of Z at these corner points are,

Corner Points	$z=8x+7y$
O	0
B	280000
G	325500
H	188000
A	160000

The maximum value of Z is 325500 which is attained at G (10500, 34500)
Thus the maximum profit is Rs.325500 obtained when 10500 bottles of A and 34500 bottles of B are manufactured.

Linear Programming exercise 29.4 question 22

Answer:
Max Profit = Rs.116000 when 20 first class tickets and 180 economy class tickets are sold.
Hint:
Form Linear Equation and solve graphically.
Given:
An aero plane can carry a maximum of 200 passengers. A profit of Rs.400 is made on each first class ticket and a profit of Rs.600 is made on each economy class ticket. The airline reserves at least 20 seats of first class. However, at least 4 times as many passengers prefer to travel by economy class to the first class.
Solution:
Let required number of first class and economy class tickets be x and y respectively.
Each ticket of first class and economy class make profit of Rs.400 and Rs.600 respectively.
So, x ticket of first class and y tickets of economy class make profit of Rs.400x and Rs.600y respectively.
Let total profit be Z=400x+600y
Given, aero plane can carry a minimum of 200 passengers, so $x+y \leq 200$
Given, airline reserves at least 20 seats for first class, so $x \geq 20$
Also, at least 4 times as many passengers prefer to travel by economy class to the first class, so
$y \geq 4 x$
Hence the mathematical formulation of the LPP is
Max Z=400x+600y
Subject to constraints
$\begin{aligned} &x+y \leq 200 \\ & \end{aligned}$
$x \geq 20 \text { And } y \geq 4 x \\$
$x, y \geq 0$ {Seats in both the classes cannot be 0}
Region represented by $x+y \leq 200$ : the line x + y = 200 meets the axes at A(200,0), B(0,200). Region containing origin represents $x+y \leq 200$ as (0,0) satisfies $x+y \leq 200$
Region represented by $x \geq 20$ : line x=20 passes through (20,0) and is parallel to y-axis. The region to the right of the line x=20 will satisfy the in equation $x \geq 20$
Region represented by $y \geq 4 x$ : line y=4x passes through (0,). The region above the line y=4x will satisfy the in equation $y \geq 4 x$
Region $x, y \geq 0$ : it represents the first quadrant.
Scale: On y-axis, 1 big Division=100 units
On x-axis, 1 big Division=50 units

The corner points are C(20,80), D(40,160), E(20,180)
The values of Z at these corner as follows

Corner Points	$z=400x+600y$
O	0
C	56000
D	112000
E	116000

The maximum value of Z is attained at E(20,180).
Thus, the max profit is Rs.116000 obtained when 20 first class tickets and 180 economy class tickets sold.

Linear Programming exercise 29.4 question 23

Answer:
Minimum cost = Rs.92 when 100kg of type I fertilizer and 80 kg of Type II fertilizer is supplied.
Hint:
Form Linear Equation and solve graphically.
Given:
A gardener supply fertilizer of type I which consists of 10% of nitrogen and 6% phosphoric acid and Type II fertilizer which consist of 5% nitrogen and 10% of phosphoric acid. After testing the soil conditions, he finds that he needs at least 14 kg of nitrogen and 14kg of phosphoric acid for two crop of the type I fertilizer cost 60P/Kg and types II fertilizer 40P/Kg
Solution:
Let x kg of Type I fertilizer and y kg of Type II fertilizers are supplied.
The quantity of fertilizers cannot be negative.
So, $x, y \geq 0$
A gardener has a supply of fertilizer of type I which consists of 10% nitrogen and type II consists of 5% nitrogen, and he needs at least 14kg for his crop.
So,
$(10 \times 100)+(5 \times 100) \geq 14 \\$
Or $\begin{aligned} & &10 x+5 y \geq 1400 \end{aligned}$
A gardener has a supply of fertilizer of type I which consists of 6% phosphoric acid and type II consists of 10% phosphoric acid, and he needs at least 14 kg of phosphoric acid for his crop.
Sp,
$(6 \times 100)+(10 \times 100) \geq 14 \\$
Or $\begin{aligned} & &6 x+10 y \geq 1400 \end{aligned}$
Therefore, A/Q, constraint is,
$\begin{aligned} &10 x+5 y \geq 1400 \\ &6 x+10 y \geq 1400 \end{aligned}$
If the type I fertilizer costs 60 paise per kg and Type II fertilizer costs 40 paise per kg. Therefore, the cost of x kg of Type 1 fertilizer and y kg of Type II fertilizer is Rs.0.60x and Rs.0.40y respectively.
Total cost=Z(let)=0.6x + 0.4y is to be minimized.
Thus the mathematical formulation of the given LPP ism,
Min Z=0.6 + 0.4y
Subject to the constraints,
$\begin{aligned} &10 x+5 y \geq 1400 \\ &6 x+10 y \geq 1400 \\ &x, y \geq 0 \end{aligned}$
Region represented by $6 x+10 y \geq 1400$ : the line $6 x+10 y=1400$ passes through $\mathrm{A}\left(\frac{700}{3}, 0\right)$ , B(0,140). The region which does not contains origin represents solutions of the in equation $6 x+10 y \geq 1400$ as (0,0) doesn’t satisfy the in equation $6 x+10 y \geq 1400$
Region represented by $10 x+5 y \geq 1400$ : the line $10 x+5 y \geq 1400$ passes through C(140,0) and D(0,280). The region which does not contains origin represents solutions of the in equation $10 x+5 y \geq 1400$ as (0,0) doesn’t satisfy the in equation $10 x+5 y \geq 1400$
The region $x, y \geq 0$ : represents the first quadrant.

The corner points are D(0,280),E(100,80), A(700/3,0)
The values of Z at these points are as follows:

Corner Points	$z=0.6x+0.4y$
O	0
D	112
E	92
F	140

The minimum value of Z is Rs.92 which is attained at E(100,80)
Thus, the minimum cost is Rs.92 obtained when 100kg of Type I fertilizer and 80 kg of Type II fertilizer is supplied.

Linear Programming exercise 29.4 question 24

Answer:
Maximum Earning Rs.1160 when Rs.2000 was invested in SC and Rs.10000 in NSB
Hint:
Form Linear Equation and solve graphically.
Given:
Anil wants to invest at most Rs.12000 in saving certificates and National saving Bonds. According to rules, he has to invest at least Rs.2000 in saving certificate and at least Rs.4000 in National saving Bonds. If the rate of interest on saving certificate is 8% per annum and the rate of interest on National Saving Bond is 10% per annum.
Solution:
Let Anil invests Rsx and Rs.y in saving certificate (SC) and National Saving Band (NSB) respectively.
Since, the rate of interest on SC is 8% annual and on NSB is 10% annual. So, interest on Rs.x of SC is $\frac{8 x}{100}$ and Rs.y of NSB is $\frac{10 x}{100}$ per annum.
Let Z be total interest earned so,
$Z=\frac{8 x}{100}+\frac{10 x}{100}$
Given he wants to invest Rs.12000 is total
$x+y \leq 12000$
According to the rules he has to invest at least Rs.2000 in SC and at least Rs.4000 in NSB
$\begin{aligned} &x \geq 2000 \\ &y \geq 4000 \\ &x+y \leq 12000 \\ &x, y \geq 0 \end{aligned}$
Region represented by $x \geq 2000$ : the line x=2000 is parallel to the y-axis and passes through (2000,0).
The region which does not contains origin represents $x \geq 2000$ as (0,0) doesn’t satisfy the in equation $x \geq 2000$
Region represented by $y \geq 4000$ : the line y=4000 is parallel to the x-axis and passes through (0,4000).
The region which does not contains origin represents $y \geq 4000$ as (0,0) doesn’t satisfy the in equation $y \geq 4000$
Region represented by $x+y \leq 12000$ : the line $x+y=12000$ meets axes at A(12000,0) and B(0,12000)respectively. The region which contains origin represents the solution set of $x+y \leq 12000$ as (0,0) satisfies the in equation $x+y \leq 12000$
Region $x, y \geq 0$ is represented by the first quadrant.
Scale: On x-axis, 1 Big Division=2000 units
On y-axis, 1 Big Division=2000 units

The corner points are E(2000,10000), C(2000,4000),D(8000,4000)
The values of Z at these corner points are as follows

Corner Points	$z=\frac{8 x}{100}+\frac{10 x}{100}$
O	0
E	1160
D	1040
C	560

The maximum value of Z is rs.1160 which is attained at E(2000,10000)
Thus the maximum earning is Rs.1160 obtained when Rs.2000 were invested in SC and Rs.10000 in NSB.

