第 12 类 RD Sharma 解决方案 – 第 30 章线性规划 – 练习 30.4 |设置 3

问题 35。一家制造商生产两种产品，A 和 B。产品 A 的售价为每个 200 卢比，制作时间为 1/2 小时。产品 B 的售价为每件 300 卢比，制作时间为 1 小时。有 14 件产品 A 和 16 件产品 B 的永久订单。一个工作周包括 40 小时的生产，每周营业额不得低于 10000 卢比。如果每个产品 A 的利润为 20 卢比产品 B 是 30 卢比，那么每种产品应该生产多少才能使利润最大化？此外，找到最大的利润。

解决方案：

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

Number of units cannot be negative.

Therefore,

x, y ≥ 0 (always)

As per the question, the given information can be shown as

Also, the availability of time is 40 hours and the revenue should be at least Rs 10000.

So, 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

200x + 300y ≥ 100000,

5x + y ≤ 40

x ≥ 14

y ≥ 16

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 respectively.

Total profit = Z = 20x + 30y which is to be maximised

Therefore, the formula of the given linear programming problem is

Max Z = 20x + 30y

Subject to constraints

2x + 3y ≥ 100

x + 2y ≤ 80

x ≥ 14

y ≥ 16

x, y ≥ 0

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

2x + 3y = 100,

x + 2y = 80,

x = 14,

y = 16,

x = 0 and

y = 0.

Area shows by 2x + 3y ≥ 100

The line 2x + 3y = 100 connects the coordinate axes at A(50, 0) and B(0, 100/3) respectively.

By connecting these points we will get the line 2x + 3y = 100 .

So, (0, 0) does not assure the 2x + 3y = 100.

So, the area which does not have the origin shows the solution set of the in-equation 2x + 3y ≥ 100.

Area shows by x + 2y ≤ 80:

The line x + 2y = 80 connects the coordinate axes at C(80, 0) and D(0, 40) respectively.

By connecting these points we will get the line x + 2y = 80.

So, (0, 0) assure the in-equation x + 2y ≤ 80.

So, the area which haves the origin shows the solution set of the in-equation x + 2y ≤ 80.

Area shown by x ≥ 14

x = 14 is the line passes through 14, 0 and is parallel to the Y axis.

The area to the right of the line x = 14 will assure the in-equation.

Area shown by y ≥ 16

y = 16 is the line passes through 0, 16 and is parallel to the X axis.

The area above the line y = 16 will assure the in-equation.

Area shows by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant assure these in-equations.

So, the first quadrant is the area shows by the in-equations x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints

2x + 3y ≥ 100,

x + 2y ≤ 80,

x ≥ 14,

y ≥ 16,

x ≥ 0 and

y ≥ 0 are so follows

The corner points of the suitable area are E(26, 16), F(48, 16), G(14, 33) and H(14, 24)

The values of Z at these corner points are so follow

The maximum value of Z is Rs 1440 which is obtained at F(48, 16).

Therefore,

The maximum profit is Rs 1440 will get when 48 units of product A and 16 units of product B were manufactured.

问题 36. 如果一个年轻人以 25 公里/小时的速度驾驶他的车辆，他必须每公里花费 2 美元在汽油上。如果他以 40 公里/小时的更快速度驾驶它，汽油成本会增加到每公里 5 美元。他有 100 美元可用于购买汽油和一小时内的旅行。将此表达给 LPP 并解决相同的问题。

解决方案：

Assume that the man travels x km when the speed is 25 km/hour and y km when the speed is 40 km/hour.

Therefore, the total distance traveled is (x + y) km.

Now, it is given that the man has Rs 100 to spend on petrol.

Total cost of petrol = 2x + 5y ≤ 100

Now, time taken to travel x km = x/25 h

Time taken to travel y km = y/40 h

Now, it is given that the maximum time is 1 hour.

So,

x/25 + y/40 ≤ 1

⇒ 8x + 5y ≤ 200

Therefore, the formula of the given linear programming problem is

Maximise Z = x + y

Subject to the constraints

2x + 5y ≤ 100

8x + 5y ≤ 200

x ≥ 0, y ≥ 0

The suitable area determined by the given constraints can be diagrammatically shown as,

The coordinates of the corner points of the suitable area are O(0, 0), A(25, 0), B(50/3, 40/3) and C(0, 20).

The value of the objective function at these points are given in the following table

So, the maximum value of Z is 30 at x = 50/3 , y = 40/3.

Therefore, the maximum distance that the man can travel in one hour is 30 km.

Therefore, The distance travelled by the man at the speed of 25 km/hour is 50/3 km, and the distance travelled by him at the speed of 40 km/hour is 40/3 km.

问题 37. 一家石油公司有两个油库，A 和 B，容量分别为 7000 升和 4000 升。该公司将向三个汽油泵 D、E、F 供油，其要求分别为 4500、3000 和 3500 升。加油站与加油站之间的距离（公里）如下表所示：

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

假设每公里的运输成本为每升 1.00 卢比，应如何安排交货以使运输成本最低？

解决方案：

Assume x and y litres of oil be supplied from A to the petrol pumps D and E.

Then, (7000 − x − y) L will be supplied from A to petrol pump F.

The requirement at petrol pump D is 4500 L.

Since, x L are transported from depot A, the remaining (4500 −x) L will be transported from petrol pump B

Similarly, (3000 − y) L and [3500 − (7000 − x − y)] L i.e. (x + y − 3500) L will be transported from depot B to petrol pump E and F. respectively.

