第 12 类 RD Sharma 解决方案 – 第 30 章线性规划 – 练习 30.4 |设置 1

问题 1. 如果一个年轻人以 25 公里/小时的速度开车，他必须每公里花费 2 卢比购买汽油。如果他以 40 公里/小时的更快速度驾驶它，汽油成本将增加到 5 卢比/每公里。他有 100 卢比可用于购买汽油和一小时内的旅行。将其表示为 LPP 并解决相同的问题。

解决方案：

Assume the young man drives x km at a speed of 25 km/hr and y km at a speed of 40 km/hr. 

Thus, x, y ≥ 0

Given: If he spends Rs 2/km if he drives at a speed of 25 km/hr 

Rs 5/km if he drives at a speed of 40 km/hr. 

Hence, 

Money spent by him when he traveled x km and y km will be Rs 2x and Rs 5y respectively.

Thus, here it is given that he can spend Rs100 maximum.

Therefore, 

2x + 5y ≤ 100

So, for traveling with a speed of 25 km/hr the time that he will be spending = x/25 hr 

So, for traveling with a speed of 40km/hr the time that he will be spending = y/40 hr 

And also, 

The time he is having is 1 hour.

x/25 + y/40 ≤ 1

40x + 25y ≤ 1000

The distance covered Z = x + y 

That is to be maximized.

Therefore, the formula of the given linear programming problem is 

Max Z = x + y 

Subject to constraints

2x + 5y ≤ 100

40x + 25y ≤ 1000

x, y ≥ 0

First we will convert in-equations as follows:

2x + 5y = 100

40x + 25y = 1000

x = 0 and y = 0.

The area shown by 2x + 5y ≤ 100

The line 2x + 5y = 100 connects the coordinate axes at A(50, 0) and B(0, 20) respectively. 

After connecting these points, we will get the line 2x + 5y = 100. 

Thus, (0, 0) assure the 2x + 5y = 100. 

As, the area that consists the origin shows the solution set of the in-equation 2x + 5y ≤ 100

The area shown by 40x + 25y ≤ 1000

The line 40x + 25y = 1000 connects the coordinate axes at C(25, 0) and D(0, 40) respectively. 

After connecting these points, we will get the line 2x + y = 12. 

Thus, (0, 0) assure the 40x + 25y = 1000. 

As, the area that consists of the origin shows the solution set of the in-equation 40x + 25y ≤ 1000

The area shown by x ≥ 0, 

y ≥ 0 :

Since, all the points in the first quadrant assure these in-equations. 

As, the first quadrant is the area shown by the in-equations x ≥ 0 

and 

y ≥ 0.

The suitable area is determined by the system of constraints

2x + 5y ≤ 100, 

40x + 25y ≤ 1000, 

x ≥ 0 

and 

y ≥ 0 are as follows

The corner points are O(0, 0), B(0, 20), E(50/3, 40/3), and C(25, 0). The value of Z at these corner points are as follows:

The maximum value of Z is 30 that is acquired at E.

Therefore, 

The maximum distance traveled by the young man is 30 kms, if he drives 50/3 km at a speed of 25 km/hr and 40/3 km at a speed of 40 km/hr.

问题 2. 制造商在其工厂安装了三台机器。机器 I 和 II 最多可以运行 12 小时，而机器 III 必须每天至少运行 5 小时。他只生产两种产品，每一种都需要使用三台机器。下表给出了在三台机器上生产每件物品所需的小时数：

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

他在商品 A 上获利 6.00 卢比，在商品 B 上获利 4.00 卢比。假设他可以卖掉他生产的所有产品，他应该生产每件商品多少才能使利润最大化？确定他的最大利润。以数学方式制定此 LPP，然后求解。

解决方案：

Assume x units of item A and y units of item B be produced. 

Therefore, x, y ≥ 0.

Given:

Machines I and II are efficient of being operated for at most 12 hours whereas Machine III should operate at least for 5 hours a day.

As per in the question, 

The constraints are

x + 2y ≤ 12

2x + y ≤ 12

x + 5/4

y ≥ 5

He makes a profit of Rs 6.00 on item A and Rs. 4.00 on item B. Profit made by him in manufacturing x items of A and y items of B is 6x + 4y.

Total profit is Z 

Z = 6x + 4y that is to be maximized

Therefore, the formula of the given linear programming problem is 

Max Z = 6x + 4y, 

Subject to constraints

x + 2y ≤ 12

2x + y ≤ 12

x + 5/4

y ≥ 5x, 

y ≥ 0

First, we have to convert the in-equations into equations as follows:

x + 2y = 12, 

2x + y = 12, 

x + 5/4

y = 5, 

x = 0 and 

y = 0.

The area shown by x + 2y ≤ 12

The line x + 2y = 12 connects the coordinate axes at A(12, 0) and B(0, 6) respectively. 

After connecting these points, we will get the line x + y = 12. 

As (0, 0) assure the x + 2y = 12. 

As, the area that consists the origin shows the solution set of the in-equation x + 2y ≤ 12

The area shown by 2x + y ≤ 12

The line 2x + y = 12 connects the coordinate axes at C(6, 0) and D(0, 12) respectively. 

After connecting these points, we will connect the line 2x + y = 12. 

As (0, 0) assure the 2x + y = 12. 

As, the area that consists the origin shown the solution set of the in-equation 2x + y ≤ 12

The area shown by x + 5/4y ≥ 5

The line x + 5/4y ≥ 5 connects the coordinate axes at E(5, 0) and F(0, 4) respectively. 

After connecting these points, 

We will get the line x + 5/4y = 5. 