Linear Programming Exercise 29.4 Question 25

Answer:
Maximum Profit Rs.1800 when Rs.20 were involved in Type A and Rs.40 were involved in Type B.
Hint:
Form Linear Equation and solve graphically.
Given:
A men owns a field of area 1000 m2. He wants to plant fruit trees in it. He has a sum of Rs.1400 to purchase young trees. He has the choice of two types of trees. Type A require 10 m2 of ground per tree and costs Rs.20 per tree and type B requires 20m2 of ground per tree and costs Rs.25 per tree. When fully grown type A produces an average of 20 kg of fruit which can be sold at a profit of Rs.2.00 per kg and type B produces an average of 40 kg of fruit which can be sold at a profit of Rs.1.50 per kg.
Solution:
Let the required number of trees of Type A and B be Rs.x and Rs.y respectively.
Number of trees cannot be negative.
$x, y \geq 0$
To plant tree of Type A requires 10sq.m and Type B requires 20sq.m of ground per tree. And it is given that a man owns a field of are 1000sq.m.
Therefore,
$\begin{aligned} &10 x+20 y \leq 1000 \\ &x+2 y \leq 100 \end{aligned}$
Type A costs Rs.20 per tree and type B costs Rs.25 per tree. Therefore, x trees of type A and y trees of type B cost Rs.20x and Rs.25y respectively. A man has a sum of Rs.1400 to purchase young trrs
$\begin{aligned} &20 x+25 y \leq 1400 \\ &4 x+5 y \leq 280 \end{aligned}$
Thus the mathematical formulation of the given LPP is
Max Z = 40x -20x + 60y - 25y = 20x + 35y
Subject to,
$x+2 y \leq 100$
$\begin{aligned} &4 x+5 y \leq 280 \\ &x, y \geq 0 \end{aligned}$
Region $4 x+5 y \leq 280$ : the line $4 x+5 y \leq 280$ meets axes at A1(70,0) and B1(0,56)respectively.
The region which contains origin represents $4 x+5 y \leq 280$ as (0,0) satisfies $4 x+5 y \leq 280$ .
Region $x+2 y \leq 100$ : the line $x+2 y \leq 100$ meets axes at A2(70,0) and B2(0,56)respectively.
The region which contains origin represents $x+2 y \leq 100$ as (0,0) satisfies $x+2 y \leq 100$ .
Region $x, y \geq 0$ : It represents by first quadrant.

The maximum value of Z is 1800 which is attained at P(20,40) as
The value of Z at these corner points are as follows

Corner Points	$Z=20 x+35 y$
O	0
A1	1750
P	1800
B2	1400

Thus the, max profit is Rs.1800 obtained when Rs.20 were involved in Type A and Rs.40 were involved in Type B.

Linear Programming Exercise 29.4 Question 26

Answer:
4 pedestal lamps and 4 wooden shades
Hint:
By using the mathematical formulation of the given Linear programming is Max Z= ax+by
Given:
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of grinding/ cutting machine and a sprayer. It takes 2 hours on the grinding/ cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp while it takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, line sprayer is available for at most 20 hours and the grinding/ cutting machine for at most 12 hours. The profit from the sale of a lamp is Rs.5.00 and a shade is Rs.3.00. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximize his profit?
Solution:
Let the cottage industry manufactures x pedestal lamps & y wooden shades
Therefore,
$x \geq 0, y \geq 0$
The given information is as follow:

	Lamps	Shades	Availability
Grinding machine	2	1	12
Sprayer	3	2	20

The profit on a lamp is Rs.5 and on the shades is Rs.3
Max Z = 5x + 3y … (i)
Subject to constraints:
$\begin{aligned} &2 x+y \leq 12 \\ & \end{aligned}$ … (ii)
$3 x+2 y \leq 20 \\$ … (iii)
$x, y \geq 0$ … (iv)
The feasible region is as follows:

Corner points	$z= 5x+3y$
$\left ( 6,0 \right )$	$30$
$\left ( 4,4 \right )$	$\\32$
$\left ( 0,10 \right )$	$\\30$

Maximum value of z= 32 at (4,4)

Linear Programming Exercise 29.4 Question 27

Answer:
Maximum Revenue is Rs.1260 obtained when 3 units of x and 8 units of y were produced is as follows.
Hint:
By using the mathematical formulation of the given Linear programming is Max Z= ax+by
Given:
A producer has 30 and 17 units of labour and capital respectively which he can use to produce two types of goods x and y. To produce one unit of x, 2 units of labour and 3 units of capital are required. Similarly, 3 units of labour and I unit of capital is required to produces one unit of y
Solution:
Let x1 and y1 units of goods x and y were produced respectively.
Number of units of goods cannot be negative.
Therefore, $x_{1}, y_{1} \geq 0$
To produce one unit of x, 3 units of Capital is required and 1 unit of capital is required to produce one unit of y.
$3 x_{1}+y_{1} \leq 17$
If x and y are priced at Rs.100 and Rs.120 per unit respectively. Therefore, cost of x1 and y1 units of goods x and y is Rs.100 x1 and Rs.120 y1
Total revenue = Z = 100 x1 +120 y1 which B to be maximized.
Thus the mathematical formulation of the given linear programming problem is
Max Z = 100 x1 +120 y1
Subject to
$2 x_{1}+3 y_{1} \leq 30 \\$
$3 x_{1}+y_{1} \leq 17 \\$
$\begin{aligned} & &x, y \geq 0 \end{aligned}$
First, we will convert in equation into equations as follows $2 \mathrm{x}_{1}+3 \mathrm{y}_{1}=30,3 \mathrm{x}_{1}+\mathrm{y}_{1}=17, \mathrm{x}=0 \& \mathrm{y}=0$
Region represented by $2 \mathrm{x}_{1}+3 \mathrm{y}_{1} \leq 30$ : the line $2 x_{1}+3 y_{1}=30$ meets axes at A1(15,0) and B1(0,10)respectively.
By joining these points we obtain the line $2 x_{1}+3 y_{1}=30$ . Clearly (0,0) satisfies the $2 x_{1}+3 y_{1}=30$ . So
the region which contains origin represents the solution set of the in equation $2 x_{1}+3 y_{1} \leq 30$ .
Region represented by $3 x_{1}+y_{1} \leq 17$ : the line $3 x_{1}+y_{1}=17$ meets axes at $C\left(\frac{17}{3}, 0\right) \& D(0,17)$ respectively. By joining these points we obtain the line $3 x_{1}+y_{1}=17$ . Clearly (0,0) satisfies the in equation $3 x_{1}+y_{1} \leq 17$ . So the region which contains origin represents the solution set of the in equation $3 x_{1}+y_{1} \leq 17$ .
Region represented by $x_{1} \geq 0 \& y_{1} \geq 0$ : Since, every point in the first quadrant satisfies these in equations. So, the first quadrant is the region represented by the in equation $x \geq 0 \& y \geq 0$
The feasible region determined by the system of constraints $2 x_{1}+3 y_{1} \leq 30,3 x+y \leq 17$ $,x \geq 0 \& y \geq 0$ as follows.

The corner points are B(0,10), E(3,8) and C(17/3,0)
The values of Z at these corner points.

Corner Points	$z=100 x_{1}+120 y_{1}$
B	1200
E	1260
C	$\frac{1700}{3}$

The maximum value of Z is 1260 which is attained at E(3,8).
Thus, the maximum revenue is Rs.1260 obtained when 3 units of x and 8 units of y were produces are as follows.