The given problem can be shown diagrammatically, so follows

Since the quantity of oil is non-negative, therefore,

x ≥ 0, y ≥ 0, and (7000 – x – y) ≥ 0

x ≥ 0,

y ≥ 0 and

x + y ≤ 7000

4500 – x ≥ 0, 3000 – y ≥ 0, and x + y – 3500 ≥ 0

x ≤ 4500,

y ≤ 3000, and

x + y ≥ 3500

Cost of transporting 10 L of petrol = Re 1

Cost of transporting 1 L of petrol = Rs 1/10

Thus, total transportation cost is given by,

z = 7/10 × x + 6/10y + 3/10(7000 – x – y) + 3/10(4500 – x) + 4/10(3000 – y) + 2/10(x + y – 3500)

= 0.3x + 0.1y + 3950

Therefore, the formula of the given linear programming problem is

Minimize Z = 0.3x + 0.1y + 3950

Subject to the constraints,

x + y ≤ 7000

x ≤ 4500

y ≤ 3000

x + y ≥ 3500

x, y ≥ 0

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

x + y = 7000,

x = 4500,

y = 3000,

x + y = 3500,

x = 0 and

y = 0

Area shows by x + y ≤ 7000:

The line x + y = 7000 connects the coordinate axes at A(7000, 0) and B(0, 7000) respectively.

By connecting these points we will get the line x + y = 7000 .

So, (0, 0) assure the x + y = 7000 .

So, the area which haves the origin shows the solution set of the in-equation x + y ≤ 7000.

Area shown by x ≤ 4500:

The line x = 4500 is the line posses through C(4500, 0) and is parallel to Y axis.

The area to the left of the line x = 4500 will assure the in-equation x ≤ 4500.

Area shows by y ≤ 3000:

The line y = 3000 is the line passes through D(0, 3000) and is parallel to X axis.

The area below the line y = 3000 will assure the in-equation y ≤ 3000.

Area shows by x + y ≥ 3500:

The line x + y = 7000 connects the coordinate axes at E13500, 0 and F₁(0, 3500) respectively.

By connecting these points we will get the line x + y = 3500 .

So, (0, 0) assure the x + y = 3500.

So, the area which have the origin shows the solution set of the in-equation

x + y ≥ 3500.

Area shown by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant assure these in-equations.

So, the first quadrant is the area shows by the in-equations x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints

x + y ≤ 7000,

x ≤ 4500,

y ≤ 3000,

x + y ≥ 3500,

x ≥ 0 and

y ≥ 0 are so follows.

The corner points of the suitable area are E(3500, 0), C(4500, 0), I(4500, 2500), H(4000, 3000), and G(500, 3000).

The values of Z at these corner points are so follows.

Corner point	Z = 0.3x + 0.1y + 3950
E (3500, 0)	5000
C (4500, 0)	5300
I (4500, 2500)	5550
H (4000, 3000)	5450
G (500, 3000)	4400

The minimum value of Z is 4400 at G (500, 3000).

Therefore, the oil supplied from depot A is 500 L, 3000 L, and 3500 L and from depot B is 4000 L, 0 L, and 0 L to petrol pumps D, E, and F respectively.

So, the minimum transportation cost is Rs 4400.

问题 38. 一家小公司生产金戒指和金链。每天制造的戒指和链条的总数最多为24个。制作一个戒指需要1小时，制作一条链条需要30分钟。每天的最大可用小时数是 16。如果戒指的利润是 300 卢比，链条的利润是 190 卢比，请找出每天应该制造的戒指和链条的数量，这样才能获得最大的利润.使它成为一个 LPP 并以图形方式解决它。

解决方案：

Assume the firm manufacture x gold rings and y chains per day.

Number of gold rings and chains cannot be negative.

Therefore,

x, y ≥0 (always)

It is given that the total number of gold rings and chains manufactured per day is at most 24.

Therefore,

x + y ≤ 24

The gold ring takes 1 hour to make and chain takes 30 min, that is, 0.5 hour to make.

The maximum number of hours available per day is 16 hours.

Therefore,

x + 0.5y ≤ 16

The profit on a ring is Rs 300 and on a chain is Rs 190.

Therefore, profit made from x gold rings and y chains is Rs 300x and Rs 190y respectively.

Therefore,

Total profit, Z = 300x + 190y

Therefore, the formula of the given linear programming problem is

Maximise Z = 300x + 190y

Subject to the constraints

x + y ≤ 24

x + 0.5y ≤ 16

x, y ≥0

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

x + y = 24,

x + 0.5y = 16,

x = 0 and y = 0

Area shown by x + y ≤ 24:

The line x + y = 24 connects the coordinate axes at A(24, 0) and B(0, 24) respectively.

By connecting these points we will get the line x + y = 24.

So, (0, 0) assure the x + y = 24. So, the area which have the origin shows the solution set of the in-equation

x + y ≤ 24.

Area shown by x + 0.5y ≤ 16:

The line x + 0.5y = 16 connects the coordinate axes at C(16, 0) and D(0, 32) respectively.

By connecting these points we will get the line x + 0.5y = 16.

So, (0, 0) assure the in-equation x + 0.5y ≤ 16.

So,the area which haves the origin shows the solution set of the in-equation x + 0.5y ≤ 16.

Area shows by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant assure these in-equations.

So, the first quadrant is the area shows by the in-equations x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints

x + y ≤ 24,

x + 0.5y ≤ 16,

x ≥ 0

and y ≥ 0 are so follows.