As (0, 0) assure the x + 5/4y ≥ 5. 

As, the area that does not consist the origin shows the solution set of the in-equation x + 5/4y ≥ 5

The area shown by x ≥ 0, 

y ≥ 0 :

As all the points in the first quadrant assure these in-equations. 

Thus, the first quadrant is the area shown by the in-equations x ≥ 0 

and 

y ≥ 0.

The suitable area determined by the system of constraints

x + 2y ≤ 12, 

2x + y ≤ 12, 

x + 5/4y ≥ 5, x, 

y ≥ 0 are as follows.

Therefore, the maximum profit is of Rs 40 gained when 4 units each of item A and B are manufactured.

The corner points are D(0, 6), I(4, 4), C(6, 0), G(5, 0), and H(0, 4). 

The values of Z at these corner points are as follows:

Hence, the maximum value of Z is 40 which is acquired at I(4, 4).

问题 3. 两名裁缝 A 和 B 每天分别赚取 15 卢比和 20 卢比。 A每天可以缝6件衬衫和4条裤子，而B每天可以缝10件衬衫和4条裤子。如果希望以最低的人工成本生产至少 60 件衬衫和 32 条裤子，每个人需要工作多少天？

解决方案：

Assume thet tailor A work for x days and tailor B work for y days.

In a day; A can stitch 6 shirts and 4 pants whereas B can stitch 10 shirts and 4 pants.

Therefore, in x days; A can stitch 6x shirts and 4x pants whereas in y days B can stitch 10y shirts and 4y pants.

Given: The minimum need of the shirt and pants are respectively 60 and 32.

Therefore,

6x + 10y ≥ 60

4x + 4y ≥ 32

So further, it is given that A and B earn Rs 15 and Rs 20 per day respectively. 

Therefore, A earn Rs 15x and B earns Rs 20y.

Assume the total cost is Z

Z = 15x + 20y

As days cannot be negative.

x, y ≥ 0 (always)

Min Z = 15x + 20y

Subject to constraints

6x + 10y ≥ 60

4x + 4y ≥ 32

x, y ≥ 0

First we will convert in-equations into equations as follows:

6x + 10y = 60, 

4x + 4y = 32, 

x = 0, 

y = 0

The area shown by 6x + 10y ≥ 60

The line 6x + 10y = 60 connects the coordinate axes at A(10,0) and B(0,6) respectively. 

After connecting these points we will get the line 6x + 10y = 60. 

As, (0, 0) assure the 6x + 10y ≥ 60. 

Since, the area that does not consist the origin shown the solution set of the in-equation 6x + 10y ≥ 60

Area shown by 4x + 4y ≥ 32

The line 4x + 4y = 32 connects the coordinate axes at C(8,0) and D(0,8) respectively. 

After connecting these points we will get the line 4x + 4y = 32. 

As, (0, 0) assure the 4x + 4y ≥ 32. 

As, the area that does not consist the origin shows the solution set of the in-equation 4x + 4y ≥ 32.

The area shown by x ≥ 0, 

y ≥ 0 :

Since, all the points in the first quadrant assure these in-equations. 

As, the first quadrant is the area shown by the in-equations x ≥ 0 

and 

y ≥ 0.

The suitable area determined by the system of constraints 6x + 10y ≥ 60, 

4x + 4y ≥ 32

x, y ≥ 0 are as follows.

The corner points are D(0, 8), E(5, 3), A(10, 0). 

The values of Z at these corner points are as follows:

Corner Points	Z = 15x + 20y
D	160
E	135
A	150

The minimum value of Z is 135 which is acquired at E(5, 3).

Therefore, For minimum labor cost, A must work for 5 days and B must work for 3 days.

问题 4. 一家工厂生产两种类型的螺钉，A 和 B，每种类型都需要使用两台机器——自动和手动。制造一包螺丝'A'，自动机4分钟，手动机6分钟，自动机6分钟，手动机3分钟制造一包螺丝'乙'。每台机器在任何一天最多可用 4 小时。制造商可以以 70 P 的利润出售一包螺丝“A”，以 1 卢比的利润出售螺丝“B”。假设他可以出售所有他能制造的螺丝，工厂应该每种类型多少包所有者为了最大化他的利润而在一天内生产？确定最大利润。

解决方案：

Assume the factory manufacture x screws of type A and y screws of type B on each day,

Thus, 

x, y ≥  0.

Given information in a table as;

	Score A	Score B	Score C
Automatic Machine (min)	4	6	4 × 60 = 240
Hand Operated Machine (min)	6	3	4 × 60 = 240

4x + 6y ≤ 240

6x + 3y ≤ 240

The manufacturer can sell a package of screws ‘A’ at a profit of Rs 0.7 and screws ‘B’ at a profit of Re 1.

Assume the total profit is Z, 

Z = 0.7x + 1y

Therefore, the formula of the given linear programming problem is 

Maximize Z = 0.7x + 1y

Subject to the constraints,

4x + 6y ≤ 240

6x + 3y ≤ 240

x, y ≥ 0

First, we will convert the in-equations into equations as follows:

4x + 6y = 240, 6x + 3y = 240, x = 0, y = 0.

Area shown by 4x + 6y ≥  240

The line 4x + 6y = 240 connects the coordinate axes at A(60, 0) and B(0, 40) respectively.

After connecting these points we will get the line 4x + 6y = 240. 

As, (0, 0) assure the 4x + 6y ≥  240. 

As, the area that consists the origin showss the solution set of the in-equation 4x + 6y ≥  240.

