第 12 类 RD Sharma 解决方案 – 第 30 章线性规划 – 练习 30.4 |设置 2

问题 18. 一家公司生产两种由胶合板制成的新颖纪念品。 A类纪念品每个切割需要5分钟，组装需要10分钟。 B类纪念品每个切割需要8分钟，组装需要8分钟。切割时间为 3 小时 20 分钟，组装时间为 4 小时。 A类纪念品每人50派萨，B类纪念品每人60派萨。为了使利润最大化，公司应该生产每种类型的纪念品多少？

解决方案：

Assume the company manufacture x souvenirs of Type A and y souvenirs of Type B.

Hence, 

x,  y ≥ 0

Given: The information can be shown 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 paisa and on Type B souvenirs is 60 paisa. 

Hence, profit gained on x souvenirs of Type A and y souvenirs of Type B is Rs 0.50x and Rs 0.60y respectively.

Total Profit, 

Z = 0.5x + 0.6y

Therefore, the formula of the given linear programming problem is 

Max Z = 0.5x + 0.6y

Subject to constraints,

5x + 8y ≤ 200

10x + 8y ≤ 240

x ≥ 0, y ≥ 0

Area 5x + 8y ≤ 200: line 5x + 8y = 200 connects axes at A(40, 0), B(0, 25) respectively. 

Area having origin shows the solution of the in-equation 5x + 8y ≤ 200 as (0, 0) assure 5x + 8y ≤ 200.

Area 10x + 8y ≤ 240: line 10x + 8y = 240 connects axes at C(24, 0), D(0, 30) respectively. 

Area having origin shows the solution of the in-equation 10x + 8y ≤ 240 as (0, 0) assures 10x + 8y ≤ 240.

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

The corner points of the suitable area are O(0, 0), B(0, 25), E(8, 20), C(24, 0).

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

Corner Points	Z = 20x + 30y
O	0
B	150
E	160
C	120

The maximum value of Z is obtained at E(8, 20).

Therefore, 8 souvenirs of Type A and 20 souvenirs of Type B must be produced each

day to get the maximum profit of Rs 160.

问题 19。制造商生产两种产品 A 和 B。产品 A 的售价为 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 个产品 A 和 16 个产品 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，那么产品B 是 30 卢比，那么每个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 are manufactured.

The number of units cannot be negative.

So, 

x, y ≥ 0  (always)

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

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

Here, the availability of time is 40 hours and the revenue must be atleast Rs 10000.

Hence further, given, there is a permanent order for 14 units of Product A and 16 units of product B.

Thus, 

The constraints are:

200x + 300y ≥ 10000,

0.5x + y ≤ 40

x ≥ 14

y ≥ 16.

If the profit on every of product A is Rs 20 and on product B is Rs 30. 

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

Total profit = 20x + 30y which is to be maximized.

Therefore, the formula of the given linear programming problem is 

Max Z = 20x + 30y

Subject to constraints,

200x + 300y ≥ 10000,

0.5x + y ≤ 40

x ≥ 14

y ≥ 16

x, y ≥ 0.

Area 200x + 300y≥  10000: line 200x + 300y = 10000 connects the axes at A(50,0), B(0, 100/3) respectively.

Area not having origin shows 200x + 300y ≥ 10000 as (0, 0) does not assure 200x + 300y ≥ 10000.

Area 0.5x + y ≤ 40: line 0.5x + y = 40 connects the axes at C(80, 0), D(0, 40) respectively.

Area having origin shows 0.5x + y ≤ 40 as (0, 0) assure 0.5x + y ≤ 40.

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 = 14 is the line passes through (16, 0) and is parallel to the X – axis. 

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

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

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

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

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

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

Therefore, the maximum profit is Rs 1440 attained when 48 units of product A and

16 units of product B are manufactured.

问题 20. 一家制造商生产两种类型的钢制行李箱。他有两台机器 A 和 B。为了完成，第一种类型的树干需要机器 A 3 小时和机器 B 3 小时，而第二种类型的树干需要机器 A 3 小时和机器 B 2 小时。机器 A 和 B 每天最多可以工作 18 小时和 15 小时。他的第一类和第二类树干分别赚取 30 卢比和 25 卢比的利润。为了获得最大的利润，他每天必须生产多少种类型的树干？

解决方案：

Assume x trunks of first type and y trunks of second type were manufactured. 

Number of trunks cannot be negative.

Thus, 

x, y ≥ 0 (always)

Given: Information can be shown as

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

Hence, 

The constraints are,

3x + 3y ≤ 18

3x + 2y ≤ 15.

He earns a profit of Rs 30 and Rs 25 per trunk of the first type and the second type respectively. 

Hence, 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 that is to be maximized.

Therefore, the formula of the given linear programming problem is 

Max Z = 30x + 25y

Subject to

3x + 3y ≤ 18

3x + 2y ≤ 15

x, y ≥ 0

Area 3x + 3y ≤ 18: line 3x + 3y = 18 connects axes at A(6, 0), B(0, 6) respectively. 

Area having origin shows the solution of the in-equation 3x + 3y ≤ 18 as (0, 0) assures 3x + 3y ≤ 18.