The suitable area determined by the system constraints is so follows:

The corner points are O(0, 0), C(16, 0), E(8, 16) and B(0, 24).

The value of Z at these corner points are so follows

Corner Point	Z = 300x + 190y
O(0, 0)	0
C(16, 0)	4800
E(8, 16)	5440
B(0, 24)	4560

Therefore, the maximum value of Z is 5440 at E(8, 16).

Therefore, the 8 gold rings and 16 chains should be manufactured per day to maximise the profits.

问题 39。图书馆必须在一个书架上容纳两种不同类型的书籍。这些书有 6 厘米和 4 厘米厚，每本分别重 1 公斤和 12 公斤。架子长 96 厘米，几乎可以支撑 21 公斤的重量。书架应该如何装满两种类型的书才能容纳最多的书？使它成为一个 LPP 并以图形方式解决它。

解决方案：

Assume x books of first type and y books of second type were accommodated. Number of books cannot be negative.

Therefore, x, y ≥ 0

As per the question, the given information can be shown as:

	Thickness cm	Weight kg
First type(x)	6	1
Second type(y)	4	1.5
Capacity of shelf	96	21

Therefore, the constraints are

6x + 4y ≤ 96

x + 1. 5y ≤ 21

Number of books = Z = x + y which is to be maximised

Therefore,

Therefore, the formula of the given linear programming problem is

Maximize Z = x + y

Subject to constraints

6x + 4y ≤ 96

x + 1. 5y ≤ 21

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

6x + 4y = 96,

x + 1.5y = 21,

x = 0 and

y = 0

Area shows by 6x + 4y ≤ 96:

The line 6x + 4y = 96 connects the coordinate axes at A16, 0 and B(0, 24) respectively.

By connecting these points we will get the line 6x + 4y = 96.

So, (0, 0) assure the 6x + 4y = 96.

So, the area which haves the origin shows the solution set of the in-equation 6x + 4y ≤ 96.

Area shows by x + 1.5y ≤ 21:

The line x + 1.5y = 21 connects the coordinate axes at C(21, 0) and D(0, 14) respectively.

By connecting these points we will get the line x + 1.5y = 21.

So, (0, 0) assure the in-equation x + 1.5y ≤ 21.

So,the area which haves the origin shows the solution set of the in-equation x + 1.5y ≤ 21.

Area shown by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant assure these in-equations.

So, the first quadrant is the area shows by the in-equations x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints

6x + 4y ≤ 96,

x + 1.5y ≤ 21,

x ≥ 0 and

y ≥ 0 are so follows

The corner points are O(0, 0), D(0, 14), E(12, 6), A(16, 0) The values of Z at these corner points are so follows

Corner point	Z= x + y
O	0
D	14
E	18
A	16

The maximum value of Z is 18 which is obtained at E(12, 6).

Therefore, the maximum number of books that can be arranged on shelf is 18 where 12 books are of first type and 6 books are the other type.

问题 40. 一家工厂生产网球拍和板球拍。网球拍需要 1.5 小时的机器时间和 3 小时的工匠时间来制作，而板球拍需要 3 小时的机器时间和 1 小时的工匠时间。一天，工厂的机器时间不超过42小时，工匠时间不超过24小时。如果球拍和球拍的利润分别为 20 卢比和 10 卢比，请找出工厂必须生产的网球拍和板球拍的数量才能获得最大利润。使它成为一个 LPP 并以图形方式解决它。

解决方案：

Assume x number of tennis rackets and y number of cricket bats were sold.

Number of tennis rackets and cricket balls cannot be negative.

Therefore,

x, y ≥ 0 (always)

It is given that a tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s

time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman’s time.

Also, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman’s time.

Therefore, 1. 5x + 3y ≤ 423x + y ≤ 24

If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively.

Therefore, profit made on x tennis rackets and y cricket bats is Rs 20x and Rs 10y respectively.

Total profit = Z = 20x + 10y

Therefore, the formula of the given linear programming problem is

Maximize Z = 20x + 10y

Subject to constraints:

1. 5x + 3y ≤ 423

x + y ≤ 24

x ≥ 0, y ≥ 0

First we will convert in-equations into equations, so follows

1.5x + 3y = 42, 3x + y = 24, x = 0 and y = 0

Area shows by 1.5x + 3y ≤ 42:

The line 1.5x + 3y = 42 connects the coordinate axes at A(28, 0) and B(0, 14) respectively.

By connecting these points we will get the line 1.5x + 3y = 42. so, (0, 0) assure the 1.5x + 3y = 42.

So, the area which has the origin shows the solution set of the in-equation 1.5x + 3y ≤ 42.

Area shows by 3x + y ≤ 24:

The line 3x + y = 24 connects the coordinate axes at C18, 0 and D₁(0, 24) respectively.

By connecting these points we will get the line 3x + y =24.

So, (0, 0) assure the in-equation 3x + y ≤ 24.

So, the area which have the origin shows the solution set of the in-equation 3x + y ≤ 24.

Area shown by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant assure these in-equations.

So, the first quadrant is the area shows by the in-equations x ≥ 0, and y ≥ 0.

The suitable area determined by the system of constraints

1.5x + 3y ≤ 42,

3x + y ≤ 24,

x ≥ 0 and

y ≥ 0 are so follows

In the above graph, the shaded area is the suitable area.

The corner points are O(0, 0), B(0, 14), E(4, 12), and C(8, 0).