The area shown by 6x + 3y ≥  240

The line 6x + 3y = 240 connects the coordinate axes at C(40, 0) and d(0, 80) respectively. 

After connecting these points we will get the line 6x + 3y = 240. 

As, (0, 0) assure the 6x + 3y ≥  240. 

Since, the area that consist the origin shows the solution set of the in-equation 6x + 3y ≥  240.

The area shown by 

x ≥  0, 

y ≥ 0 :

As, all the points in the first quadrant assure these in-equations. 

Since, the first quadrant is the area shown by the in-equations 

x ≥  0 

and 

y ≥  0.

The suitable are determined by the system of constraints 

4x + 6y ≤ 240, 

6x + 3y ≤ 240, 

x ≥  0,

y ≥  0 are as follows.

The corner points are C(40, 0), E(30, 20), B(0, 40). The values of Z at these corner points are as follows

Corner Point	Z = 7x + 10y
C(40, 0)	280
E(30, 20)	410
B(0, 40)	400

The maximum value of Z is 410 at (30, 20).

Therefore, the factory must manufacture 30 packages of screws A and 20 packages 

of screws b to get the maximum profit of Rs 410.

问题 5. 一家公司生产两种类型的皮带，比如 A 型和 B 型。皮带 A 质量上乘，皮带 B 质量低。每种皮带的利润分别为每条皮带 2 卢比和 1.50 卢比。每条 A 型皮带所需的时间是 B 型皮带所需时间的两倍。如果所有皮带都是 B 型皮带，该公司每天可以生产 1000 条皮带。但皮革供应仅够每天 800 条皮带（A 和 B 加起来）。腰带 A 需要花式搭扣，每天只有 400 个花式搭扣可用。对于 B 型皮带，每天只能提供 700 个搭扣。公司应该如何生产这两种类型的皮带才能获得最大的整体利润？

解决方案：

Assume the company manufactures x belts of types A and y belts of type B. 

Number of belts cannot be negative. 

Thus, 

x, y ≥ 0 (always)

It is given that leather is enough only for 800 belts per day (both A and B combined).

Thus,

x + y ≤ 800

It is given that the rate of manufacture of belts of type B is 1000 per day. 

Therefore, the time taken to produce y belts of type B is y/1000.

And, 

As each belt of type A needs twice as much time as a belt of type B, the rate of production of belts of type A is 500 per day and thus, total time taken to produce x belts of type A is x/500 

Therefore, 

We have,

x/500 + y/1000 ≤ 1

Or, 

2x + y ≤ 1000

Belt A needs fancy buckle and only 400 fancy buckles are accessible for this per day.

x ≤ 400

For Belt of type B only 700 buckles are accessible per day.

y ≤ 700

Profits on each type of belt are Rs 2 and Rs 1.50 per belt, respectively. 

Thus, profit gained on x belts of type A and y belts of type B is Rs 2x and Rs 1.50y respectively. 

Therefore, the total profit would be Rs(2x + 1.50y). 

Assume Z denote the total profit

Z = 2x + 1.50y

Therefore, the mathematical formulation of the given linear programming problem is;

Max Z = 2x + 1.50y subject to

x + y ≤ 800

2x + y ≤ 1000

x ≤ 400

y ≤ 700

First we will convert these in-equations into equations as follows:

x + y = 800

2x + y = 1000

x = 400

y = 700

The area shown by x + y = 800

The line x + y = 800 meets the coordinate axes at A(800, 0) and B(0, 800) respectively. 

After connecting these points we obtain the line x + y = 800. 

As (0, 0) assure the x + y ≥ 800. 

As, the area that consist the origin shows the solution set of the in-equation x + y ≥ 800.

The area shown by 2x + y ≥  1000

The line 2x + y = 1000 connects the coordinate axes at C(500, 0) and D(0, 1000) respectively. 

After connecting these points we will get the line 2x + y = 1000. 

As, (0, 0) assure the 2x + y ≥ 1000. 

As, the area that consists the origin shows the solution set of the in-equation 2x + y ≥ 1000.

The area shown by x ≤ 400

The line x = 400 will pass through (400, 0). 

The area to the left of the line x = 400 will assure the in-equation x ≤ 400

The area shown by y ≤ 700

The line y = 700 will pass through (0, 700). 

The area to the left of the line y = 700 will assure the in-equation y ≤ 700.

The area shown by x ≥ 0, 

y ≥ 0 :

As, all the points in the first quadrant assure these in-equations. 

As, the first quadrant is the area shown by the in-equations x ≥ 0 

and y ≥ 0.

The suitable area determined by the system of constraints x + y ≤ 800, 

2x + y ≤ 1000, 

x ≤ 400,

y ≤ 700

The corner points are F(0, 700), G(200, 600), H(400, 200), E(400, 0). 

The values of Z at these corner points are as follows

Corner Points	Z = 2x + 1.5y
F(0, 700)	1050
G(200, 600)	1300
H(400, 200)	1100
E(400, 0)	800

The maximum value of Z is 1300 which is attained at G(200, 600).

Therefore, the maximum profit acquires is Rs 1300 when 200 belts of type A and 600 belts of type B are produced.

问题 6. 一家小型制造商雇佣了 5 名技术人员和 10 名半技术人员，并制作了两种质量豪华型和普通型的产品。制作豪华模型需要2小时。由技术熟练的人工作 2 小时。由一个半熟练的人工作。普通模型需要技术人员1小时和3小时。由一个半熟练的人。根据工会规定，任何人每天的工作时间不得超过 8 小时。制造商在豪华车型上的明确利润为 15 卢比，普通车型为 10 卢比。每种类型应该生产多少才能使他的每日总利润最大化。

解决方案：

Assume x articles of deluxe model and y articles of an ordinary model be made.