Area 3x + 2y ≤ 15: line 3x + 2y = 15 connects axes at C(5, 0), D(0, 15/2) respectively. 

Area having origin shows the solution of the in-equation 3x + 2y ≤ 15 as (0, 0) assures 3x + 2y ≤ 15.

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

The corner points are O(0, 0), B(0, 6), E(3, 3), and C(5, 0).

The values of Z at these 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 obtained at E(3, 3).

Therefore, the maximum profit is of Rs 165 attained when 3 units of each type of trunk is manufactured.

问题 21. 某专利药生产企业正在制定 A 药和 B 药的生产计划，有足够的原料可以生产 20000 瓶 A 和 40000 瓶 B，但只有 45000 瓶其中任何一种药都可以生产。可以放。此外，准备足够的材料以填充 1000 瓶 A 需要 3 小时，准备足够的材料以填充 1000 瓶 B 需要 1 小时，并且有 66 小时可用于此操作。 A 的利润为每瓶 8 卢比，B 的利润为每瓶 7 卢比。制造商应如何安排生产以使利润最大化？

解决方案：

Assume manufacturing of each bottle of A and B are x and y respectively.

As profits on each bottle of A and B are Rs 8 and Rs 7 per bottle respectively. 

Thus, profit on x bottles of A and y bottles of of B are 8x and 7y respectively. 

Assume Z be total profit on bottles 

So,

Z = 8x + 7y

As, it takes 3 hours and 1 hour to prepare enough material to fill 1000 bottles of Type A and Type B respectively, thus, x bottles of A and y bottles of B are preparing is 3x/1000 hours and y/1000 hours respectively, but only 66 hours are available, so, 

3x/1000 + y/1000 ≤ 66

3x + y ≤ 66000

Hence, raw materials accessible 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,

x ≤ 20000

y ≤ 40000

x + y ≤ 45000

x, y ≥ 0. {Since production of bottles can not be negative}

Therefore, the formula of the given linear programming problem is 

Max Z = 8x + 7y

Subject to constraints,

3x + y ≤ 66000

x ≤ 20000

y ≤ 40000

x + y ≤ 45000

x, y ≥ 0

Area 3x + y ≤ 66000: line 3x + y = 66000 connects the axes at A(22000, 0), B(0, 66000) respectively.

Area containing origin shows 3x + y ≤ 10000 as (0, 0) satisfy 3x + y ≤ 66000

Area x + y ≤ 45000: line x + y = 45000 connects the axes at C(45000, 0), D(0, 45000) respectively.

Area towards the origin will assure the in-equation as (0, 0) assures the in-equation

Area shown by x ≤ 20000,

x = 20000 is the line passes through (20000, 0) and is parallel to the Y – axis. 

The area towards the origin will assure the in-equation.

Area shown by y ≤ 40000,

y = 40000 is the line passes through (0,40000) and is parallel to the X – axis. 

The area towards the origin will assure the in-equation.

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

The corner points are O(0, 0), B(0, 40000), G(10500, 34500), H(20000, 6000), A(20000, 0).

The values 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 obtained at G(10500, 34500).

Therefore, the maximum profit is Rs 325500 attained 

when 10500 bottles of A and 34500 bottles of B are manufactured.

问题 22。一架飞机最多可搭载 200 名乘客。每张头等舱机票可赚取 400 卢比的利润，每张经济舱机票可赚取 600 卢比的利润。该航空公司至少保留 20 个头等舱座位。然而，选择乘坐经济舱到头等舱的乘客至少是头等舱的 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 needed number of first class and economy class tickets be x and y respectively.

Each ticket of first class and economy class made profit of Rs 400 and Rs 600 respectively.

Thus, x ticket of first class and y tickets of economy class make profit of Rs 400x and Rs 600y respectively.

Assume the total profit be Z and it is shown by

Z = 400x + 600y

Given, aeroplane can carry a minimum of 200 passengers, 

So,

x + y ≤ 200

Given airline reserves at least 20 seats for first class, 

So,

x ≥ 20

Also, at the minimum 4 times as many passengers choose to travel by economy class to the first class, 

So

y ≥ 4x

Therefore, the formula of the given linear programming problem is 

Max Z = 400x + 600y

Subject to constraints

x + y ≤ 200

y ≥ 4x

x ≥ 20

x, y ≥ 0 {since seats in both the classes can not be zero}

Area shown by x + y ≤ 200: the line x + y = 200 connects the axes at A(200, 0), B(0, 200). 

Area having origin shows x + y ≤ 200 as (0, 0) assures x + y ≤ 200.

Area shown by x ≥ 20: line x = 20 passes through (20, 0) and is parallel to y-axis. 

The area to the right of the line x = 20 will assure the in-equation x ≥ 20

Area shown by y ≥ 4x: line y = 4x passes through (0, 0). 

The area over the line y = 4x will assure the in-equation y ≥ 4x

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

The corner points are C(20, 80), D(40, 160), E(20, 180).

The values of Z at these corner points are 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).

Therefore, the maximum profit is Rs 116000 attained when 20 first class tickets and 180 economy class tickets are sold.