The values of the objective function Z at corner points of the suitable area are given in the following table:

Corner Points	Z = 20x +10y
O(0, 0)	0
B(0, 14)	140
E(4, 12)	200 Maximum
C(8, 0)	160

So, Z is maximum at x = 4 and y = 12 and the maximum value of Z at this point is 200.

Therefore, the maximum profit is of Rs 200 will get when 4 tennis rackets and 12 cricket bats were sold.

问题 41. 某商家计划销售两种类型的个人电脑，台式电脑和便携式电脑，售价分别为 25,000 卢比和 40,000 卢比。他估计，计算机的每月总需求量不会超过 250 台。如果商家不想投资超过 700 万卢比，并且他在台式机型号上的利润为 4500 卢比，而在便携式型号上的利润为 5000 卢比，则确定商家应库存的每种计算机的数量以获得最大利润。制作 LPP 并以图形方式求解。

解决方案：

Assume x and y be the number of desktop model and portable model respectively.

Number of desktop model and portable model cannot be negative.

Therefore,

It is given that the monthly demand will not exist 250 units.

Therefore,

x + y ≤ 250

Cost of desktop and portable model is Rs 25,000 and Rs 40,000 respectively.

Therefore, cost of x desktop model and y portable model is Rs 25,000 and Rs 40,000 respectively, and he does not want to invest more than Rs 70 lakhs.

25000x + 40000y ≤ 7000000

Profit on the desktop model is Rs 4500 and on the portable model is Rs 5000.

Therefore, profit made by x desktop model and y portable model is Rs 4500x and Rs 5000y respectively.

Total profit = Z = 4500x + 5000y

Therefore, the formula of the given linear programming problem is

Maximize Z = 4500x + 5000y

Subject to constraints:

x + y ≤ 250

25000x + 40000y ≤ 7000000

x, y ≥ 0

First we will transform in-equations into equations as follows

x + y = 250,

25000x + 40000y = 7000000,

x = 0 and

y = 0

Area shown by x + y ≤ 250:

The line x + y = 250 connects the coordinate axes at A250, 0 and B(0, 250) respectively.

By connecting these points we will get the line x + y = 250.

So, (0, 0) assures the x + y = 250.

So, the Area which contains the origin shows the solution set of the in-equation x + y ≤ 250.

Area shown by 25000x + 40000y ≤ 7000000:

The line 25000x + 40000y = 7000000 connects the coordinate axes at C(280, 0) and D(0, 175) respectively.

By connecting these points we will get the line 25000x + 40000y = 7000000.

So, (0, 0) assures the in-equation 25000x + 40000y ≤ 7000000.

So, the Area which contains the origin shows the solution set of the in-equation 25000x + 40000y ≤ 7000000.

Area shown by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant assures these in-equations.

So, 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 ≤ 250, 25000x + 40000y ≤ 7000000, x ≥ 0 and y ≥ 0 are as follows

The corner points are O(0, 0), D(0, 175), E(200, 50) and A(250, 0).

The values of the objective function Z at corner points of the suitable Area are given in the following table

Corner Points	Z = 4500x + 5000y
O(0, 0)	0
D(0, 175)	875000
E2(00, 50)	1150000 Maximum
A(250, 0)	1125000

So, Z is maximum at x = 200 and y = 50 and the maximum value of Z at this point is 1150000.

Therefore, 200 desktop models and 50 portable units should be sold to maximize the profit.

问题 42. 一个农民合作社有 50 公顷的土地种植两种作物 X 和 Y。每公顷作物 X 和 Y 的利润估计分别为 10,500 和 9,000。为了控制杂草，必须对作物 X 和 Y 分别使用每公顷 20 升和 10 升的液体除草剂。此外，应使用不超过 800 升的除草剂，以使用收集这片土地的排水的池塘来保护鱼类和野生动物。每种作物应该分配多少土地才能使社会的总利润最大化？

解决方案：

Assume the land allocated for crop X be x hectares and crop Y be y hectares.

Maximum area of the land available for two crops is 50 hectares.

Therefore,

x + y ≤ 50

The liquid herbicide to be used for crops X and Y are at the rate of 20 litres and 10 litres per hectare, respectively.

Maximum amount of herbicide to be used is 800 litres.

Therefore,

20x + 10 y ≤ 800

⇒ 2x + y ≤ 80

The profits from crops X and Y per hectare are 10,500 and 9,000, respectively.

Therefore, total profit = (10,500x + 9,000y)

Therefore, the formula of the given linear programming problem is

Maximise Z = 10500x + 9000y

Subject to the constraints

x + y ≤ 50

2x + y ≤ 80

x ≥ 0,

y ≥ 0

The suitable Area determined by the given constraints can be diagrammatically shown as,

The coordinates of the corner points of the suitable Area are O(0, 0), A(40, 0), B(30, 20) and C(0, 50).

The value of the objective function at these points are given in the following table

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

So, the maximum value of Z is 495000 at x = 30 and y = 20.

Therefore, 30 hectares of land should be allocated for crop X and 20 hectares

of land should be allocated for crop Y.

So, the maximum profit of the society is 4,95,000.

问题 43. 一家制造公司生产两种型号 A 和 B 的产品。 A型每件制作需要9工时，精加工需要1工时。每件 B 型需要 12 个工时的制作和 3 个工时的精加工。对于制造和精加工，可用的最大工时分别为 180 和 30。公司A型每件利润8000，B型每件利润12000，每周生产A型和B型多少件才能实现最大利润？每周最大利润是多少？

解决方案：

Let x pieces of model A and y pieces of model B are manufactured per week.