Numbers cannot be negative.

Therefore,

x, y ≥ 0 (alwaya)

As per the question, 

The profit on each model of deluxe and ordinary type model are Rs 15 and Rs 10.

Thus, profits on x deluxe model and y ordinary models are 15x and 10y.

Assume Z be total profit,

So,

Z = 15x + 10y

As, the making of a deluxe and ordinary model needs 2 hrs. and 1 hr work by skilled men, 

so, x deluxe and y ordinary models needs 2x and y hours of skilled men but time needed 

by skilled men is 5 × 8 = 40 hours.

Thus,

2x + y ≤ 40 {First Constraint}

As, the making of a deluxe and ordinary model needs 2 hrs. and 3 hrs work by semi-skilled men, 

Since, x deluxe and y ordinary models need 2x and 3y hours of skilled men but time 

need by skilled men is 10 × 8 = 80 hours.

As,

2x + 3y ≤ 80 {Second constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 15x + 10y

Subject to constraints,

2x + y ≤ 40

2x + 3y ≤ 80

x, y ≥ 0

Area 2x + y ≤ 40: 

Line 2x + 4y = 40 connects axes at 

A₁(20, 0), B₁(0, 40) respectively. 

The area having origin shows 2x + 3y ≤ 40 as (0, 0) assure 2x + y ≤ 40

Area 2x + 3y ≤ 80:

Line 2x + 3y = 80 connects axes at 

A₂(40, 0), B₂(0, 80/3) respectively. 

The area having origin shows 2x + 3y ≤ 80.

The corner points are A₁(20, 0), P(10, 20), B₂(0, 80/3).

The value of Z = 15x + 10y at these corner points are

Corner Points	Z = 15x + 10y
A1	300
P	350
B2	800/3

The maximum value of Z is 350 which is acquired at P(10, 20).

Therefore, maximum profit is attained when 10 units of deluxe model and 20 units of ordinary model is produced.

问题 7. 一家制造商生产 A 型和 B 型两种茶杯。制造需要三台机器，机器上每个杯子所需的时间（以分钟为单位）如下所示：

Item	Number of hours required by the machine
	I	II	III
A	1	2	1
B	2	1

每台机器每天最多可使用 6 小时。如果 A 杯的利润是 75 派萨，B 杯的利润是 50 派萨，则说明一天应该生产 15 个 A 型茶杯和 30 个 B 型茶杯，以获得最大的利润。

解决方案：

Assume the needed number of tea cups of Type A and B are x and y respectively.

As, the profit on each cup A is 75 paise and that on each cup B is 50 paise. 

So, the profit on x tea cup of type A and y tea cup of type B are 75x and 50y respectively.

Assume total profit on tea cups be Z, so

Z = 75x + 50y

Since, every tea cup of type A and B needs to work machine I for 12 and 6 minutes respectively. 

So, x tea cups of Type A and y tea cups of Type B needs to work on machine I for 12x and 6y minutes.

Total time available on machine I is 6 × 60 = 360 minutes. 

As

12x + 6y ≤ 360 {First Constraint}

Thus, every tea cup of type A and B need to work machine II for 18 and 0 minutes respectively. 

So, x tea cups of Type A and y tea cups of Type B need to work on machine IIII for

18x and 0y minutes respectively.

Total time available on machine I is 6 × 60 = 360 minutes. 

So,

18x + 0y ≥ 360

x ≤ 20 {Second Constraint}

As, every tea cup of type A and B need to work machine III for 6 and 9 minutes respectively 

So, x tea cups of Type A and y tea cups of Type B need to work on machine I 

for 6x and 9y minutes respectively.

Total time available on machine I is 6 × 60 = 360 minutes. 

So,

6x + 9y ≤ 360 {Third Constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 75x + 50y

Subject to constraints,

12x + 6y ≤ 360

x ≤ 20

6x + 9y ≤ 360

x, y ≥ 0 {Since production of tea cups can not be less than zero}

Area 12x + 6y ≤ 360: line 12x + 6y = 360 connects axes at A(30, 0), B(0, 60) 

The area having origin shows 12x + 6y ≤ 360 as (0, 0) assure 12x + 6y ≤ 360

Area x ≤ 20: line x = 20 is parallel to y-axis and connects x-axes at C(20, 0). 

The area having origin shows x ≤ 20

As (0, 0) assure x ≤ 20.

Area 6x + 9y ≤ 360: line 6x + 9y = 360 connects axes at E(60, 0), F(0, 40) respectively. 

The area having origin shows 6x + 9y ≤ 360 as (0, 0) assure 6x + 9y ≤ 360.

Area x, y ≥ 0: it shows the first quadrant.

The shaded area is the suitable area determined by the constraints,

12x + 6y ≤ 360

x ≤ 20

6x + 9y ≤ 360

x, y ≥ 0

The corner points are F(0, 40), G(15, 30), H(20, 20), C(20, 0).

The values of Z at these corner points are as follows

Corner Points	Z = 75x + 50y
F	2000
G	2624
H	2500
C	1500

Here Z is maximum at G(15, 30).

Hence, 15 teacups of Type A and 30 tea cups of Type B are required to maximize the profit.

问题 8. 一位工厂主为他的工厂购买了两种类型的机器 A 和 B。机器的要求和限制如下：

Item	Number of hours required by the machine
	I	II	III
A	1	2	1
B	2	1	$\frac{5}{4}$

他拥有7600平方米的可用面积和72名可以操作机器的技术人员。他应该购买每种类型的机器多少台才能使日产量最大化？

解决方案：

Assume needed number of machine A and B are x and y respectively.