问题 23. 一位园丁供应了由 10% 的氮和 6% 的磷酸组成的 I 型肥料和由 5% 的氮和 10% 的磷酸组成的 II 型肥料。在测试土壤条件后，他发现他的作物至少需要 14 公斤的氮和 14 公斤的磷酸。如果 I 型肥料每公斤 60 派萨，II 型肥料每公斤 40 派萨，请确定每种肥料应使用多少公斤，以便以最低成本满足养分需求。最低成本是多少？

解决方案：

Assume x kg of Type I fertilizer and y kg of Type II fertilizers are supplied.

The quantity of fertilizers can not be negative.

So, 

x, y ≥ 0 (always)

A gardener has a supply of fertilizer of type I which contains of 10% nitrogen and 

Type II contains of 5% nitrogen, and he requires at least 14 kg of nitrogen for his crop.

Thus,

(10 × 100) + (5 × 100) ≥ 14

Or, 

10x + 5y ≥ 1400

A gardener has a supply of fertilizer of type I which contains of 6% phosphoric acid and 

Type II contains of 10% phosphoric acid, and he requires at least 14 kg of phosphoric acid for his crop.

Thus,

(6 × 100) + (10 × 100) ≥ 14

Or, 

6x + 10y ≥ 1400

Hence,

Subject to constraints are,

10x + 5y ≥ 1400

6x + 10y ≥ 1400

Assuming that the Type I fertilizer costs 60 paise per kg and Type II fertilizer costs 40 paise per kg. 

Thus, the cost of x kg of Type I fertilizer and y kg of Type II fertilizer is Rs0.60x and Rs 0.40y respectively.

Assume the Total cost is Z

So,

Z = 0.6x + 0.4y is to be minimized.

Therefore, the formula of the given linear programming problem is 

Min Z = 0.6x + 0.4y

Subject to the constraints,

10x + 5y ≥ 1400

6x + 10y ≥ 1400

x, y ≥ 0

The area shown by 6x + 10y ≥ 1400: line 6x + 10y = 1400 passes through A(700/3, 0) and B(0, 140). 

The area which doesn’t consist the origin shows the solution of the in-equation 6x + 10y ≥ 1400

Since, (0,0) doesn’t assure the in-equation 6x + 10y ≥ 1400

Area shown by 10x + 5y ≥ 1400: line 10x + 5y = 1400 passes through C(140, 0) and D(0, 280). 

The area that doesn’t consist the origin shows the solution of the in-equation 10x + 5y ≥ 1400

As (0,0) doesn’t assure the in-equation 10x + 5y ≥ 1400

The area, x,y ≥ 0: shows 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 obtained at E(100, 80)

Thus, the minimum cost is Rs92 attained when 100 kg of Type I fertilizer and 80 kg of Type II fertilizer is supplied.

问题 24. Anil 想在储蓄证书和国民储蓄债券上最多投资 12000 卢比。根据规定，他必须投资至少 2000 卢比的储蓄证书和至少 4000 卢比的国民储蓄债券。如果储蓄凭证的年利率为 8%，国民储蓄债券的年利率为 10%，他应该投资多少才能获得最大的年收入？求他的最大年收入。

解决方案：

Let Anil invests Rs x and Rs y in saving certificate (SC) and National saving bond (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 8x/100 and Rs y of NSB is 10y/100 per annum.

Assure Z be total interest earned so,

Z = 8x/100 + 10y/100

Given: He wants to invest Rs 12000 is total

x + y ≤ 12000

According to the rules he has to invest at least Rs 2000 in SC and at least Rs 4000 in NSB.

x ≥ 2000

y ≥ 4000

Therefore, the formula of the given linear programming problem is 

Max Z = 8x/100+ 10y/100  

Subject to constraints

x ≥ 2000

y ≥ 4000

x + y ≤ 12000

x,y ≥ 0

The area shown by x ≥ 2000: line x = 2000 is parallel to the y – axis and passes through (2000, 0).

The area not having the origin shows x ≥ 2000

As (0, 0) doesn’t assure the in-equation x ≥ 2000

The area shown by y ≥ 4000: line y = 4000 is parallel to the x – axis and passes through (0, 4000).

The region not containing the origin represents y ≥ 4000

As (0, 0) doesn’t assure the in-equation y ≥ 4000

Area shown by x + y ≤ 12000: line x + y = 12000 connects axes at A(12000, 0) and B(0, 12000) respectively. 

The area which have the origin shows the solution set of x + y ≤ 12000

As (0, 0) assure the in-equality x + y ≤ 12000.

Area x, y ≥ 0 is shown by the first quadrant.

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 = 8x/100+ 10y/100
O	0
E	1160
D	1040
C	560

The maximum value of Z is Rs 1160 which is obtained at E(2000,10000).

Thus, the maximum earning is Rs1160 attained when Rs 2000 were invested in SC and Rs 10000 in NSB.