Since each piece of model A requires 9 labour hours and each piece of model B requires 12 labour hours for fabricating.

Therefore, x pieces of model A and y pieces of model B require (9x + 12y) labour hours for fabricating.

But, the maximum labour hours available for fabricating are 180.

Therefore,

9x + 12y ≤ 180

⇒ 3x + 4y ≤ 60

Similarly, each piece of model A requires 1 labour hour and each piece of model B

requires 3 labour hours for finishing,

Therefore, x pieces of model A and y pieces of model B require (x + 3y) labour hours for finishing.

But, the maximum labour hours available for finishing are 30.

Therefore,

x + 3y ≤ 30

The profit from each piece of model A is 8,000 and from each piece of model B is 12,000.

Therefore, the total profit from x pieces of model A and y pieces of model B is (8,000x + 12,000y).

Therefore, the formula of the given linear programming problem is

Maximise Z = 8000x + 12000y

Subject to the constraints

3x + 4y ≤ 60

x + 3y ≤ 30

x, y ≥ 0

The suitable Area determined by the given constraints can be diagrammatically shown as

The coordinates of the corner points of the suitable Area are O(0, 0), A(20, 0), B(12, 6) and C(0, 10).

The value of the objective function at these points are given in the following table.

Corner Point	Z = 8000x + 12000y
O(0, 0)	8000 × 0 + 12000 × 0 = 0
A(20, 0)	8000 × 20 + 12000 × 0 = 160000
B(12, 6)	8000 × 12 + 12000 × 6 = 168000 → Maximum
C(0, 10)	8000 × 0 + 12000 × 10 = 120000

The maximum value of Z is 168000 at x = 12, y = 6.

Therefore, the manufacturing company should produce 12 pieces of model A and

6 pieces of model B to realise maximum profit.

So, the maximum profit is 1,68,000.

问题 44. 一家工厂生产网球拍和板球拍。网球拍需要 1.5 小时的机器时间和 3 小时的工匠时间来制作，而板球拍需要 3 小时的机器时间和 1 小时的工匠时间。一天，工厂的机器时间不超过42小时，工匠时间不超过24小时。

(i) 如果工厂要满负荷生产，必须制作多少球拍和球拍？

(ii) 如果球拍和球拍的利润分别为 20 卢比和 10 卢比，求工厂满负荷生产时的最大利润。

解决方案：

Assume x number of tennis rackets and y number of cricket bats were sold.

Number of tennis rackets and cricket balls cannot be negative.

Therefore,

x ≥ 0, y ≥ 0

It is given that a tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time

in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman’s time.

Also, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman’s time.

Therefore, 1 · 5x + 3y ≤ 423x + y ≤ 24

If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively.

Therefore, profit made on x tennis rackets and y cricket bats is Rs 20x and Rs 10y respectively.

Total profit = Z = 20x + 10y

Therefore, the formula of the given linear programming problem is

Maximize Z = 20x + 10y

Subject to constraints:

1 · 5x + 3y ≤ 42

3x + y ≤ 24

x ≥ 0, y ≥ 0

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

1.5x + 3y = 42,

3x + y = 24,

x = 0 and

y = 0

Area shown by 1.5x + 3y ≤ 42:

The line 1.5x + 3y = 42 connects the coordinate axes at A(28, 0) and B(0, 14) respectively.

By connecting these points we will get the line 1.5x + 3y = 42.

So, (0, 0) assures the 1.5x + 3y = 42.

So, the Area which contains the origin shows the solution set of the in-equation 1.5x + 3y ≤ 42.

Area shown by 3x + y ≤ 24:

The line 3x + y = 24 connects the coordinate axes at C(8, 0) and D(0, 24) respectively.

By connecting these points we will get the line 3x + y = 24. So, 0, 0 assures the in-equation 3x + y ≤ 24.

So, the Area which contains the origin shows the solution set of the in-equation 3x + y ≤ 24.

Area shown by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant assures these in-equations.

So, 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

1.5x + 3y ≤ 42,

3x + y ≤ 24,

x ≥ 0 and

y ≥ 0 are as follows

In the above graph, the shaded Area is the suitable Area.

The corner points are O(0, 0), B(0, 14), E(4, 12), and C(8, 0).

The values of the objective function Z at corner points of the suitable Area are given in the following table

Corner Points	Z = 20x +10y
O(0, 0)	0
B (0, 14)	140
E (4, 12)	200 Maximum
C(8, 0)	160

So, Z is maximum at x = 4 and y = 12 and the maximum value of Z at this point is 200.

(i) 4 tennis rackets and 12 cricket bats must be made if the factory is to work at full capacity.

(ii) The maximum profit of the factory when it works at full capacity is Rs 200.

问题 45. 某商家计划销售两种类型的个人电脑，台式电脑和便携式电脑，售价分别为 25,000 卢比和 40,000 卢比。他估计，计算机的每月总需求量不会超过 250 台。如果商家不想投资超过 700 万卢比，而他在台式机型号上的利润为 4500 卢比，在便携式型号上为 5000 卢比，则确定商家应库存的每种计算机的数量以获得最大利润。

解决方案：

Assume x and y be the number of desktop model and portable model respectively.

Number of desktop model and portable model cannot be negative.

Therefore,

x, y ≥ 0 (always)

It is given that the monthly demand will not exist 250 units.