Thus, products of each machine A and B are 60 and 40 units daily respectively. 

As, production by x number of machine A and y number of machine B are 60x and 40y respectively.

Assume Z is total output daily, 

So,

Z = 60x + 40y

As, every machine of type A and B needs 1000sq. m and 1200 s. m area so, 

x machine of type A and y machine of type B require 1000x and 1200y sq. m area but,

Total available area for machine is 7600 sq. m. 

So,

1000x + 1200y ≤ 7600

or, 

5x + 6y ≤ 38. {First Constraint}

As, every machine of type A and B needs 12 men and 8 men to work respectively. 

So, x machine of type A and y machine of type B need 12x and 8y men to work respectively.

But total men available for work are 72.

So,

12x + 8y ≤ 72

3x + 2y ≤ 18 {Second Constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 50x + 40y

Subject to constraints,

5x + 6y ≤ 38

3x + 2y ≤ 18

x, y ≥ 0 {Since number of machines can not be less than zero}

Area 5x + 6y ≤ 38: line 5x + 6y = 38 connects the axes at A(38/5, 0), B(0, 19/3) respectively.

Area having the origin shows 5x + 6y ≤ 38 as origin satisfies 5x + 6y ≤ 38

Area 3x + 2y ≤ 18: line 3x + 2y = 18 connects the axes at C(6, 0), D(0, 9) respectively.

Area having the origin shows 3x + 2y ≤ 18 as origin assure 3x + 2y ≤ 18.

Area x, y ≥ 0: it represents the first quadrant.

Shaded region shows the suitable area.

The corner points are O(0, 0), B(0, 19/3), E(4, 3), C(6, 0).

Therefore, the values of Z at these corner points are as follows:

Corner Points	Z = 60x + 40y
O	0
B	253.3
E	360
C	360

The maximum value of Z is 360 which is acquired at E(4, 3), C(6, 0).

Therefore, the maximum output is Rs 360 archive when 4 units of type A and 

3 units of type B or 6 units of type A and 0 units of type B are manufactured.

问题 9. 一家公司生产两种需要黄金和白银的商品 A 和 B。 A 型的每个单位需要 3 克银和 1 克金，而 B 型需要 1 克银和 2 克金。该公司可生产 9 克白银和 8 克黄金。如果 A 类的每个单位带来 40 卢比的利润，B 类的每个单位带来 50 卢比的利润，请找出公司应该生产的每种类型的单位数量以使利润最大化。最大利润是多少？

解决方案：

Assume needed number of goods A and B are x and y respectively.

As, profits of each A and B are Rs. 40 and Rs. 50 respectively. 

Thus, profits on x number of type A and y number of type B are 40x and 50y respectively.

Assume Z is total output daily, 

So,

Z = 40x + 50y

So, every A and B needs 3 grams and 1 gram of silver respectively. 

Thus, x of type A and y of type B need 3x and y of silver respectively. 

But,

Total silver available is 9 grams. So,

3x + y 

9 {First Constraint}

So, every A and B needs 1 gram and 2 grams of gold respectively. 

Thus, x of type A and y of type B require x and 2y respectively.

But total gold available is 8 grams.

Thus,

x + 2y ≤ 8 {Second Constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 40x + 50y

Subject to constraints,

3x + y ≤ 9

x + 2y ≤ 8

x, y ≥ 0 {Since production of A and B can not be less than zero}

Area 3x + y ≤ 9: line 3x + y = 9 connects the axes at A(3, 0), B(0, 9) respectively.

The area having the origin shows 3x + y ≤ 9

As, origin assure 3x + y ≤ 9.

Area x + 2y ≤ 8: line x + 2y = 8 connects the axes at C(8, 0), D(0, 4) respectively.

Area having the origin shows x + 2y ≤ 8 as origin assure x + 2y ≤ 8.

Area x, y ≥ 0: it shows the first quadrant.

The corner points are O(0, 0), D(0, 4), E(2, 3), A(3, 0)

The values of Z at these corner points are as follows

Corner Points	Z = 40x + 50y
O	0
D	200
E	230
A	120

The maximum value of Z is 230 which is obtained at E(2, 3).

Therefore, the maximum profit is of Rs 230 when 2 units of Type A 3 units of Type B are produced.

问题 10. 家具制造商生产两种产品：椅子和桌子。这些产品的加工在两台机器 A 和 B 上完成。椅子需要在机器 A 上 2 小时，在机器 B 上需要 6 小时。在机器 A 上需要 4 小时，在机器 B 上需要 2 小时。每人需要 16 小时 0 时间机器 A 上可用一天，机器 B 上可用 30 小时。制造商从椅子和桌子上获得的利润分别为 3 卢比和 5 卢比。借助图表找出这两种产品的每日产量应该是多少，以使他的利润最大化。

解决方案：

Assume daily production of chairs and tables be x and y respectively.

So, profits of every chair and table is Rs. 3 and Rs. 5 respectively. 

Thus, profits on x number of type A and y number of type B are 3x and 5y respectively.

Assume Z is total output daily, 

So,

Z = 3x + 5y

As, every chair and table needs 2 hrs and 3 hrs on machine A respectively. 

So, x number of chair and y number of table need 2x and 4y hrs on machine A respectively. 

But,

Total time available on Machine A is 16 hours. 