Linear Programmig Excercise 29.4 Question 28

Answer:
Maximum Value of Z is Ra.1020 which is attained at $\mathrm{B}_{1}(60,240)$ . Thus the maximum profit is Rs.1020 obtained when 60 units of product A and 240 units of product B were manufactured.
Hint:
By using the mathematical formulation of the given Linear programming is Max Z=ax+by
Given:
Firm manufacturer’s two types of Products A and B and sells them at a profit of Rs.5 per unit of Type A and Rs.3 per unit of Type B.
Solution:
Let x units of Product A and y units of Product B were manufactured.
Number of products cannot be negative.
Therefore, $x, y \geq 0$
According to question, the given information can be tabulated as

	Time on M1(Minutes)	Time on M2(Minutes)
Product A(x)	1	2
Product B(y)	1	1
Availability	300	360

The constraints are
$\begin{aligned} &x+y \leq 300 \\ &2 x+y \leq 360 \end{aligned}$
Firm manufactures two types of Products A and B and sells them at a profit of Rs.5 per unit of type A and Rs. 3 per unit of the type B. Therefore x unit of product A and y units of product B costs Rs.5x and Rs.3y respectively.
Total Profit = Z = 5x + 3y which is to be maximized
Thus, the mathematical formulations of the given linear programming problem is,
Max Z = 5x + 3y
Subject to
$\begin{aligned} &x+y \leq 300 \\ &2 x+y \leq 360 \\ &x, y \geq 0 \end{aligned}$
First we will convert in equations as follows $x+y=300,2 x+y=360, x=0 \& y=0$
Region represented by $x+y \leq 300$ : the line $x+y=300$ meets the coordinate axes at A1(300,0) and B1(0,300)respectively.
By joining these points we obtain the line $x+y=300$ . Clearly (0,0) satisfies the $x+y=300$ . So
the region which contains origin represents the solution set of the in equation $x+y \leq 300$ .
Region represented by $2 x+y \leq 360$ : the line meets the coordinate axes at C1(180,0) and D1(0,360) respectively.
By joining these points we obtain the line $2 x+y=360$ . Clearly (0,0) satisfies the $2 x+y \leq 360$ . So
the region which contains origin represents the solution set of the in equation $2 x+y \leq 360$ .
Region represented by $x_{1} \geq 0 \& y_{1} \geq 0$ : Since, every point in the first quadrant satisfies these in equations. So, the first quadrant is the region represented by the in equation $x \geq 0 \& y \geq 0$
The feasible region determined by the system of constraints $x+y \leq 300,2 x+y \leq 360_{\ell} x \geq 0 \& y \geq 0$ are as follows.

Scale: On x-axis: 1 Big Division = 100 units
On y- axis: 1 Big Division = 100 units

The corner points are $Q(0,0), B_{1}(0,300), E_{1}(60,240) \text { and } C_{1}(180,0)$
The values of Z at these corner points are as follows

Corner Points	$z=5 x+3 y$
$O$	0
$B_{1}$	900
$E_{1}$	1020
$C_{1}$	900

The maximum value of X is Rs. 1020 which B attained at B1(60,240)
Thus, the maximum profit is Rs.1020 obtained when 60 units of Product A and 240 units of Product B were manufactured.

Linear Programmig Excercise 29.4 Question 29

Answer:
The answer of the given question B that the firm should produce 8A products and 16B products to earn maximum profit of Rs.4960
Hint:
By using the mathematical formulation of the given Linear programming is Max Z=ax + by
Given:
A small firm manufactures item A and B- The total number of items that it can manufacture in a day B at most 24. Item A takes one hour while items B takes only half an hour.
Solution:
Let the firm manufactures x members of A and y members of B products.
According to the question
$x+y \leq 24, x+0.5 y \leq 16, x \geq 0, y \geq 0$
Maximize Z = 300x + 160y
The feasible region determined by $x+y \leq 24, x+0.5 y \leq 16, x \geq 0, y \geq 0$ given by

The corner points of feasible region are A(0,0), B(0,24), C(8,16) and D(16,0)
The value of Z at corner point is

Corner Points	Z=300x+160y
A(0,0)	0
B(0,24)	3840
C(8,16)	4960	Maximum
D(16,0)	4800

The maximum value of Z is 4960 and occurs at point (8,16)
The firm should produce 8A products and 16 B products to earn maximum profit of Rs.4960.

Linear Programmig Excercise 29.4 Question 30

Answer:
The answer of the given question is that the maximum profit is Rs.1500 at E(12,15)
Hint:
By using the mathematical formulation of the given Linear programming is Max Z=ax + by
Given:
Two types of toys A and B. A toy of type A requires 5 minutes for cutting and 10 minutes for assembling. A toy of type B requires 8 minutes for cutting and 8 minutes for assembling.
Solution:

	Toy A	Toy B	Times in a day
Cutting time	5 min	8 min	180 min
Assembly time	10 min	8 min	240 min
Profit	50	60
Assumed Quantity	x	y

Profit function Z=50x+60y
$\begin{aligned} &x \geq 0, y \geq 0 \\ &5 x+8 y \leq 180 \\ &10 x+8 y \leq 240 \text { or } 5 x+4 y \leq 120 \end{aligned}$

$5 x+8 y=180$

	A	B
x	0	36
y	22.5	8

$5 x+4 y \leq 120$

	C	D
x	0	24
y	30	0

Corner Points	$z=50 x+60 y$
At O(0,0)	0
At D(24,0)	1200
At E(12,15)	1500
At A(0,22.5)	1350

Hence the maximum profit is Rs.1500 at E(12,15)

Linear Programming Exercise 29.4 Question 31

Answer:.
The answer of the given question B that the maximum profit is Rs.120 obtained when 12 units of articles A and 6 units of articles B were manufactured.
Hint:
By using the mathematical formulation of the given Linear programming is Max Z=ax + by
Given:
The maximum capacity of first department is 60 hours a week and that of other department is 48 hours per week. The product of each unit of article A requires 4 hours in assembly and 2 hours in finishing and that of each unit of B requires 2 hours in assembly and 4 hours in finishing.
Solution:
Let x units and y units of articles A and B are produced respectively.
Number of articles can’t be negative.
Therefore, $x, y \geq 0$
The product of each unit of article A requires 4 hours in assembly and that of articles B requires 2 hours in assembly and the maximum capacity of the assembly department in 60 hours a week.
$4 x+2 y \leq 60$
The product of each unit of article A requires 2 hours in finishing and that of articles B requires 4 hours in assembly and the maximum capacity of the finishing department in 48 hours a week.
$2 x+4 y \leq 48$
If the profit is Rs.6 for each unit of A and Rs.8 for each unit of B. Therefore, profit gained from x units and y units of articles A and B respectively is Rs.6x and Rs.8y respectively.
Total revenue = Z = 6x + 8y which is to be maximized.
Thus, the mathematical formulation of the given linear programming problem is
Max Z = 6x + 8y
Subject to
$\begin{aligned} &2 x+4 y \leq 48 \\ &4 x+2 y \leq 60 \\ &x, y \geq 0 \end{aligned}$
First, we will convert in in equations into equations as follows:
$2 x+4 y=48,4 x+2 y=60, x=0 \& y=0$
Region represented by $2 x+4 y \leq 48$ : the line $2 x+4 y=48$ meets the coordinate axes at A1(24,0) and B1(0,12)respectively.
By joining these points we obtain the line $2 x+4 y=48$ . Clearly (0,0) satisfies the $2 x+4 y=48$ . So
The region which contains origin represents the solution set of the in equation $2 x+4 y \leq 48$ .
Region represented by $14 x+2 y \leq 60$ : the line $4 x+2 y=60$ meets the coordinate axes at C1(15,0) and B1(0,30)respectively.
By joining these points we obtain the line $4 x+2 y=60$ . Clearly (0,0) satisfies the $14 x+2 y \leq 60$ . So
The region which contains origin represents the solution set of the in equation $14 x+2 y \leq 60$ .
Region represented by $x_{1} \geq 0 \& y_{1} \geq 0$ : Since, every point in the first quadrant satisfies these in equations. So, the first quadrant is the region represented by the in equation $x \geq 0 \& y \geq 0$
The feasible region determined by the system of constraints $2 x+4 y \leq 48,4 x+2 y \leq 60, x \geq 0 \& y \geq 0$ are as follows.

The corner points are O(0,0), B(0,12),E1(12,6) and C1(15,0)
The values of Z at these corner points as follows.

Corner Points	$z=6 x+8 y$
O	0
B1	96
E1	120
C1	90

The maximum value of Z is 120 which is attained at E1(12,6)
Thus, the maximum profit is Rs. 120 obtained when 12 units of articles A and 6 units of Article B were manufactured.