问题 25. 一个人拥有一块 1000 平方米的土地。他想在里面种果树。他有一笔 1400 卢比的资金来购买幼树。他可以选择两种类型的树。 A 型每棵树需要 10 平方米的地面，每棵树的成本为 20 卢比，B 型需要每棵树的 20 平方米的地面，每棵树的成本为 25 卢比。完全成熟后，A 型平均生产 20 公斤水果，每公斤可获利 2.00 卢比，B 型平均可生产 40 公斤水果，每公斤可获利 2.00 卢比。每公斤1.50。当树木完全长大时，每种类型应该种植多少才能获得最大的利润？最大利润是多少？

解决方案：

Assume the needed number of trees of Type A and B be Rs x and Rs y respectively.

Number of trees cannot be negative.

x, y ≥ 0. (always)

To plant tree of Type A requires 10 sq. m and Type B needs 20 sq. m of ground per tree. 

And it is given that a man owns a field of area 1000 sq. m. 

Therefore,

10x + 20y ≤ 1000

x + 2y ≤ 100

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 trees.

20x + 25y ≤ 1400

4x + 5y ≤ 280

Therefore, the mathematical formulation of the given LPP is

Max Z = 40x – 20x + 60y – 25y = 20x + 35y

Subject to constraints;

x + 2y ≤ 100

4x + 5y ≤ 280

x, y ≥ 0

Area 4x + 5y ≤ 280: line 4x + 5y ≤ 280 connects axes at A₁(70, 0), B₁(0, 56) respectively.

The area having origin shows 4x + 5y ≤ 280 as (0, 0) assure 4x + 5y ≤ 280.

Area x + 2y ≤ 100: line x + 2y = 100 connects axes at A₂(100, 0), B₂(0, 50) respectively.

Area having origin shows x + 2y ≤ 100 as (0, 0) assure x + 2y ≤ 100

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

The corner points are A₁(70, 0), P(20, 40), B₂(0, 50)

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

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

The maximum value of Z is 1800 which is obtained at P(20, 40).

Thus, the maximum profit is Rs 1800 attained when Rs 20 were 

involved in Type A and Rs 40 were involved in Type II.

问题 26. 家庭手工业生产底座灯和木制灯罩，每个都需要使用研磨/切割机和喷雾器。制作台灯需要磨/切割机2小时，喷涂机3小时，制造灯罩需要磨/切割机1小时，喷涂机2小时。在任何一天，喷雾机最多可用 20 小时，研磨/切割机最多可用 12 小时。销售一盏灯的利润是 5.00，一个灯罩是 3.00。假设制造商销售他生产的所有灯具和窗帘，他应该如何安排他的日常生产以最大化他的利润？

解决方案：

Assume x units of pedestal lamps and y units of wooden shades are manufactured on a day to maximise the profit.

As a pedestal lamp needs 2 hours on the grinding/cutting machine and a wooden shade needs 1 hour on the grinding/cutting machine, 

Hence, the total hours needs for grinding/cutting x units of pedestal lamps and 

y units of wooden shades are (2x + y). 

Thus, 

The grinding/cutting machine is accessible for at maximum 12 hours on a day.

Therefore,

2x + y ≤ 12 

As well as,

A pedestal lamp needs 3 hours on the sprayer and a wooden shade needs 2 hours on the sprayer.  

Therefore, the total hours needed for spraying x units of pedestal lamps and

y units of wooden shades are (3x + 2y). 

As, the sprayer is accessible for at most 20 hours on a day.

Therefore,

3x + 2y ≤ 20

The profit from the sale of a pedestal lamp is 5.00 and a wooden shade is 3.00. 

Thus, the total profit from the sale of x units of pedestal lamps and y units of wooden shades is (5x + 3y). 

Therefore, the mathematical formulation of the given LPP is

Maximise Z = 5x + 3y

Subject to the constraints 

2x + x ≤ 12

3x + 2y ≤ 20

 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(6, 0), B(4, 4) and C(0, 10). 

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

Corner Point	Z = 5x + 3y
0, 0	5 × 0 + 3 × 0 = 0
6, 0	5 × 6 + 3 × 0 = 30
4, 4	5 × 4 + 3 × 4 = 32   → Maximum
0, 10	5 × 0 + 3 × 10 = 30

The maximum value of Z is 32 at x = 4, y = 4.

Therefore, the manufacturer must manufacture 4 pedestal lamps and 4 wooden shades to maximise his profit. 

The maximum profit of the manufacturer is 32 on a day.

问题 27。一个生产者分别有 30 和 17 个劳动力和资本单位，他可以用来生产两种类型的商品 x 和 y。要生产 1 个单位的 x，需要 2 个单位的劳动力和 3 个单位的资本。同样，生产一单位 y 需要 3 单位劳动和 1 单位资本。如果 x 和 y 的价格分别为每单位 100 卢比和 120 卢比，生产者应该如何利用他的资源来最大化总收入？以图形方式解决问题。

解决方案：

Let x₁ and y₁ units of goods x and y were produced respectively.

Number of units of goods cannot be negative.

Therefore, x₁, y₁ ≥ 0

To manufacture one unit of x, 2 units of labour and for one unit of y, 3 units of labour are needed.

2x₁ + 3y₁ ≤ 30

To manufacture one unit of x, 3 units of capital is needed and 

1 unit of capital is needed to manufacture one unit of y.