So,

2x + 3y ≤ 16

x + 2y ≤ 8 {First Constraint}

So, every chair and table needs 6 hrs and 2 hrs on machine B respectively. 

Since, x number of chair and y number of table need 6x and 2y hrs on machine B respectively. 

But,

Total time need on Machine B is 30 hours. So,

6x + 2y ≤ 30

3x + y ≤ 15 {Second Constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 3x + 5y

Subject to constraints,

x + 2y ≤ 8

3x + y ≤ 15

x, y ≥ 0 {Since production of chairs and tables can not be less than zero}

Area x + 2y ≤ 8: line x + 2y = 8 connects the axes at A(8, 0), B(0, 4) respectively.

Area having the origin shows x + 2y ≤ 8

As, origin assure x + 2y ≤ 8.

Area 3x + y ≤ 15: line 3x + y = 15 connects the axes at C(5, 0), D(0, 15) respectively.

Area having the origin shows 3x + y ≤ 15 as origin assure 3x + y ≤ 15

Area x, y ≥ 0: it shows the first quadrant.

The corner points are O(0, 0), B(0, 4), E(22/5, 9/5), and C(5, 0).

The values of Z at these corner points are as follows,

Corner Points	Z = 3x + 5y
O	0
B	20
E	22.2
C	15

The maximum value of Z is 22.2 which is obtained at E(22/5, 9/5).

Therefore, the maximum profit of Rs 22.2 when 22/5 units of chair and 9/5 units of table are produced.

问题 11. 一家家具制造公司计划生产两种产品：椅子和桌子。从其可用资源包括 400 平方英尺到柚木和 450 个工时。众所周知，制作一把椅子需要 5 平方英尺的木材和 10 个工时，利润为 45 卢比，而每张桌子使用 20 平方英尺的木材和 25 个工时，利润为 80 卢比。公司应该生产多少个产品才能使利润最大化？

解决方案：

Assume needed production of chairs and tables be x and y respectively.

As, profits of every chair and table is Rs. 45 and Rs. 80 respectively. 

Thus, profits on x number of type A and y number of type B are 45x and 80y respectively.

Assume Z is total output daily, 

So, Z = 45x + 80y

As, every chair and table needs 5 sq. ft and 80 sq. ft of wood respectively. 

As, x number of chair and y number of table need 5x and 80y sq. ft of wood respectively. 

But, 400 sq. ft of wood is accessible. 

So, 5x + 80y ≤ 400

x + 4y ≤ 80 {First Constraint}

As, every chair and table needs 10 and 25 men – hours respectively. 

As, x number of chair and y number of table need 10x and 25y men – hours respectively. 

But, 

Only 450 hours are available . 

So,

10x + 25y ≤ 450

2x + 5y ≤ 90 {Second Constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 45x + 80y

Subject to constraints,

x + 4y ≤ 80

2x + 5y ≤ 90

x, y ≥ 0 {Since production of chairs and tables can not be less than zero}

Area x + 4y ≤ 80: line x + 4y = 80 connects the axes at A(80, 0), B(0, 20) respectively.

Area having the origin shows x + 4y ≤ 80 as origin assure x + 4y ≤ 80

Area 2x + 5y ≤ 90: line 2x + 5y = 90 connects the axes at C(45, 0), D(0, 20) respectively.

Area having the origin shows 2x + 5y ≤ 90

As origin assure 2x + 5y ≤ 90

Area x, y ≥ 0: it shows the first quadrant.

The corner points are O(0, 0), D(0, 18), C(45, 0).

The values of Z at these corner points are as follows:

Corner Points	Z = 45x + 80y
O	0
D	1440
C	2025

The maximum value of Z is 2025 which is obtained at C(45, 0).

Therefore, the maximum profit of Rs 2025 is attained when 45 units of chairs and no units of tables are produced.

问题 12。一家公司生产两种产品 A 和 B。每种产品在两台机器 M ₁和 M ₂上加工。产品 A 在 M ₁和 8 分钟上需要 4 分钟的处理时间。在 M ₂上；产品 B 在 M ₁和 4 分钟上需要 4 分钟。在 M ₂上。机器 M ₁的可用时间不超过 8 小时 20 分钟。而机器 M ₂可用 10 小时。在任何工作日。产品 A 和 B 的销售利润分别为 3 卢比和 4 卢比。将问题表述为线性规划问题，并找出公司每天应该生产多少每种类型的产品才能获得最大利润并以图形方式解决它。

解决方案：

Assume needed production of product A and B be x and y respectively.

As, profit on every product A and B are Rs. 3 and Rs. 4 respectively. 

Thus, profits on x number of type A and y number of type B are 3x and 4y respectively.

Assume Z is total output daily, 

So,

Z = 3x + 4y

Thus, every A and B needs 4 minutes each on machine M₁. 

As, x of type A and y of type B need 4x and 4y minutes respectively. 

But,

Total time accessible on machine M₁ is 8 hours 20 minutes = 500 minutes.

Thus,

4x + 4y ≤ 500

x + y ≤ 125 {First Constraint}

So, every A and B needs 8 minutes and 4 minutes on machine M₂ respectively. 

Thus, x of type A and y of type B need 8x and 4y minutes respectively. 

But,

Total time accessible on machine M₁ is 10 hours = 600 minutes.

Thus,

8x + 4y ≤ 600

2x + y ≤ 150 {Second Constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 3x + 4y

Subject to constraints,

x + y ≤ 125

2x + y ≤ 150

x, y ≥ 0 {Since production of A and B can not be less than zero}

Area x + y ≤ 125: line x + y = 125 connects the axes at A(125, 0), B(0, 125) respectively.