Linear Programming Exercise 29.4 Question 32

Answer:.
The answer of the given questions is that 8 items of type A and 16 of type B should be produced for max profit.
Hint:
By using the mathematical formulation of the given Linear programming is Max Z=ax + by
Given:
A firm makes items A and B and the total number of items it can make in a day is m. It takes one hour to make an item of A and only half an hour to make an item of B. The maximum time available per day is 16 hours.
Solution:

Let the number of items of type A and B produce be x and y respectively.
The LPP is maximize Z=300x+160y
Subject to the constraints.
$\begin{aligned} &x+y \leq 24 \\ &x \cdot 1+y \cdot \frac{1}{2} \leq 16 \\ &x \geq 0, y \geq 0 \end{aligned}$
Draw the lines
$x+y=24$ … (i)
$\begin{aligned} &\\ &x+\frac{1}{2} y=16 \end{aligned}$ … (ii)
These meet at P (8, 16)
The feasible region is OCPB
The value of Z = 300x + 160y at 0 is zero
At C(16,0) is 4800
At B(0,24) is 3840
At P(8,16) is 4960
Clearly, value is max at P(8,16)
$\Rightarrow$ 8 items type A and 16 type B should be produced for max. Profit

Linear Programming Exercise 29.4 Question 33

Answer:The answer of the given question is that maximum profit is Rs.2375, 25 units of A and 125 units of B should be manufactured.
Hint:
By using the mathematical formulation of the given Linear programming is Max Z=ax + by
Given:
Total capacity of 500man-hour. It takes 5 hours to produce unit A and 3 hours to produce unit B.
Solution:
Let x units of Product A and y units of Product B were manufactured.
Clearly, $x \geq 0, y \geq 0$
It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The two products are produced in a common production process, which has total capacity of 500 man-hours.
$5 x+3 y \leq 500$
The maximum number of unit of A that can be sold is 70 and that for B is 125.
$\begin{aligned} &x \leq 70 \\ &y \leq 125 \end{aligned}$
If the profit is Rs.20 per unit for the product A and Rs.15 per unit for the product B. Therefore, profit x units of product A and y units of product B is Rs.20x and 15y respectively.
Total Profit $\Rightarrow z=20 x+15 y$
The mathematical formulation of the given problem is
Max $z=20 x+15 y$
Subject to
$\begin{aligned} &5 x+3 y \leq 500 \\ &x \leq 70 \\ &y \leq 125 \\ &x \geq 0, y \geq 0 \end{aligned}$
First we will convert in equation into equations as follows:
$5 x+3 y=500, x=70, y=125, x=0 \& y=0$
Region represented by $5 x+3 y \leq 500$ : the line $5 x+3 y=500$ meets the coordinate axes at A1(100,0) and B1 $1\left(0, \frac{500}{3}\right)$ respectively.
By joining these points we obtain the line $5 x+3 y=500$ . Clearly (0,0) satisfies the $5 x+3 y=500$ . So
The region which contains origin represents the solution set of the in equation $5 x+3 y \leq 500$ .
Region represented by $x \leq 70$ .
The line x = 70 is the line passes through C1(70,0) and is parallel to y-axis. The region to the left of the line x = 70 will satisfy the in equation $x \leq 70$ .
Region represented by $y \leq 125$
The line y = 125 is the line passes through D1(0,125) and is parallel to x-axis. The region below the line
y=125 will satisfy the in equation $y \leq 125$ .
Region represented by $x \geq 0 \& y \geq 0$
Since, every point in the first quadrant satisfies these in equations. So the first quadrant is the region represented by the in equation $x \geq 0 \& y \geq 0$
The feasible region determined by the system of constraints $5 x+3 y \leq 500, x \leq 70, y \leq 125$ $,x \geq 0 \& y \geq 0$ are as follows:

The corner points are O(0,0), D(0,125), E(25,125), F(70,50) and C(70,0). The values of Z at the corner points are:

Corner Points	$z=20 x+15 y$
O	0
$D_{1}$	1875
$E_{1}$	2375
$F_{1}$	2150
$C_{1}$	1400

The maximum value of Z is 2375 which is at E1(25,125)
Thus, the maximum profit is Rs.2375. 25 units of A and 125 units of B should be manufactured.

Linear Programming exercise 29.4 question 34

Answer:
No. of box=6, small box=12
Maximum profit=Rs.42
Hint:
Let required quantity of large and small boxes are x and y respectively.
Given:
Profits on each unit of large and small boxes are Rs.3 and Rs.2 respectively.
Solution:
Let Z be total profit
Z=3x+2y [Where x and y are the quantity of large and small boxes]
Since each large and small box require 4sq.m and 3sq.m cardboard, but only 60sq.m of cardboard is available
$4 x+3 y \leq 60$
Since manufacture is required to make at least three large boxes.
So,
$x \geq 3$
Since manufacture is required to make at least twice as many small boxes as large boxes.
So,
$y \geq 2$
The required LPP is Max Z = 20x + 5y

Linear Programming exercise 29.4 question 35

Answer: 48 units of product A and 16 units of product. Maximum profit = Rs.1440.
Hint:
Let x be the number of units A and y be the number of units of B
Given:
The given data can be written in the tabular form

Product	A	B	Working week	Turn over
Time	0.5	1	40
Price	200	300		10000
Profit	20	30
Permanent order	14	16

Solution:
The mathematical model of the LPP as follows.
Max Z = 20x + 30y
Subject to:
$\begin{aligned} &0.5 x+y \leq 40 \\ &200 x+300 y \geq 10000 \\ &x \geq 14, y \geq 16 \end{aligned}$
Subject to constraint:
$\begin{aligned} &4 x+3 y \geq 60 \\ & \end{aligned}$ … (i)
$x \geq 3 \\$ … (ii)
$y \geq 2 x \\$ … (iii)
$x, y \geq 0$
The feasible region determined by the subject of constraints.

No of large box=6, small box=12
Maximum profit=Rs.42
The coordinates of the vertices (corner points) of shaded region ABCD are A(26,16),B(48,16),C(14,33) and D(14,24)

Points	Z=20x+30y
A(26,16)	Z=1000
B(48,16)	Z=1440
C(14,33)	Z=1270
D(14,24)	Z=60

48 units of Product A and 16 units of Product B should be produced to earn the maximum profit of Rs.1440.

Linear Programming exercise 29.4 question 36

Answer: The maximum that the man can travel in 1 hr. is 30km. Distance travelled at the speed of 25km/hr is $\frac{50}{3}$ km and 40 km/hr. is $\frac{40}{3}$ km
Hint:
Let us assume that the man travels xkm/hr. and y km when the speed is 40km/hr.
Given:
Speed of vehicle=25km/hr.
Per km cost on petrol=2Rs/km
Young man carries only =100 Rs.
Solution:
Let us assume that the man travels x km when the speed is 25km/hour and y km when the speed is 40km/hour.
Thus, the total distance travelled is (x+y) km.
Now, it’s given that the man has 100Rs to spend on petrol.
Total cost of petrol= $2 x+5 y \leq 100$
Now, Time taken to travel x km = $\frac{x}{25}$ hour
Time taken to travel y km = $\frac{y}{40}$ hour
Now it’s given that the maximum time is 1 hour. So,
$\begin{gathered} \frac{x}{25}+\frac{y}{40} \leq 1 \\ \Rightarrow 8 x+5 y \leq 200 \end{gathered}$
Thus, the given linear programming problem is
Maximize Z=x+y
Subject to the constraints
$\begin{aligned} &2 x+5 y \leq 100 \\ &8 x+5 y \leq 200 \\ &x \geq 0, y \geq 0 \end{aligned}$
The feasible region determined by the given constraints can be diagrammatically represented as,

The coordinates of the corner points of region are O(0,0), A(25,0), B(50/3,40/3) and C(0,20).
The value of objective function at these points are given in the following

Corner Points	Z=x+y
(0,0)	0+0=0
(25,0)	25+0=25
(50/3,40/3)	50/3+50/3=30
(0,20)	0+20=20

So the maximum value of Z is 30 at $x=\frac{50}{3}, y=\frac{40}{3}$
Thus, the maximum distance that the man can travel in one hour 30 km
Hence, the distance travelled by the man at the speed of 225 km/hr is 50/3 km and the distance travelled by him at the speed of the 40km/hour is 40/3 km.

Linear Programming Exercise 29.4 Question 37

Answer:
The minimum transportation cost is 4400.
Hint:
Assuming that the transportation cost of 50 liters of oil is Rs.1 per km.
Given:

	Distance	(in km)
To/From	A	B
D	7	3
E	6	4
F	3	2

Solution:
Let x and y liters of oil be supplied from A to the petrol pumps, D and E. Then, (7000-x-y) will be supplied from A to petrol pump F. The requirement at petrol pump D is 4500l since xl are transported from depot A, the remaining (7000-x) will be transported from petrol pump B.
Similarly, (3000-y)l and 35000-(7000-x-y)=(x+y-3500)l will be transported from depot B to petrol pump E and F respectively. The given problem can be represented diagrammatically as follows.