Therefore, x + y ≤ 250

Cost of desktop and portable model is Rs 25,000 and Rs 40,000 respectively.

Therefore, cost of x desktop model and y portable model is Rs 25,000 and Rs 40,000 respectively and

he does not want to invest more than Rs 70 lakhs.

25000x + 40000y ≤ 7000000

Profit on the desktop model is Rs 4500 and on the portable model is Rs 5000.

Therefore, profit made by x desktop model and y portable model is Rs 4500x and Rs 5000y respectively.

Total profit = Z = 4500x + 5000y

Therefore, the formula of the given linear programming problem is

Maximize Z = 4500x + 5000y

Subject to constraints:

x + y ≤ 250

25000x + 40000y ≤ 7000000

x, y ≥ 0

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

x + y = 250,

25000x + 40000y = 7000000,

x = 0 and

y = 0

Area shown by x + y ≤ 250:

The line x + y = 250 connects the coordinate axes at A(250, 0) and B(0, 250) respectively.

By connecting these points we will get the linex + y = 250.

So, 0, 0 assures the x + y = 250.

So, the Area which contains the origin shows the solution set of the in-equation x + y ≤ 250.

Area shown by 25000x + 40000y ≤ 7000000:

The line 25000x + 40000y = 7000000 connects the coordinate axes at C(280, 0) and D(0, 175) respectively.

By connecting these points we will get the line 25000x + 40000y = 7000000.

So, (0, 0) assures the in-equation 25000x + 40000y ≤ 7000000.

So, the Area which contains the origin shows the solution set of the in-equation 25000x + 40000y ≤ 7000000.

Area shown by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant assures these in-equations.

So, 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 ≤ 250, 25000x + 40000y ≤ 7000000, x ≥ 0 and y ≥ 0 are as follows

The corner points are O(0, 0), D(0, 175), E(200, 50) and A(250, 0).

The values of the objective function Z at corner points of the suitable Area are given in the following table:

Corner Points	Z = 4500x + 5000y
O(0, 0)	0
D(0, 175)	875000
E(200, 50)	1150000 Maximum
A(250, 0)	1125000

So, Z is maximum at x = 200 and y = 50 and the maximum value of Z at this point is 1150000.

Therefore, 200 desktop models and 50 portable units should be sold to maximize the profit.

问题 46. 一家玩具公司生产 A 和 B 两种类型的玩偶。市场测试和现有资源表明，每周总产量不应超过 1200 个玩偶，B 类型玩偶的需求最多为A型玩偶。此外，A型玩偶的生产水平可以超过其他类型玩偶产量的三倍，最多600个。如果公司在娃娃 A 和 B 上每个娃娃分别赚取 12 和 16 的利润，那么每个娃娃每周应该生产多少才能使利润最大化？

解决方案：

Assume x units of doll A and y units of doll B are manufactured to will get the maximum profit.

The mathematical formulation of the above problem is as follows:

Maximize Z = 12x + 16y Subject to

x + y ≤ 1200y ≤ x/12,

x – 3y ≤ 600,

x, y ≥ 0

The shaded Area shows the set of suitable solutions.

The coordinates of the corner points of the suitable Area are O(0, 0), A(800, 400), B(1050, 150) and C(600, 0).

The value of Z at O0, 0 = 12(0) + 16(0) = 0

The value of Z at A800, 400 = 12(800) + 16(400) = 16000 Maximum

The value of Z at B1050, 150 = 12(1050) + 16(150) = 15000

The value of Z at C600, 0 = 12(600) + 16(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.

问题 47. 有两种肥料 F ₁和 F ₂ 。 F1 由 10% 的氮和 6% 的磷酸组成，F ₂由 5% 的氮和 10% 的磷酸组成。在测试土壤条件后，一位农民发现她的作物至少需要 14 公斤的氮和 14 公斤的磷酸。如果 F ₁的价格为 6/kg，F ₂的价格为 5/kg，请确定每种肥料的用量，以便以最低的成本满足养分需求。最低成本是多少？

解决方案：

Let 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.

But, the farmer needs atleast 14 kg of nitrogen for the crops.

Therefore,

10% of x kg + 5% of y kg ≥ 14 kg

=> x/10 + y/20 ≥ 14

=> 2x + y ≥ 280

Similarly, F₁ consists of 6% phosphoric acid and F₂ consists of 10% phosphoric acid.

But, the farmer needs atleast 14 kg of phosphoric acid for the crops.

Therefore,

6% of x kg + 10% of y kg ≥ 14 kg

=> 6x/100 + 10y/200 ≤ 14

=> 3x + 5y ≤ 700

The cost of fertilizer F₁ is 6/kg and fertilizer F₂ is 5/kg,

Therefore, total cost of x kg of fertilizer F₁ and y kg of fertilizer F₂ is (6x + 5y).

Therefore, the formula of the given linear programming problem is

Minimise Z = 6x + 5y

Subject to the constraints:

2x + y ≥ 280

3x + 5y ≥ 700

x, y ≥ 0

The suitable Area determined by the given constraints can be diagrammatically shown as

The coordinates of the corner points of the suitable Area are A(700/3, 0) , B(100, 80) and C(0, 280).

The value of the objective function at these points are given in the following table.

Corner Point	Z = 6x + 5y
700/3, 0	6 × 700/3 + 5 × 0 = 1400
100, 80	6 × 100 + 5 × 80 = 1000 → Minimum
0, 280	6 × 0 + 5 × 280 = 1400

The smallest value of Z is 1000 which is acquired at x = 100, y = 80.