Area having the origin shows x + y ≤ 125 as origin assure x + y ≤ 125.

Area 2x + y ≤ 150: line 2x + y = 150 connects the axes at C(75, 0), D(0, 150) respectively.

Area having the origin shows 2x + y ≤ 150 as origin assure 2x + y ≤ 150.

Area x, y ≥ 0: it shows the first quadrant.

The corner points are O(0, 0), B(0, 125), E(25, 100), and C(75, 0).

The values of Z at these corner points are as follows:

Corner Points	Z = 3x + 4y
O	0
B	500
E	475
C	225

The maximum value of Z is 500 which is obtained at B(0, 125).

Therefore, the maximum profit is Rs 500 attained when no units of product A and 

125 units of product B are manufactured.

问题 13. 一家生产 A 和 B 两种电气产品的公司，每单位 A 的利润为 20 卢比，每单位 B 的利润为 30 卢比。每单位 A 需要 3 台电机和 4 台变压器，每单位 B需要2个电机和4个变压器。这些产品每月的总供应量仅限于 210 台电机和 300 台变压器。 B 型是需要稳压器的出口型号，其供应限制为每月 65 台。制定线性规划问题以获得最大利润并以图形方式求解。

解决方案：

Assume x units of item A and y units of item B were manufactured.

Numbers of items cannot be negative. 

Therefore,

x, y ≥ 0 (always)

Given: The information can be tabulated as follows:

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

Thus, after this it is given that type B is an export model, whose supply is restricted to 65 units per month.

So, 

The constraints are

3x + 2y ≤ 210

4x + 4y ≤ 300

y ≤ 65

A and B can make profit of Rs 20 and Rs 30 per unit respectively.

Hence, profit gained from x units of item A and y units of item B is Rs 20x and 30y respectively.

Assume total profit be Z,

So,

Total Profit = Z = 20x + 30y as per the question is to be maximised.

Therefore, the formula of the given linear programming problem is 

Max Z = 20x + 30y

Subject to constraints

3x + 2y ≤ 210

4x + 4y ≤ 300

y ≤ 65

x, y ≥ 0

Area shown by 3x + 2y ≤ 210: The line 3x + 2y = 210 connects the axes at A(70, 0), B(0, 105) respectively.

Area having the origin shows 3x + 2y ≤ 210 as origin assure 3x + 2y ≤ 210.

Area shown by 4x + 4y ≤ 300: The line 4x + 4y = 300 connects the axes at C(75, 0), D(0, 75) respectively.

Area having the origin shows 4x + 4y ≤ 300 as origin assure 4x + 4y ≤ 300

y = 65 is the line passing through the point E(0, 65) and is parallel to x – axis.

Area x,y ≥ 0: it shows the first quadrant.

The corner points are O(0, 0), E(0, 65), G(10, 65), F(60, 15) and A(70, 0).

The values of Z at these corner points are as follows:

Corner Points	Z = 20x + 30y
O	0
E	1950
G	2150
F	1650
A	1400

The maximum value of Z is 2150 which is obtained at G(10, 65).

Therefore, the maximum profit is Rs. 2150 attained when 10 units of item A and 

65 units of item B are manufactured.

问题 14. 一家工厂使用三种不同的资源来制造两种不同的产品，20 单位的资源 A、12 单位的 B 和 16 单位的 C 可用。第一个产品的 1 个单位需要 2、2 和 4 个单位的相应资源，而第二个产品的 1 个单位需要 4、2 和 0 个单位的相应资源。已知第一个产品每单位的利润为 2 个货币单位，第二个产品为 3。 制定线性规划问题。每种产品应该生产多少单位才能使利润最大化？以图形方式解决它。

解决方案：

Assume number of product I and product II are x and y respectively.

As, profits on every product I and II are 2 and 3 monetary unit. 

As, profits on x number of Product I and y number of Product II are 2x and 3y respectively.

Assume Z is total output daily, 

So,

Z = 2x + 3y

Here, every I and II needs 2 and 4 units of resources A. 

So, x units of product I and y units of product II needs 2x and 4y minutes respectively. 

But, maximum available quantity of resources A is 20 units.

So,

2x + 4y ≤ 20

x + 2y ≤ 10 {First Constraint}

So, each I and II needs 2 and 2 units of resources B. 

So, x units of product I and y units of product II needs 2x and 2y minutes respectively. 

But, maximum available quantity of resources A is 12 units.

So,

2x + 2y ≤ 12

x + y ≤ 6 {Second Constraint}

Here, every unit of product I require 4 units of resources C. 

It is not needed by product II. Thus, x units of product I require 4x units of resource C. 

But, maximum available quantity of resources C is 16 units.

Thus,

4x ≤ 16

x ≤ 4 {Third Constraint}

Therefore, the formula of the given linear programming problem is 

Max Z = 2x + 3y

Subject to constraints,

x + 2y ≤ 10

x + y ≤ 6

x ≤ 4

x, y ≥ 0  {Since production for I and II can not be less than zero}

Area shown by x + 2y ≤ 10: The line x + 2y = 10 connects the axes at A(10, 0), B(0, 5) respectively.

Area having the origin shows x + 2y ≤ 10 as origin assure x + 2y ≤ 10.

Area shown by x + y ≤ 6: The line x + y = 6 connects the axes at C(6, 0), D(0, 6) respectively. 

Area having the origin shows x + y ≤ 6 as origin assure x + y ≤ 6

Area x, y ≥ 0: it shows the first quadrant.

The corner points are O(0, 0), B(0, 5), G(2, 4), F(4, 2), and E(4, 0).