$\begin{aligned} & x \geq 0, y \geq 0, \text { and }(7000-x-y) \geq 0 \\ \Rightarrow \quad & x \geq 0, y \geq 0 \text { and } x+y \leq 7000 \\ & 4500-x \geq 0,3000-y \geq 0, \text { and } x+y-3500 \geq 0 \\ \Rightarrow & x \leq 4500, y \leq 3000 \text { and } x+y \geq 3500 \end{aligned}$

Cost of transporting 10 litres of petrol = Rs.1

Cost of transporting 1 litre of petrol = $\frac{1}{10} \mathrm{Rs}$
Therefore, total transportation cost is given by,
$\! \! \! \! \! \! \! \! z=\! \frac{1}{10}, x+\frac{6}{10} y+\frac{3}{10}(7000-\! x-y)+\frac{3}{10}(4500-x)+\! \frac{4}{10}(3000-\! y)+\frac{2}{10}(x+y-3500)$
= 0.3x + 0.1y + 3950
The problem can be formulated as follows.
Minimize Z=0.3x + 0.1y + 3950 …(i)
Subject to the constraints
$x+y \leq 7000$ … (ii)
$\begin{aligned} &\\ &x \leq 4500 \end{aligned}$ … (iii)
$\begin{aligned} &y \leq 3000 \\ & \end{aligned}$ … (iv)
$x+y \geq 3500 \\$ … (v)
$x \geq 0, y \geq 0$ … (vi)
The feasible region determined by the constraints as follows

The corner points of the feasible region are A(3500,0), B(4500,0), C(4500,2500), D(4000,3000) and E(500,3000)
The values at the corner points as follows:

Corner Points	Z=0.3x+0.1y+3950
A(3500,0)	5000
B(4500,0)	5300
C(4500,2500)	5550
D(4000,3000)	0+20=20
E(500,3000)	4400	Minimum

The minimum value of Z is 4400 at (500, 3000)
Thus the oil supplied from depot A is 500L, 3000L and 3500L and from depot B is 400L, 0L and 0L petrol pumps D,E and F respectively.
The minimum transportation cost is Rs.4400.

Linear Programming Exercise 29.4 Question 38

Answer: 8 gold rings and 16 chains must be manufactured per day.
Hint:
Let, gold ring manufactured per day= x
Chains manufactured per day= y
Given:
Total number of rings and chain manufactured per day is almost 24.
Time taken to make a ring=1 hour
Time taken to make a chain=30 minutes
Maximum number of hours available per day=16 hrs
Profit on a chain =190 and profit on ring=300
Solution:
Let gold rings manufactured per day=x
Chain manufacture per day=y
LPP is:

Maximize Z= 300x + 190y
Subject to $x \geq 0, y \geq 0$
$x+y \leq 24, x+\frac{1}{2} \leq 16$
Possible point for maximum Z is (16, 0), (8, 16) and (0, 24)
Hence Z MAXIMUM at (8, 16)
8 gold rings and 16 chains must be manufactured per day

Linear Programming Exercise 29.4 Question 39

Answer: The number of book of I type is 12 and II type is 6
Hint:
Let, two types of books be x and y.
Given:
Thickness of the books=6cm and 4 cm
Weight of the books=1 kg and $1 \frac{1}{2}$ kg each
Shelf is 9 cm and at most can support a weight of 21 kg
Solution:
Let two types of books be x and y respectively.
The required LPP is maximized.
Z=x+y
Subject to the constraints,
$6 x+4 y \leq 96\; Or \; 3 x+2 y \leq 48$
$x+\frac{3}{2} y \leq 21 \text { Or } 2 x+3 y \leq 42$
And $x, y \geq 0$
On considering the inequalities as equations,
We get,
$3 x+2 y=48$ …(i)
$\begin{aligned} &\\ &2 x+3 y=42 \end{aligned}$ … (ii)
Table for line $3 x+2 y=48$ is

x	0	16
y	24	0

So, it passes through (0, 24) and (16,0)
On putting (0, 0) in $3 x+2 y \leq 48$ ,
We get $0 \leq 48$ [Which is true]
Table for $2 x+3 y=42$

x	0	21
y	14	0

So it passes through (0, 14) and (21, 0)
On putting (0, 0) in $2 x+3 y \leq 42$ ,
We get $0 \leq 42$ [Which is true]
On solving equation (i) and (ii), we get
X = 12 and y = 6
Thus, the intersection point is B (12, 6)

From the graph, OABCD is the feasible region which is bounded. The corner points are O(0,0) A(0,14), B(12,6) and C(16,0)
The values of Z at corner points are as follows.

Corner Points	Z=x+y
O(0,0)	Z=0+0=0
A(0,14)	Z=0+14=14
B(12,6)	Z=12+6=18
C(16,0)	Z=16+0=16

From the table, the maximum value of Z is 18 at B(12,6)
Hence, the maximum number of books is 18 and number of books of I type is 12 and books of II type is 6.

Linear Programming Exercise 29.4 Question 40

Answer: The factory makes 4 tennis racket and 12 cricket bats. Maximum profit is 200.
Hint:
Let the number of tennis rackets and cricket bats be x and y.
Given:
A tennis racket takes 1.5 hours if machine time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftsman’s time. In a day factory has availability of not more than 42 hours of machine time and 24 hours of craftsman’s time. If the profit on a rackets and bat is 20Rs and 10Rs.
Solution:
Let the number of tennis racket and cricket bats manufactured by factory be x and y.
Hence, Profit is the objective function Z.
Z = 20x + 10y … (i)
We have to maximize Z subject to the constraints.
$\begin{aligned} &1.5 x+3 y \leq 42 \\ & \end{aligned}$ … (ii) [Constraints for machine hour]
$3 x+y \leq 24 \\$ … (iii) [Constraints for craftman’s hour]
$x \geq 0 \& y \geq 0$

Graph of x = 0 and y = 0 is the y-axis and x-axis.
Graph of $x \geq 0 \& y \geq 0$ is the 1st quadrant.
Graph of $1.5 x+3 y=42$

x	0	28
y	14	0

Graph for $1.5 x+3 y \leq 42$ is the part of 1st quadrant which contains the origin.
Graph for $\begin{aligned} &3 x+y \leq 24 \\ & \end{aligned}$
$3 x+y=24$

x	0	8
y	24	0

Graph for $3 x+y \leq 24$ is the part of st quadrant in which origin lies.
Hence, shaded area OACB is the feasible region for coordinate of C equation
$\begin{aligned} &1.5 x+3 y=42 \\ & \end{aligned}$ … (iv)
$3 x+y=24 \\$ … (v)
$\begin{aligned} 2 \times(\mathrm{iv})-(\mathrm{v}) \Rightarrow 3 \mathrm{x}+6 \mathrm{y}=84 \\ & \end{aligned}$
$3 \mathrm{x}+\mathrm{y}=24 \\$
$\overline{5 y=60}\; \Rightarrow y=12$
X = 4 (Substituting y = 12 in (iv))
Now the value of objective function Z at each corner of feasible region is

Corner Points	$Z=20x+10y$
O(0,0)	$Z=0+0=0$
A(8,0)	$20 \times 8+10 \times 0=160$
B(0,14)	$20 \times 0+10 \times 14=140$
C(4,12)	$20 \times 4+10 \times 12=200$

Therefore, maximum profit is Rs.200 when factory makes 4 tennis rackets and 12 cricket bats.

Linear Programming Exercise 29.4 Question 41

Answer: The merchant should stock 200 desktop models and 50 portable models to get maximum profit.
Hint:
Let merchant plans has personal computers x desktop model and y portable model.
Given:
Cost of desktop model computer=25000
Cost of portable model computer=40000
Total monthly demand will not exceed 250 units.
Profit on desktop model=4500Rs.
Profit on portable model=5000Rs.
Solution:
Let merchant plans has personal computers x desktops model and y portable model.
Thus, $x \geq 0 \& y \geq 0$
The cost of desktop model is cost Rs.25000 and portable model is Rs.40000
Merchant can invest Rs.70 lakhs maximum
$\begin{aligned} &25000 x+40000 y \leq 700000 \\ &5 x+8 y \leq 1400 \end{aligned}$
The total monthly demand will not exceed 250 units.
$x+y \leq 250$
Profit on desktop model is 4500 and on portable model is Rs.5000
Total Profit = Z,
$z=4500 x+500 y$
The feasible region determined by constraints is as follows.

The corner points of feasible region are A(25,0), B(250,50), C(0,175) , D(0,0)
The value of Z corner points is as shown

Corner Points	Z=4500x+500y
A(250,0)	1125000
B(200,50)	1150000(Maximum)
C(0,175)	875000
D(0,0)	0

The maximum value of Z is 1150000 at B (200, 50)
Thus, merchant should stock 200 desktop models and 50 portable models to get maximum profit.