3x₁ + y₁ ≤ 17

If x and y are priced at Rs 100 and Rs 120 per unit respectively, 

Thus, price of x₁ and y₁ units of goods x and y is Rs 100x₁ and Rs 120y₁ respectively.

Total revenue

Z = 100x₁ + 120y₁ which is to be maximised.

Therefore, the mathematical formulation of the given LPP is

Maximise Z = 100x₁ + 120y₁

Subject to 

2x₁ + 3y₁ ≤ 303x₁ + y₁ ≤ 17

x, y ≥ 0

In the starting we will convert in-equations into equations as follows:

2x₁ + 3y₁ = 30, 

3x₁ + y1 = 17, 

x = 0 and 

y = 0

Area shown by 2x₁ + 3y₁ ≤ 30:

The line 2x₁ + 3y₁ = 30 meets the coordinate axes at A(15, 0) and B(0, 10) respectively. 

After connecting these points we will get the line 2x₁ + 3y₁ = 30. 

Thus, (0, 0) assures the 2x₁ + 3y₁ = 30. 

Hence, the area that have the origin shows the solution set of the in-equation 2x₁ + 3y₁ ≤ 30.

Area shown by 3x₁ + y₁ ≤ 17:

The line 3x₁ + y₁ = 17 connects the coordinate axes at C ( 17/3 , 0) and D(0, 17) respectively. 

After connecting these points we will get the line

3x₁ + y₁ = 17. 

Here, (0, 0) assures the in-equation 3x₁ + y1 ≤ 17. 

Thus, the area that have the origin shows the solution set of the in-equation 3x₁ + y₁ ≤ 17.

Area shown by x₁ ≥ 0 and y1 ≥ 0:

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

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

The suitable area is determined by the system of constraints 

2x₁ + 3y₁ ≤ 30, 

3x₁ + y₁ ≤ 17, 

x ≥ 0 and 

y ≥ 0 are 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 are as follows

Corner point	Z= 100x1 + 120y1
B	1200
E	1260
C	1700/3

The maximum value of Z is 1260 that is obtained at E(3, 8).

Therefore, the maximum revenue is Rs 1260 attained when 3 units of x and 8 units of y were manufactured.

问题 28。一家公司生产两种产品 A 和 B，并以每单位 A 型产品 5 卢比和每单位 B 型产品 3 卢比的利润销售它们。每种产品在两台机器 M ₁和 M ₂上加工。一个类型 A 的单元在 M ₁上需要一分钟的处理时间，在 M ₂上需要两分钟的处理时间，而一个类型 B 的单元在 M ₁上需要一分钟的处理时间，在 M ₂上需要一分钟的处理时间 机器 M ₁和 M ₂个每天最多可使用5小时和6小时。找出公司每天应该生产多少单位的每种产品才能使利润最大化。以图形方式解决问题。

解决方案：

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

Number of products cannot be negative.

Thus, 

x, y ≥ 0 (always) 

As in the question, the given information can be shown 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

x + y ≤ 300, 

2x + y ≤ 360

Firm produces two types of products A and B and sells them at a profit of Rs 5 per unit oftype A and 

Rs 3 per unit of type B.

Thus, x units of product A and y units of product B costs Rs 5x and Rs3y respectively. 

Total profit is Z = 5x + 3y which has to be maximised

Therefore, the mathematical formulation of the given LPP is

Maximise Z = 5x + 3y

Subject to

x + y ≤ 300, 

2x + y ≤ 360 

x, y ≥ 0

In the starting we will convert in-equations into equations as follows:

x + y = 300,  

2x + y = 360, 

x = 0 and 

y = 0

The area shown by x + y ≤ 300:

The line x + y = 300 connects the coordinate axes at A₁(300, 0) and B₁(10, 300) respectively. 

After connecting these points we will get the line x + y = 30. 

As (0, 0) assures the x + y = 30. 

Hence, the area that have the origin shows he solution set of the in-equation

x + y ≤ 300.

The area shown by 2x + y ≤ 360:

The line 2x + y = 360 connects the coordinate axes at C₁(180, 0) and D₁(10, 360) respectively. 

After connecting these points we will get the line

2x + y = 360. 

As (0, 0) assures the in-equation 2x + y ≤ 360. 

Thus, the area which have the origin shows the solution set of the in-equation 2x + y ≤ 360.

The area shown by x ≥ 0 and y ≥ 0:

Thus, all the points in the first quadrant assures these in-equations. 

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

The suitable area shown by the system of constraints x + y ≤ 300, 

2x + y ≤ 360, 

x ≥ 0 and 

y ≥ 0 are as follows;

The corner points are O(0, 0), B₁(0, 300), E₁(60, 240) and C₁(180, 0). The values of Z at these corner points are as follows

Corner point	Z= 5x + 3y
O	0
B1	900
E1	1020
C1	900

The maximum value of Z is Rs 1020 that is obtained at B₁(60, 240).

Therefore, the maximum profit is Rs 1020 attained when 60 units of product A and 

240 units of product B were produced.