The values of Z at these corner points are as follows:

Corner Points	Z = 2x + 3y
O	0
B	15
G	16
F	14
E	8

The maximum value of Z is 16 which is obtained at G (12, 4).

Therefore, the maximum profit is 16 monetary units attained when 2 units of first product and 

4 units of second product were manufactured.

问题 15。出版商以 72.00 卢比的价格出售一本教科书的精装版，以 40.00 卢比的价格出售相同分机的平装版。除了每周 9600.00 卢比的成本外，出版商的成本分别为每本书 56.00 卢比和 28.00 卢比。两种类型都需要 5 分钟的印刷时间，虽然精装本需要 10 分钟装订时间，而平装本只需 2 分钟。印刷和装订操作每周都有 4,800 分钟的可用时间。为了最大化利润，每种类型的书应该生产多少？

解决方案：

Assume the sale of hand cover edition be ‘h’ and that of paperback editions be ‘t’.

SP of a hard cover edition of the textbook = Rs 72

SP of a paperback edition of the textbook = Rs 40

Cost to the publisher for hard cover edition = Rs 56

Cost to the publisher for a paperback edition = Rs 28

Weekly cost to the publisher = Rs 9600

Profit to be maximized, Z = (72 – 56)h + (40 – 28)t – 9600

Z = 16h + 12t – 9600

5(h + t) ≤ 4800

10h + 2t ≤ 4800.

The corner points are O(0, 0), B1(0, 960), E₁ (360, 600) and F₁ (480, 00).

The values of Z at these corner points are as follows:

Corner Points	Z = 16h + 12t – 9600
O	-9600
B	1920
E	3360
F	-1920

The maximum value of Z is 3360 which is obtained at E₁ (360, 600).

Hence, the maximum profit is 3360 which is attained by selling 360 copies of hardcover edition and 

600 copies paperback edition.

问题 16. 一家公司生产 A 和 B 两种尺寸的头痛药。A 尺寸包含 2 粒阿司匹林、5 粒碳酸氢盐和 1 粒可待因；尺寸 B 包含 1 粒阿司匹林、8 粒碳酸氢盐和 66 粒可待因。用户发现，它需要至少 12 粒阿司匹林、7.4 粒碳酸氢盐和 24 粒可待因才能提供立竿见影的效果。以图形方式确定患者需要立即缓解的最少药片数量。还要确定患者消耗的可待因量。

解决方案：

Here the above LPP can be shown as a table below,

	Pill size A	Pill size B
	X	Y
Aspirin	2x	1y	≥ 12
Bicarbonate	5x	8y	≥ 7.4
Codeine	1x	66y	≥ 24
Relief	X	Y	Minimize

Therefore, the formula of the given linear programming problem is 

Max Z = x + y

Subject to constraints,

2x + y  ≥ 12

5x + 8y ≥ 7.4

x + 66y ≥ 24

x, y ≥ 0  {Since production can not be less than zero}

The corner points are B(0, 12), P(5.86, 0.27), E(24, 0)

The values 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 minimum value of Z is 6.13. 

But,

The area is unbounded thus check whether x + y < 6.13

Thus, we can see that it does not has any common area.

So, x = 5.86, y = 0.27

This is the least quantity of pill A and B.

Codeine quantity = x + 66y = 5.86 + (66 × 0.27) = 24(approx).

问题 17. 一家化工公司生产两种化合物 A 和 B。下表给出了每公斤化合物 A 和 B 的成分单位 C 和 D 以及 C 和 D 的最低要求以及每公斤 A 和 B 的成本. 找出以最低成本提供 C 和 D 的 A 和 B 的数量。

Corner Points	Z = 6x + 4y
D	24
I	40
C	36
G	30
H	16

解决方案：

Assume needed quantity of compound A and B are x and y kg.

As, cost of one kg of compound A and B are Rs 4 and Rs 6 per kg. So,

Cost of x kg of compound A and y kg of compound B are Rs 4x and Rs 6y

Respectively.

Assume Z be the total cost of compounds, 

So,

Z = 4x + 6y

As, compound A and B contain 1 and 2 units of ingredient C per kg respectively, 

Thus, 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 ≥ 80 {first constraint}

Hence, compound A and B have 3 and 1 units of ingredient D per kg respectively, 

Hence, x kg of compound A and y kg of compound B have 3x and y units of 

ingredient D respectively but minimum need of ingredient C is 75 units, so,

3x + y ≥ 75 {second constraint}

Therefore, the formula of the given linear programming problem is 

Min Z = 4x + 6y

Subject to constraints,

x + 2y ≥ 80

3x + y ≥ 75

x, y ≥ 0 {Since production can not be less than zero}

Area x + 2y ≥ 80: line x + 2y = 80 connects axes at A(80, 0), B(0, 40) respectively. 

Area not having origin shows x + 2y ≥ 80 as (0, 0) does not assure x + 2y ≥ 80.

Area 3x + y ≥ 75: line 3x + y = 75 connects axes at C(25, 0), D(0, 75) respectively. 

Area not having origin shows 3x + y ≥ 75 as (0, 0) does not assure 3x + y ≥ 75.

Area x,y ≥ 0: it shows first quadrant.

The corner points are D(0, 75), E(14, 33), A(80, 0).

The values at Z at these corner points are as follows:

Corner Point	Z = 4x + 6y
D	450
E	254
A	320

The minimum value of Z is 254 which is obtained at E(14, 33).

Therefore, the minimum cost is Rs 254 attained when 14 units of compound A and 

33 units compound B are produced.