Linear Programming Exercise 29.4 Question 42

Answer: The society will get the maximum profit of Rs.495000 by allocating 30 hectares for crop X and 20 hectare for Crop y.
Hint:
Let x hectare of land be allocated to crop x and y hectare to crop y.
Given:
Profit per hectare on crop x=Rs.10500
Profit per hectare on crop y = Rs.9000
Solution:
By the given profit on x crop and y crop.
Total profit=Rs.(10500x+9000y)
The mathematical formulation of the problem is as follows.
Maximize Z = 10500x + 9000y
Subject to the constraints
$x+y \leq 50$ (Constraint related to land) … (i)
$20 x+10 y \leq 800$ (Constraint related to use of herbicide) i.e $2 x+8 y \leq 80$ … (ii)
$x \geq 0 \& y \geq 0$ (Non –negative constraint) … (iii)
Let us draw the graph of the system of inequalities (i) to (iii).
The feasible region ABC is shown (shaded)

Corner Points	Z=10500x+9000y
O(0,0)	0
A(40,0)	420000
B(30,20)	495000(Maximum)
C(0,50)	450000

The coordinate of the corner points O, A, B and C are (0,0), (45,0),(30,20) and (0,50) respectively.
Let us evaluate the objection function Z = 10500x + 9000y at these vertices to find on gives the maximum profit.
Hence, the society will get the maximum profit of Rs.495000 by allocating 90 hectares for crop x and 20 hectares for crop y.

Linear Programming exercise 29.4 question 43

Corner Points	Z=10500x+9000y
O(0,0)	0
A(40,0)	420000
B(30,20)	495000(Maximum)
C(0,50)	450000

Linear Programming exercise 29.4 question 44

Answer:

4 Tennis Racket Bats must be made so that factory runs at all capacity.
Maximum profit is 200 when 4 tennis and 12 cricket balls are produced.

Hints:
Let the number of tennis racket be x number of cricket bat be y
Given:

Item	Number	Machine hours	Crafts man hours	Profit
Tennis Racket	x	1.5	3	Rs.20
Cricket Bats	y	3	1	Rs.10
Maximum time available		42	24

Solution:
Let the number of Tennis Racket be x.
Number of cricket bats be y.
According to the question,

Machine hours Craftsman’s hours

Tennis Racket – 1.5 hours Tennis Racket requires -3 hours
Cricket bat requires – 3 hours Cricket Bat requires – 1 hour
Maximum time available – 42 hours Max time available – 24 hour
$\begin{aligned} &1.5 x+3 y \leq 42 \\ & \end{aligned}$
$3 x+6 y \leq 84 \\$ $3 x+y \leq 24$
$x+2 y \leq 28$ Also $x \geq 0, y \geq 0$
As we want maximize the profit.
Hence function used here will be maximize
profit on each tennis rocket à Rs.20
Profit on each Bat à Rs.10
Max Z = 20x + 10y
Combining all the constraints:
Maximize Z = 20x + 10y
Subject to the constraints:
$\begin{aligned} &x+2 y \leq 28 \\ &3 x+y \leq 24 \\ &x \geq 0, y \geq 0 \end{aligned}$
Now:
$x+2 y \leq 28$

x	0	14
y	14	7

$3 x+y \leq 24$

x	2	8
y	18	0

Corner Points	Value of Z
(0,4)	14
(4,12)	16
(8,0)	8

Thus,

4 Tennis Racket Bats must be made so that factory runs at all capacity.
Maximum profit is 200 when 4 tennis and 2 cricket balls are produced.

Linear Programming exercise 29.4 question 45

Answer: The merchant should stock 200 desktop models and 50 portable models to get maximum profit.
Hint:
Let merchant plans has personal computers x desktop model and y portable model.
Given:
Cost of desktop model computer=25000
Cost of portable model computer=40000
Total monthly demand will not exceed 250 units.
Profit on desktop model=4500Rs.
Profit on portable model=5000Rs.
Solution:
Let merchant plans has personal computers x desktops model and y portable model.
Thus, $x \geq 0 \& y \geq 0$
The cost of desktop model is cost Rs.25000 and portable model is Rs.40000
Merchant can invest Rs.70 lakhs maximum
$\begin{aligned} &25000 x+40000 y \leq 700000 \\ &5 x+8 y \leq 1400 \end{aligned}$
The total monthly demand will not exceed 250 units.
$x+y \leq 250$
Profit on desktop model is 4500 and on portable model is Rs.5000
Total Profit = Z,
$Z=4500 x+5000 y$
The feasible region determined by constraints is as follows.

The corner points of feasible region are A(25,0), B(250,50), C(0,175) , D(0,0)
The value of Z corner points is as shown

Corner Points	Z=4500x+500y
A(250,0)	1125000
B(200,50)	1150000(Maximum)
C(0,175)	875000
D(0,0)	0

The maximum value of Z is 1150000 at B(200,50)
Thus, merchant should stock 200 desktop models and 50 portable models to get maximum profit.

Linear Programming Exercise 29.4 Question 46

Answer: Therefore, 800 units of doll A and 400 units of doll B should be produced weekly to get the maximum profit of Rs.16000
Hint:
Use properties of LPP
Given:
A toy company manufacturers two types of dolls A and B and if the company makes profit of Rs.12 and Rs.16 per doll respectively.
Solution:
Let x units of doll A and y units of doll B are manufactured to obtain the maximum profit.
The mathematical formulation of the above problem as follows.
Maximize Z = 12x + 16y
Subject to
$\begin{aligned} &x+y \leq 1200 \\ &y \leq \frac{x}{2} \\ &x-3 y \leq 600 \\ &x, y \geq 0 \end{aligned}$

The shaded region represents the set of feasible solutions.
The coordinates of the corner points of the feasible region are O(0,0), A(800,400), B(1050,150) and C(600,0)
=12(0) + 16(0) = 0
The value of Z at A(800,400)
=12(800) + 16(400) = 16000
Maximum value of Z at B(1020,150)
=12(1050) + 16(150) = 15000
The value of Z at C(600,0)
=12(600) + 6(0) = 7200
Therefore, 800 units of doll A and 400 units of doll B should be produced weekly to get the maximum profit of Rs.16000.

Linear Programming Exercise 29.4 Question 47

Answer: Therefore, the minimum cost is Rs.1000
Hint:
Use properties of LPP
Given:
Two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid if F1 consists of Rs.6/kg and F2 costs Rs.5/Kg
Solution:
Suppose x kg of fertilizer F1 and y kg of fertilizer F2 is used to meet the nutrient requirements.
F1 consists of 10% nitrogen and F2 consists of 5% nitrogen. Bu the farmer needs at least 14 kg of nitrogen for the crops.
$\begin{aligned} &10 \% \text { of } x \mathrm{~kg}+5 \% \text { of } y \geq 14 \mathrm{~kg} \\ &\Rightarrow \frac{x}{10}+\frac{y}{5} \geq 14 \\ &\Rightarrow 2 x+y \geq 280 \end{aligned}$
Similarly, F1 consists of 6% phosphoric acid and F2 consists of 10% phosphoric acid. But the farmer need at least 14 kg of phosphoric acid for the crops.
$6 \% \text { of } x \mathrm{~kg}+10 \% \text { of } y \geq 14 \mathrm{~kg}$
$\begin{aligned} &\Rightarrow \frac{6 x}{10}+\frac{10 y}{100} \geq 14 \\ &\Rightarrow 3 x+5 y \geq 700 \end{aligned}$
The cost of fertilizer F1 is Rs.6/kg and Fertilizer F2 in Rs.5/Kg, therefore total cost of x kg of fertilizer F1 and y kg of fertilizer F2 is Rs.(6x+5y)
Thus, the given linear programming problem is
Minimize Z = 6x + 5y
Subject to the constraints
$\begin{aligned} &2 x+y \geq 280 \\ &3 x+5 y \geq 700, x, y \geq 0 \end{aligned}$
The feasible region determined by the given constraints can be diagrammatically represented as,

The coordinates of the corner points of the feasible region are
$\mathrm{A}\left(\frac{700}{3}, 0\right), \mathrm{B}(100,80) \& \mathrm{C}(0,280)$
The value of the objective function at these points are given in the following table.