问题 29. 一家小公司生产 A 和 B 项，它一天最多可以生产 A 和 B 项的总数为 24 个。 A 项需要一个小时，而 B 项只需半小时。每天最多可用时间为 16 小时。如果一单位物品 A 的利润是 300 卢比，一单位物品 B 的利润是 160 卢比，那么每种类型的物品生产多少才能使利润最大化？以图形方式解决问题。

解决方案：

Assume the firm manufacture x items of A and y items of B per day. 

Number of items cannot be negative.

Thus, 

x, y ≥0 (always)

It is given that the total number of items manufactured per day is at most 24.

Therefore, 

x + y ≤ 24

Item A requires 1 hour to make and item B requires 0.5 hour to make. 

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

Therefore, 

x + 0.5y ≤ 16

If the profit on one unit of item A be Rs 300 and one unit of item B be Rs 160.

Thus, profit gained on x items of A and y items of B is Rs 300x and Rs 160y respectively.

Therefore, 

Total profit is Z = 300x + 160y

Therefore, the mathematical formulation of the given LPP is

Maximise Z = 300x + 160y 

Subject to the constraints 

x+ y ≤ 24

x + 0.5y ≤ 16

x, y ≥0

At starting, we will convert in-equations into equations as follows:

x + y = 24, 

x + 0.5y = 16, 

x = 0 and 

y = 0

The area shown by x + y ≤ 24:

The line x + y = 24 connects the coordinate axes at A₁(24, 0) and B₁(0, 24) respectively. 

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

As, (0, 0) assures the x + y = 24. 

Thus, the area having the origin shows the solution set of the in-equation

x + y ≤ 24.

The area shown by x + 0.5y ≤ 16:

The line x + 0.5y = 16 connects the coordinate axes at C₁(16, 0) and D₁(0, 32) respectively. 

After connecting these points we will get the line 

x + 0.5y = 16. 

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

Thus,the area that have the origin shows the solution set of the in-equation 

x + 0.5y ≤ 16.

The area shown by x ≥ 0 and y ≥ 0:

As, all the points in the first quadrant assures 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 + y ≤ 24, 

x + 0.5y ≤ 16, 

x ≥ 0 and 

y ≥ 0 are as follows

The corner points are O(0, 0) ,C₁(16, 0), E₁(8, 16) and B₁(0, 24). 

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

Corner Point	Z = 300x + 160y
O(0, 0)	0
C(16, 0)	4800
E1(8, 16)	4960
B1(0, 24)	3840

Therefore, the maximum value of Z is 4960 at E₁(8, 16).

Therefore, 8 units of item A and 16 units of item B should be manufactured per day to maximise the profits.

问题 30. 某公司生产 A 型和 B 型两种玩具，A 型分别需要 5 分钟切割和 10 分钟组装。 B型切割需要8分钟，组装需要8分钟。一天有3个小时的切割时间和4个小时的组装时间。利润是 A 型每个 50 卢比，B 型每个 60 卢比。公司每天应该生产多少个每种类型的玩具才能使利润最大化？

解决方案：

Assume x toys of type A and y toys of type B were manufactured. 

The given information can be shown in the tabulated as follows:

	Cutting time (minutes)	Assembling time (minutes)
Toy A(x)	5	10
Toy B(y)	8	8
Availability	180	240

The constraints are

5x + 8y ≤ 180

10x + 8y ≤ 240

The profit is Rs 50 each on type A and Rs 60 each on type B. 

Thus, profit gained on x toys of type A and y toys of type B is Rs 50x and Rs 60 y respectively.

Total profit is Z = 50x + 60y

Therefore, the mathematical formulation of the given LPP is 

Maximise Z = 50x + 60y

Subject to

5x + 8y ≤ 180

10x + 8y ≤ 240

At first we will convert in-equations into equations as follows: 

5x + 8y = 180, 

10x + 8y = 240, 

x = 0 and 

y = 0

The area shown by 5x + 8y ≤ 180:

The line 5x + 8y = 180 connects the coordinate axes at A₁(36, 0) and B₁ (0, 45/2) respectively. 

After connecting these points we will get the line 5x +8y = 180. 

As, (0, 0) assures the 5x + 8y = 180. 

Thus, the area that have the origin shows the solution set of the in-equation 5x + 8y ≤ 180.

The area shown by 10x + 8y ≤ 240:

The line 10x + 8y = 240 meets the coordinate axes at C₁(24, 0) and D₁(0, 30) respectively.

After connecting these points we will get the line 10x + 8y = 240. 

As (0, 0) assures the in-equation 10x + 8y ≤ 240.

Thus, the area that have the origin shows the solution set of the in-equation 10x + 8y ≤ 240.

The area shown by x ≥ 0 and y ≥ 0:

As, all the points in the first quadrant assures 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 

5x + 8y ≤ 180, 

10x + 8y ≤ 240, 

x ≥ 0 and 

y ≥ 0 are as follows. 

The suitable area is shown in the figure

The corner points are  B₁(0, 45/2), E₁(12, 15) and C₁(24, 0).

The values of Z at the corner points are

Corner points	Z = 50x + 60y
O	0
B1	1350
E1	1500
C1	1200

The maximum value of Z is 1500 which is at E₁(12, 15).