It can be seen that the open half-plane shown by 6x + 5y < 1000 has no common points with the suitable Area.

So, the minimum value of Z is 1000.

Therefore, 100 kg of fertilizer F₁ and and 80 kg of fertilizer F₂ should be used so that the nutrient requirements are met at minimum cost.

So, the minimum cost is 1,000.

问题 48. 某制造商在其工厂安装了三台机器 I、II、III。机器 I 和 II 最多可以运行 12 小时，而机器 III 每天必须运行至少 5 小时。她只生产两个项目 M 和 N，每个项目都需要使用所有三台机器。下表给出了在三台机器上生产 M 和 N 各 1 个单位所需的小时数：

Corner point	Z= 20x + 30y
E	1000
F	1440
G	1270
H	1000

她在项目 M 和 N 上分别获得了 600 和 400 的利润。假设她可以卖掉她生产的所有物品，她应该生产多少件才能使利润最大化？最大的利润是多少？

解决方案：

Let x units of item M and y units of item N are produced to maximise the profit.

Since each unit of item M require 1 hours on machine I and each unit of item N require 2 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.

Therefore,

2x + y ≤ 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, the 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.

Therefore,

x + 2y ≤ 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 atleast 5 hours.

Therefore,

x + 1.25y ≥ 5

The profit from each unit of item M is 600 and each unit of item N is 400.

Therefore, the total profit from x units of item M and y units of item N is (600x + 400y).

Therefore, the formula of the given linear programming problem is

Maximise Z = 600x + 400y

Subject to the constraints:

2x + y ≤ 12

x + 2y ≤ 12

x + 1.25y ≥ 5

x, y ≥ 0

The suitable Area determined by the given constraints can be diagrammatically shown as,

The coordinates of the corner points of the suitable Area are A(5, 0), B(6, 0), C(4, 4), D(0, 6) and E(0, 4).

The value of the objective function at these points are given in the following table.

Corner Point	Z = 600x + 400y
5, 0	600 × 5 + 400 × 0 = 3000
6, 0	600 × 6 + 400 × 0 = 3600
4, 4	600 × 4 + 400 × 4 = 4000 → Maximum
0, 6	600 × 0 + 400 × 6 = 2400
0, 4	600 × 0 + 400 × 4 = 1600

The maximum value of Z is 4000 at x = 4, y = 4.

Therefore, 4 units of item M and 4 units of item N should be produced to maximise the profit.

So, the maximum profit of the manufacturer is 4,000.

问题 49. 有两个工厂，一个位于 P 地，另一个位于 Q 地。从这些地点，某种商品将被运送到位于 A、B 和 C 的三个仓库中的每一个。仓库的每周需求分别为 5、5 和 4 单位商品，而 P 和 Q 工厂的生产能力分别为 8 和 6 单位。单位运输费用如下：

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

为了使运输成本最小，应该从每个工厂运输多少个单元到每个仓库。最低运输成本是多少？

解决方案：

Here, demand of the commodity 5 + 5 + 4 = 14units is equal the supply of the commodity 8 + 6 = 14units.

So, no commodity would be left at the two factories.

Assume x units and y units of the commodity be transported from the factory P to the depots at 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 factory 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 units of commodity transported are from the

factories to the depots are non-negative. Therefore,

x ≥ 0, y ≥ 0, 8 − x − y ≥ 0, 5 − x ≥ 0, 5 − y ≥ 0, x + y − 4 ≥ 0

Or

x ≥ 0, y ≥ 0, x + y ≤ 8, x ≤ 5, y ≤ 5, x + y ≥ 4

Total transportation cost = 160x + 100y + 150(8 − x − y) + 100(5 − x) +

120(5 − y) + 100(x + y − 4) = 10x − 70y + 1900

Therefore, the formula of the given linear programming problem is

Minimise Z = 10x − 70y + 1900

Subject to the constraints:

x + y ≤ 8

x ≤ 5

y ≤ 5

x + y ≥ 4

x ≥ 0, y ≥ 0

The suitable Area determined by the given constraints can be diagrammatically shown as,

The coordinates of the corner points of the suitable Area 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 are given in the following table.

Corner Point	Z = 10x − 70y + 1900
4, 0	10 × 4 − 70 × 0 + 1900 = 1940
5, 0	10 × 5 − 70 × 0 + 1900 = 1950
5, 3	10 × 5 − 70 × 3 + 1900 = 1740
3, 5	10 × 3 − 70 × 5 + 1900 = 1580
0, 5	10 × 0 − 70 × 5 + 1900 = 1550 → Minimum
0, 4	10 × 0 − 70 × 4 + 1900 = 1620

The minimum value of Z is 1550 at x = 0, y = 5.

Therefore, the 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, C respectively.

So, the minimum transportation cost is 1,550.

问题 50. 一家制造商制造了两种类型的玩具 A 和 B。为此需要三台机器，机器上每个玩具所需的时间（以分钟为单位）如下所示：

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

每台机器每天最多可使用 6 小时。如果每个 A 类玩具的利润是 7.50，每个 B 类玩具的利润是 5，则表明一天应该生产 15 个 A 类玩具和 30 个 B 类玩具，以获得最大利润。

解决方案：

Let the manufacturer makes x toys of type A and y toys of type B.

Since each toy of type A require 12 minutes on machine I and

each toy of type B require 6 minutes on machine I, therefore, x toys of typ

A and y toys of type B require (12x + 6y) minutes on machine I.