Corner Points	Z=6x+5y
$A\left(\frac{700}{3}, 0\right)$	$6 \times \frac{700}{3}+5 \times 0=1400$
$\mathrm{B}(100,80)$	$6 \times 100+5 \times 80=1000$ (minimum)
$C(0,280)$	$6 \times 0+5 \times 280=1400$

The smallest value of Z is 1000 which is obtained at x = 100, y = 80.
It can be sure that the open half-plane represented by 6x + 5y < 1000 has no common points with the feasible region.
So, the minimum value of Z is 1000.
Hence, 100 kg of fertilizer F1 and 80 kg of fertilizer F2 should be used so that the nutrient requirements are met at minimum cost.
The minimum cost is Rs.1000

Linear Programming Exercise 29.4 Question 48

Answer: The maximum profit of the manufacture is Rs.4000
Hint:
Use properties of LPP
Given:

Item	Number of hours required on machines
	I II III
M	1 2 1
N	2 1 1.25

She makes a profit of Rs.600 and Rs.400 on items M and N respectively.
Solution:
Suppose x units of item M and y units of item N are produced to maximize the profit. Since each unit of item M require hour on machine I and each unit of item N require hours on Machine I, therefore, the total hours required for producing x units of item M and y units of item N on machine I are (2x +y). But machines I is capable of being operated for at most 12 hours.
$2 x+y \leq 12$
Similarly, each unit of item M require 2 hours on machine II and each unit of item N require 1 hour on machine II, therefore, total hours required for producing x units of item M and y units of item N on machine II are (x + 2y). But machines II is capable of being operated for at most 12 hours.
$x+2 y \leq 12$
Also, each unit of item M require 1 hour on machine III and each unit of item N require 1.25 hour on machine III, therefore, the total hours required for producing x units of item M and y units of item N on machine III are (x + 1.25y). But, machines III must be operated for at least 5 hours.
$x+1.25 y \geq 5$
The profit from each unit of item M is Rs.600 and each unit of item N is Rs.400, Therefore the total profit from x units of item M and y units of item N is (600x + 400y).
Thus, the given linear programming problem is
Maximize Z = 600x + 400y
Subject to the constraints,
$\begin{aligned} &2 x+y \leq 12 \\ &x+2 y \leq 12 \\ &x+1.25 y \geq 5 \\ &x, y \geq 0 \end{aligned}$
The feasible region determined by the given constraints can be diagrammatically represented as,

$12 x+y=12 \quad x+1.25 y=5$
The coordinates of the corner points of the feasible region are A(5,0), B(6,0), C(4,4), D(0,6) and E(0,4)
The value of the object function at these points are given in the following table.

Corner Points	Z=600x+400y
(5,0)	3000
(6,0)	$600 \times 6+400 \times 0=3600$
(4,4)	$600 \times 4+400 \times 4=4000$ (maximum)
(0,6)	$600 \times 0+400 \times 6=2400$
(0,4)	$600 \times 0+400 \times 4=1600$

The maximum value of Z is 4000 at x=4,y=4.
Hence, 4 units of item M and 4 units of item N should be produced to maximize the profit.
The maximum profit of the manufacturer is Rs.4000.

Linear Programming Exercise 29.4 Question 49

Answer: The minimum transportation cost is Rs.1550
Hint:
Use properties of LPP
Given:

From To	Cost (in Rs.)
	A B C
P	160 100 150
Q	100 120 100

Solution:
Here, demand of the commodity (5 + 5 +4 = 14 units) is equal the supply of the commodity (8 + 6 = 14 units). So, no commodity could be left at the two factories.
Let x units and y units of the commodity be transported from the factory P to the depots A and B respectively.

Then (8-x-y) units of the commodity will be transported from the factory P to the depot C.
Now, the weekly requirement of depot A is 5 units of the commodity. Now, x units of the commodity are transported from factory P so the remaining (5-x) units of the commodity are transported from the factor Q to the depot A.
The weekly requirement of depot B is 5 units of the commodity. Now, y units of the commodity are transported from factory P. So the remaining (5-y) units of the commodity are transported from the factory Q to the depot B.
Similarly, $6-(5-x)-(5-y)=(x+y-4)$ units of the commodity will be transported from the factory Q to the depot C.
Since the number of the units of commodity transported are from the factories to the depots are non-negative, therefore,
$\begin{aligned} &x \geq 0, y \geq 0,8-x-y \geq 0,5-x \geq 0,5-y \geq 0, x+y-4 \geq 0 \\ & \end{aligned}$
$x \geq 0, y \geq 0, x+y \leq 8, x \leq 5, y \leq 5, x+y \geq 4$
Total transportation cost =
$\! \! \! \! \! \! \! \! \! 160 x+100 y+150(8-x-y)+100(5-x)+120(5-y)+100(x+y-4)=100 x-70 y+1900$
Thus, the given linear programming problem is
Minimize Z = 10x – 70y + 1900
Subject to constraints:
$\begin{aligned} &x+y \leq 8 \\ &x \leq 5 \\ &y \leq 5 \\ &x+y \geq 4 \\ &x \geq 0, y \geq 0 \end{aligned}$
The feasible region determined by the given constraints can be diagrammatically represented as,

The coordinates of the corner points of the feasible region are A(4,0),B(5,0),C(5,3), D(3,5), E(0,5) and F(0,4).
The value of the objective function at these points is given in the follow table.

Corner Points	Z=10x-70y+1900
(4,0)	$10 \times 4-70 \times 0+1900=1940$
(5,0)	$10 \times 5-70 \times 0+1900=1950$
(5,3)	$10 \times 5-70 \times 3+1900=1740$
(3,5)	$10 \times 3-70 \times 5+1900=1580$
(0,5)	$10 \times 0-70 \times 5+1900=1550$ (minimum)
(0,4)	$10 \times 0-70 \times 4+1900=1620$

The minimum value of Z is 1550 at x = 0, y = 5
Hence, for minimum transportation cost factory P should supply 0, 5, 3 units of commodity to depots A,B,C respectively and factory Q should supply 5,0,1 units of commodity to depots A,B and C respectively.
The minimum transportation cost is Rs.1550.

Linear Programming Exercise 29.4 Question 50

Answer: Therefore, the maximum profit is Rs.262.50
Hint:
Use properties of LPP
Given:

Types of toys	Machines
	I II III
A	12 18 6
B	16 0 9

Solution:
Suppose the manufacturer makes x toys of type A and y types of toy B.
Since each toy of type A requires 12 minutes on machine I and each toy of type B require 6 minutes on machine I, therefore, x toys of type A and y toys of type B require (12x + 6y) minutes on machine I.But, machines I is available for at most 6 hours.
$\begin{aligned} &12 x+6 y \leq 360 \\ &2 x+y \leq 60 \end{aligned}$
Similarly, each toy of type A requires 18 minutes on machine II and each toy of type B require 0 minutes on machine II, therefore, x toys of type A and y toys of type B require (18x + 0y) minutes on machine II. But, machines II is available for at most 6 hours.
$\begin{aligned} &18 x+0 y \leq 360 \\ &\Rightarrow x \leq 20 \end{aligned}$
Also, each toy of type A requires 6 minutes on machine III and each toy of type B require 9 minutes on machine III, therefore, x toys of type A and y toys of type B require (6x + 9y) minutes on machine III. But, machines III is available for at most 6 hours.
$\begin{aligned} &6 x+9 y \leq 360 \\ &\Rightarrow 2 x+3 y \leq 120 \end{aligned}$
The profit on each toy of type A is Rs.7.50 and each toy of type B is Rs.5. Therefore, the total profit from x toys of type A and y toys of type B is Rs.(7.50x+5y)
Thus the given linear programming problem is
Maximize Z=7.5x+5y
Subject to the constraints:
$\begin{aligned} &2 x+y \leq 60 \\ &x \leq 20 \\ &2 x+3 y \leq 120 \\ &x, y \geq 0 \end{aligned}$
The feasible region determined by the given constraints can be diagrammatically represented as.

The coordinates of the corner points of the feasible region are O(0,0), A(20,0), B(20,20), C(15,30) and D(0,40).
The value of the object function at these points is given in the following table.

Corner Points	Z=7.5x+5y
(0,0)	$7.5 \times 0+5 \times 0=0$
(20,0)	$7.5 \times 20+5 \times 0=150$
(20,20)	$7.5 \times 20+5 \times 20=250$
(15,30)	$262.5$ (maximum)
(0,40)	$7.5 \times 0+5 \times 40=200$

The maximum value of Z is 262.5 at x = 15, y = 30.
Hence, 15 toys of type A and 30 toys of type B should be manufactured in a day to get maximum profit.
The maximum profit is Rs.262.5

Linear Programming exercise 29.4 question 51

Answer:
Max Profit = Rs.1, 36,000 when 40 first class tickets and 160 economy class tickets are sold.
Hint:
Form Linear Equation and solve graphically.
Given:
An aero plane can carry a maximum of 200 passengers. A profit of Rs.1000 is made on each first class ticket and a profit of Rs.600 is made on each economy class ticket. The airline reserves at least 20 seats of first class. However, at least 4 times as many passengers prefer to travel by economy class to the first class.
Solution:
Let required number of first class and economy class tickets be x and y respectively.
Each ticket of first class and economy class make profit of Rs.1000 and Rs.600 respectively.
So, x ticket of first class and y tickets of economy class make profit of Rs.1000x and Rs.600y respectively.
Let total profit be Z=1000x+600y
Given, aero plane can carry a minimum of 200 passengers, so $x+y \leq 200$
Given, airline reserves at least 20 seats for first class, so $x \geq 20$
Also, at least 4 times as many passengers prefer to travel by economy class to the first class, so
$y \geq 4 x$
Hence the mathematical formulation of the LPP is
Max Z = 1000x+600y
Subject to constraints
$\begin{aligned} &x+y \leq 200 \\ & \end{aligned}$
$x \geq 20 \text { And } y \geq 4 x \\$
$x, y \geq 0$ {Seats in both the classes cannot be 0}

The feasible region determined by the given constraints can be diagrammatically represented as.