Therefore, for maximum profit, 12 units of toy A and 15 units of toy B should be manufactured.

问题 31. 一家公司生产 A 和 B 两种物品。有两个部门处理这些物品：(i) 装配和 (ii) 整理部门。第一个部门的最大容量是每周 60 小时，另一个部门的最大容量是每周 48 小时。 A 件的每个单元的产品需要 4 小时的组装和 2 小时的精加工，B 的每个单元的产品需要 2 小时的组装和 4 小时的精加工。如果 A 的每个单位的利润是 6 卢比，B 的每个单位的利润是 8 卢比，请找出每周要生产的 A 和 B 的单位数量，以便获得最大利润。

解决方案：

Assume x units and y units of articles A and B are produced respectively. 

Number of articles cannot be negative.

Therefore, 

x, y ≥ 0 (always)

The product of every unit of article A needs 4 hours in assembly and that of article B needs 2 hours in assembly and the maximum capacity of the assembly department is 60 hours a week.

4x + 2y ≤ 60

The product of each unit of article A needs 2 hours in finishing and that of article B needs 

4 hours in assembly and the maximum capacity of  the finishing department is 48 hours a week.

2x + 4y ≤ 48

If the profit is Rs 6 for every unit of A and Rs 8 for every 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 that is to be maximised.

Therefore, the mathematical formulation of the given LPP is 

Max Z = 6x + 8y

Subject to constraints

2x + 4y ≤ 48

4x + 2y ≤ 60

x, y ≥ 0

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

2x + 4y = 48, 

4x + 2y = 60, 

x = 0 and y = 0 

Area shown by 2x + 4y ≤ 48:

The line 2x + 4y = 48 connects the coordinate axes at A(24, 0) and B(0, 12) respectively. 

By connecting these points we will get the line 2x + 4y = 48. 

So, (0, 0) assure the 2x + 4y = 48. 

So, the area which haves the origin shows the solution set of the in-equation 2x + 4y ≤ 48. 

Area shows by 4x + 2y ≤ 60:

The line 4x + 2y = 60 connects the coordinate axes at C(15, 0) and D(0, 30) respectively. 

By connecting these points we will get the line 4x + 2y = 60. 

So, (0, 0) assure the in-equation 4x + 2y ≤ 60. 

So, the area which haves the origin shows the solution set of the in-equation 4x + 2y ≤ 60.

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 + 4y ≤ 48, 

4x + 2y ≤ 60, 

x ≥ 0 and 

y ≥ 0 are so follows

The corner points are O(0, 0), B(0, 12), E(12, 6) and C(15, 0). 

The values of Z at these corner points are so follows

Corner Points	Z = 6x + 8y
O	0
B	96
E	120
C	90

The maximum value of Z is 120 which is obtained at E(12, 6).

Therefore, the maximum profit is Rs 120 will get when 12 units of article A and 

6 units of article B were manufactured

问题 32. 一家公司生产 A 项和 B 项，一天总共可以生产 24 项。A 项需要 1 小时，B 项需要半小时。最长可用时间每天是 16 小时。 A 的一件商品的利润是 300 卢比，B 的一件商品的利润是 160 卢比。每种类型的商品应该生产多少才能使利润最大化？以图形方式解决问题。

解决方案：

Assume x and y be the number of items of A and B that should be produced each day to maximize the profit. 

Number of items cannot be negative.

Therefore,

 x, y ≥ 0

It is also given that the firm can produce at most 24 items in a day.

Therefore,

x+ y ≤ 24

Also, the time needed to make an item of A is one hour and time needed to make an item of B is half an hour.

Therefore, the time needed to produce x items of A and y items of B is hours. 

However, the maximum time available in a day is 16 hours.

Therefore,

x + y/2 ≤ 16

It is given that the profit on an item of A is Rs 300 and on one item of B is Rs 160. 

Therefore, the profit gained from x items of A and y items of B is Rs 300x and Rs 160y respectively. 

Total profit Z = 300x + 160y

Therefore, the mathematical formulation of the given LPP is 

Maximize Z = 300x + 160y 

Subject to constraints:

x + y ≤ 24,

$x+\frac{1}{2}y\le16$

x ≥ 0, 

y ≥ 0

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

x + y = 24, 

x + y/2 ≤ 16

x = 0 and y = 0 

Area shows 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 shows by x + y/2 ≤ 16

The line x + y/2 ≤ 16 connects the coordinate axes at C(16, 0) and D(0, 32) respectively. 

By connecting these points we will get the line x + y/2 = 16

So, (0, 0) assure the in-equation x + y/2 ≤ 16. 

So, the area which haves the origin shows the solution set of the in-equation x + y/2 ≤ 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 + y/2 ≤ 16, x ≥ 0 and 

y ≥ 0 are so follows. 

The suitable area is shown in the figure.

In the above graph, the shaded area is the suitable area. The corner points are O(0, 0), C(16, 0), E(8, 16), and B(0, 24).

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

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

So, Z is maximum at x = 8 and y = 16 and the maximum value of Z at this point is 4960.

Therefore, 8 items of A and 16 items of B should be produced in order to maximize the profit and 

the maximum profit is Rs 4960.