But, machines I is available for at most 6 hours.

Therefore,

12x + 6y ≤ 360

⇒ 2x + y ≤ 60

Similarly, each toy of type A require 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.

Therefore,

18x + 0y ≤ 360

⇒ x ≤ 20

Also, each toy of type A require 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.

Therefore,

6x + 9y ≤ 360

⇒ 2x + 3y ≤ 120

The profit on each toy of type A is 7.50 and each toy of type B is 5. Therefore, the total profit from x toys of type A and y toys of type B is (7.50x + 5y).

Therefore, the formula of the given linear programming problem is

Maximise Z = 7.5x + 5y

Subject to the constraints:

2x + y ≤ 60

x ≤ 20

2x + 3y ≤ 120

x, y ≥ 0

The suitable Area determined by the given constraints can be diagrammatically shown as

The coordinates of the corner points of the suitable Area are O(0, 0), A(20, 0), B(20, 20), C(15, 30) and D(0, 40).

The value of the objective function at these points are given in the following table.

Corner Point	Z = 7.5x + 5y
0, 0	7.5 × 0 + 5 × 0 = 0
20, 0	7.5 × 20 + 5 × 0 = 150
20, 20	7.5 × 20 + 5 × 20 = 250
15, 30	7.5 × 15 + 5 × 30 = 262.5 → Maximum
0, 40	7.5 × 0 + 5 × 40 = 200

The maximum value of Z is 262.5 at x = 15, y = 30.

Therefore, 15 toys of type A and 30 toys of type B should be manufactured in a day to get maximum profit.

So, the maximum profit is 262.50

问题 51. 一架飞机最多可搭载 200 名乘客。每张行政舱机票可赚取 1000 美元的利润，每张经济舱机票可赚取 600 美元的利润。该航空公司为行政舱预留了至少 20 个座位。然而，选择乘坐经济舱的乘客至少是乘坐行政舱的 4 倍。确定每种类型的机票必须售出多少才能使航空公司的利润最大化。最大利润是多少？

解决方案：

Assume x tickets of executive class and y tickets of economy class are sold by the airline.

The profit on each executive class ticket is 1000 and on each economy class ticket is 600.

Thus, the total profit from x executive class tickets and y economy class ticket is (1000x + 600y).

Now here, the aeroplane can carry a maximum of 200 passengers.

Therefore,

x + y ≤ 200

The airline reserves atleast 20 seats for executive class.

Therefore,

x ≥ 20

Also, at minimum 4 times as many passengers prefer to travel by economy class than by the executive class.

Therefore,

y ≥ 4x

Therefore, the formula of the given linear programming problem is

Maximise Z = 1000x + 600y

Subject to the constraints:

x + y ≤ 200

x ≥ 20

y ≥ 4x

x, y ≥ 0

The Suitable area determined by the given constraints can be diagrammatically shown as

The coordinates of the corner points of the feasible region are A(20, 80), B(40, 160) and C(20, 180).

The value of the objective function at these points are given in the following table

Corner Point	Z = 1000x + 600y
20, 80	1000 × 20 + 600 × 80 = 68000
40, 160	1000 × 40 + 600 × 160 = 136000 → Maximum
20, 180	1000 × 20 + 600 × 180 = 128000

The maximum value of Z is 136000 at x = 40, y = 160.

Thus, 40 tickets of executive class and 160 tickets of economy class should be sold to maximise the profit.

So, the maximum profit of the airline is 1,36,000.

问题 52. 制造商认为男女工人的效率相同，因此他支付相同的工资。他分别有 30 和 17 个单位的男性和女性工人和资本，他用来生产两种类型的商品 A 和 B。生产 1 个单位的 A 需要 2 个工人和 3 个单位的资本，而需要 3 个工人和 1 个单位生产一单位 B 需要资本。如果 A 和 B 的价格分别为每单位 100 和 120，他应该如何使用他的资源来最大化总收入？将上述形成为 LPP 并以图形方式求解。您是否同意制造商的这种观点，即男女工人的效率相同，因此应该按相同的比率支付工资？

解决方案：

Assume x units of A and y units of B be produced by the manufacturer.

The price of one unit of A is 100 and the price of one unit of B is 120.

Thus, the total price of x units of A and y units of B or the total revenue is (100x + 120y).

One unit of A needs 2 workers and one unit of B needs 3 workers.

Hence, x units of A and y units of B needs (2x + 3y) workers.

But, the manufacturer has 30 workers.

Therefore,

2x + 3y ≤ 30

Similarly, one unit of A needs 3 units of capital and one unit of B needs 1 unit of capital.

Thus, x units of A and y units of B needs (3x + y) units of capital.

But, the manufacturer has 17 units of capital.

Therefore,

3x + y ≤ 17

Hence, the given linear programming problem is

Maximise Z = 100x + 120y

Subject to the constraints:

2x + 3y ≤ 30

3x + y ≤ 17

x, y ≥ 0

The suitable area determined by the given constraints can be diagrammatically shown 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 are showing the following table.

Corner Point	Z = 100x + 120y
0, 0	100 × 0 + 120 × 0 = 0
0, 10	100 × 0 + 120 × 10 = 1200
(17/3, 0)	100 × 17/3 + 120 × 0 = 1700/3
3, 8	100 × 3 + 120 × 8 = 1260 → Maximum

The maximum value of Z is 1260 at x = 3, y = 8.

Therefore, the maximum total revenue is 1,260 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 a female.