The corner points are A(20,80), B(40,160), C(20,180)

The values of Z at these corner as follows

Corner Points	z=1000x+600y
O	0
A	68000
B	136000(maximum)
C	128000

The maximum value of Z is attained at B(40,160).
Thus, the max profit is Rs.136000 obtained when 40 first class tickets and 160 economy class tickets sold.

Linear Programming exercise 29.4 question 52

Answer: The maximum total revenue is Rs.1260 when 3 units of A and 8 units of B are produced. Yes, because the efficiency of a worker does not depend on whether the worker is a male or female.
Hint:
Use property of LPP
Given:
That men and women workers are equally efficient and So, he pays them at the same rate.
Solution:
Let x unit of A and y units of B produced by the manufacturer.
The price of one unit of A is Rs.100 and the price of one unit of B is Rs.120. Therefore, the total price of x unit of A and y units of B. The total revenue is Rs. (100x+120y) one unit of A requires 2 workers are one unit of B requires 3 workers. Therefore x unit of A and y units of B requires (2x + 3y) workers. But, the manufacturer has 30 workers.
$2 x+3 y \leq 30$
Similarly, one unit of A requires 3 units of capital. Therefore, x unit of A and y unit of B requires (3x+y) units of capital. Therefore, x unit of A and y unit of B requires (3x + y) units of capital. But the manufacturer has 17 units of capital.
$\begin{aligned} &2 x+3 y \leq 30 \\ &3 x+y \leq 17 \\ &x, y \geq 0 \end{aligned}$
The feasible region determined by the given constraints can be diagrammatically represented as.

The coordinates of the corner points of the feasible region are O(0,0), A(0,10), B(17/3,0) and C(3,8)
The value of the objective function at these points is given in the following table.

Corner Points	$z=100x+120y$
O	$100 \times 0+120 \times 0=0$
A	$100 \times 0+120 \times 10=1200$
B	$100 \times \frac{17}{3}+120 \times 0=\frac{1700}{3}$
C	$100 \times 3+120 \times 8=1260$ (maximum)

The maximum value of Z is 1260 at x=3 and y=8.
Hence the maximum total revenue is Rs.1260 when 3 units of A and 8 units of B are produced.
Yes, because the efficiency of a worker does not depend on whether the worker is a male or female.

Linear Programming exercise 29.4 question 53

Answer: The maximum daily profit of the manufacturer is Rs.26
Hint:
Use property of LPP
Given:
First machine is 12 hours and second machine is 9 hours per day.
Solution:
Let x units of product A and y units of Product B be manufactured by the manufacturer per dat.
It is given that one unit of product A requires 3 hours of processing time on first machine, while one unit of product B requires 2 hours of processing time on first machine.
It is also given that first machine is available for 12 hours per day.
$3 x+2 y \leq 12$
Also, one unit of product A requires 3 hours of processing time on second machine, while one unit of product B requires 1 hour of processing time on second machine.
It is also given that second machine is available for 9 hours per day.
$3 x+y \leq 9$
The profits on one unit each of Product A and B are Rs.7 and Rs.4 respectively.
So the objective function is given by,
$Z=R s .(7 x+4 y)$
Hence the mathematical formulation of the LPP is
Maximize Z=7x+4y
Subject to the constraints
$3 x+2 y \leq 12 \\$ … (i)
$3 x+y \leq 9 \\$ … (ii)
$\begin{aligned} & &x \geq 0, y \geq 0 \end{aligned}$ … (iii)
The feasible region determined by constraints (1) and (2) is graphically represented as

Here it is seen that OABCO is the feasible region and it is bounded. The value of Z at the corner points of the feasible region are represented in tabular form as follows.

Corner Points	$z=70x+4y$
O(0,0)	$7 \times 0+4 \times 0=0$
A(3,0)	$7 \times 3+4 \times 0=21$
B(2,3)	$7 \times 2+4 \times 3=26$ (maximum)
C(0,6)	$7 \times 0+4 \times 6=24$

The maximum value of Z is 26, which is obtained at x = 2 and y = 3.
Thus, 2 units of Product A and 3 units of product B. Should be manufactured by the manufacturer per day in order to maximize the profit.
Also, the maximum daily profit of the manufacturer is Rs.26.

Linear Programming exercise 29.4 question 54

Answer: The total minimum cost of the fertilizers is Rs.1980
Hint:
Use property of LPP
Given:
A consists of 12% of nitrogen and 5% of phosphoric acid at costs Rs.10/kg and B consists of 4% of nitrogen and 5% of phosphoric acid at costs Rs.8/kg
Solution:
The given information can be tabulated as follows.

Fertilizer	Nitrogen	Phosphoric Acid	Cost/Kg
A	12%	5%	10
B	4%	5%	8

Let the requirement of fertilizer A by the farmer be x kg and that of B be y kg
It is given that farmer requires at least 12 kg of nitrogen and 12 kg of phosphoric acid for his crops.
The in equations thus formed based on the given information are as follows.
$\begin{aligned} &\frac{12}{100} x+\frac{4}{100} y \geq 12 \\ &\Rightarrow 12 x+4 y \geq 1200 \\ &\Rightarrow 3 x+y \geq 300 \end{aligned}$
Also,
$\begin{aligned} &\frac{5}{100} x+\frac{5}{100} y \geq 12 \\ &\Rightarrow 5 x+5 y \geq 1200 \\ &\Rightarrow x+y \geq 240 \end{aligned}$
Total cost of the fertilizer Z = Rs.(10x+8y)
Therefore, the mathematical formulation of the given LPP is
Minimize Z=10x+8y
Subject to the constraints
$3 x+y \geq 300 \\$ … (i)
$x+y \geq 240 \\$ … (ii)
$\begin{aligned} & &x \geq 0, y \geq 0 \end{aligned}$ … (iii)
The feasible region determined by constraints (1) to (3) is graphically represented as

$X+Y=240 \\$
$3 X+Y=300 \quad 10 X+8 Y=1980$
Hence, it is seen that the feasible region is unbounded. The value of Z at the corner points of the feasible region are represented in tabular form as

Corner Points	$z=10x+8y$
(0,300)	$10 \times 0+8 \times 300=2400$
(30,210)	$10 \times 30+8 \times 210=1980$ (minimum)
B(240,0)	$10 \times 2400+8 \times 10=2400$

The open half plane determined by $10 x+8 y \leq 1980$ has no point in common with the feasible region. So, the minimum value of Z is 1980.
The minimum value of Z is 1980, which is obtained at x = 30 and y = 210.
Thus, the minimum requirement of fertilizer of type A will be 30kg and that of type B will be 210 kg.
Also, the total minimum cost of the fertilizers is Rs.1980.

Linear Programming exercise 29.4 question 55

Answer: Maximum Z=100x+300y is the required LPP.
Hint:
Use property of LPP
Given:
The maximum number of hours available per day is16, If the profit on a necklace is Rs.100 and the bracelet is Rs.300
Solution:
Let the number of necklaces manufacturer be x and the number of bracelets manufacture be y.
Since the total number of item are at most 24.
$x+y \leq 24$ … (i)
Bracelets take 1 hour to manufacture and necklaces take half an hour to manufacture.
X item take x hour to manufacture and y item take y/2 hours to manufacture and maximum time available is 16 hours
Therefore,
$\frac{x}{2}+y \leq 16$ … (ii)
The profit on one necklace is Rs.100 and the profit on one bracelet is Rs.300.
Let the profit be Z. Now we wish to maximize the profit.
So,
Maximize Z=100x + 300y … (iii)
So,
$\begin{aligned} &x+y \leq 24 \\ &\frac{x}{2}+y \leq 16 \end{aligned}$
Maximize Z = 100x + 300y is required LPP

X+y=24

Corner points	Max Z
(0,16)	1800
(24,0)	2400
(16,18)	7000

The maximum value of Z = 7000 at (16,18)

RD Sharma class 12 solutions Linear Programming 29.4 is the ideal NCERT solutions to have for board exam preparations. Chapter 29 of the NCERT is titled Linear Programming and the concepts covered are formulating problems with different conditions, objective function, constraints, optimisation problem, feasible region, Bounded and unbounded region and are based on diet problems, manufacturing problems, and transportation problems. Exercise 29.4 has 55 questions that cover concepts from the entire chapter. The RD Sharma class 12th exercise 29.4 will help you solve all these questions and improve your performance.

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