问题 33. 一家公司销售两种不同的产品，A 和 B。这两种产品在一个共同的生产流程中生产，总产能为 500 工时。生产一个A需要5个小时，生产一个B需要3个小时。经过市场调查，公司负责人认为A最多可以销售70个，B最多可以销售125个。如果产品 A 的利润为每单位 20 卢比，产品 B 的利润为每单位 15 卢比，那么每种产品应销售多少单位才能使利润最大化？

解决方案：

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

So, x ≥ 0, y ≥ 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 a total capacity of 500 man-hours

5x + 3y ≤ 500

The maximum number of unit of A that can be sold is 70 and that for B is 125.

 x ≤ 70y ≤ 125

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 Rs 15yrespectively.

Total profit = Z = 20x + 15y

Therefore, the mathematical formulation of the given LPP is 

Max Z = 20x + 15y

Subject to constraints

5x + 3y ≤ 500,

x ≤ 70,

y ≤ 125,

x ≥ 0, y ≥ 0

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

5x + 3y = 500, 

x = 70, 

y = 125, 

x = 0 and y = 0

Area shows by 5x + 3y ≤ 500

The line 5x + 3y = 500 connects the coordinate axes at A(100, 0) and B(0, 500/3) respectively. 

By connecting these points we will get the line 5x +3y = 500. 

So, (0, 0) assure the 5x + 3y = 500. 

So, the area which haves the origin shows the solution set of the in-equation 5x + 3y ≤ 500.

Area shows by x ≤ 70:

The line x = 70 is the line passes through C(70, 0) and is parallel to Y axis. 

The area to the left of the line x = 70 will assure the in-equation x ≤ 70.

Area shows by y ≤ 125:

The line y = 125 is the line passes through D(0, 125) and is parallel to X axis. 

The area below the line y = 125 will assure the in-equation y ≤ 125.

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 

5x + 3y ≤ 500, 

x ≤ 70, 

y ≤ 125, 

x ≥ 0 and y ≥ 0 are so 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 = 20x + 15y
O	0
D	1875
E	2375
F	2150
C	1400

The maximum value of Z is 2375 which is at E(25, 125).

Therefore, the maximum profit is Rs 2375, 25 units of A and 125 units of B should be manufactured

问题 34. 一个盒子制造商用一大块硬纸板制作大小盒子。大箱子每箱需要4平方米，而小箱子每箱需要3平方米。制造商需要制造至少三个大盒子和至少两倍这么大的小盒子。如果有 60 平方米的纸板库存，如果大箱子和小箱子的利润分别为每箱 3 卢比和 2 卢比，每个箱子应该做多少才能使总利润最大化？

解决方案：

Assume x large boxes and y small boxes be manufactured. Number of boxes cannot be negative.

Therefore, 

x ≥ 0, y ≥ 0 (always)

The large boxes need 4 sq. meter per box while the small boxes need 3 sq. meter per box and 

if 60 sq. meter of cardboard is in stock. 4x + 3y ≤ 60

The manufacturer is needed to make at least three large boxes and at least twice

so many small boxes so large boxes.

x ≥ 3y ≥ 2x

If the profits on the large and small boxes are Rs 3 and Rs 2 per box. 

Thus, profit gained by him on x large boxes and y small boxes is Rs 3x and Rs 2y respectively.

Total profit = Z = 3x + 2y

Therefore, the mathematical formulation of the given LPP is 

Max Z = 3x + 2y

Subject to constraints

4x + 3y ≤ 60

x ≥ 3

y ≥ 2x 

x ≥ 0, y ≥ 0

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

4x + 3y = 60, 

x = 3, 

y = 2x, 

x = 0 and 

y = 0

Area shown by 4x + 3y ≤ 60:

The line 4x + 3y = 60 connects the coordinate axes at A(15, 0) and B(0, 20) respectively. 

By connecting these points we will get the line 4x + 3y = 60. 

So, (0, 0) assure the 4x + 3y = 60. 

So, the area which haves the origin shows the solution set of the in-equation 4x + 3y ≤ 60.

Area shown by x ≥ 3:

The line x = 3 is the line passes through 3, 0 and is parallel to Y axis. 

The area to the right of the line x = 3 will assure the in-equation x ≥ 3.

Area shown by y ≥ 2x:

The line y = 2x is the line that passes through (0, 0). 

The area above the line y = 2x will assure the in-equation y ≥ 2x. 

Like if we take an example taking a point 5, 1 below the line y = 2x .

Here, 1 < 10 which does not assure the in-equation y ≥ 2x. 

Therefore, the area above the line y = 2x will assure the in-equation y ≥ 2x.

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 

4x + 3y ≤ 60, 

x ≥ 3, 

y ≥ 2x, 

x ≥ 0 and y ≥ 0 are as follows

The corner points are E(3, 16), D(6, 12) and C(3, 6). 

The values of Z at the corner points are

Corner points	Z = 3x + 2y
E	41
D	42
C	21

The maximum value of Z is 42 which is at D(6, 12).

Therefore, for maximum profit is Rs 42, 6 units of large boxes and

12 units of smaller boxes must be manufactured.