第 12 类 RD Sharma 解决方案 – 第 30 章线性规划 – 练习 30.1 |设置 1
问题 1.一家小型制造公司生产两种类型的小工具 A 和 B,它们首先在铸造厂进行加工,然后送到机加工车间进行精加工。每个车间生产 A 和 B 的每个单位所需的工时数,以及公司每周可用的工时数如下:
Gadget | Foundry | Machine-shop |
A | 10 | 5 |
B | 6 | 4 |
Firm’s capacity per week | 1000 | 60 |
销售 A 的利润是每单位 30 卢比,而每单位 B 的销售利润是 20 卢比。问题是要确定小工具 A 和 B 的每周产量,以使总利润最大化。将此问题表述为 LPP。
回答:
The given data may be put in the following tabular form:
Let the required weekly production of gadgets A and B be x and y respectively.
Given that, profit on each gadget A is Rs 30 and gadget B is Rs 20.
Profit on x gadget of type A = 30x
Profit on y gadget of Type B = 20y
Let Z denote the total profit, so
Z = 30x + 20y
Given, production of one gadget A requires 10 hours per week for foundry and gadget B requires 6 hours per week for foundry.
So, x units of gadget A requires 10x hours per week and y units of gadget B requires 6y hours per week. But the maximum capacity of foundry per week is 1000 hours. So,
10x + 6y <= 1000 (First constraint.)
Given, production of one unit gadget A requires 5 hours per week of machine shop and production off one unit of gadget B requires 4 hours per week of machine shop.
So, x units of gadget A requires 5x hours per week and y units of gadget B requires 4y hours per week, but the maximum capacity of machine shop is 600 hours per week.
So, 5x + 4y <= 600 (Second constraint.)
Hence, mathematical formulation of LPP is:
Find x and y which maximize Z = 30x + 20y
Subject to constraints,
10x + 6y <= 1000
5x + 4y <=600
And, x, y >=0 (Since production cannot be less than 0.)
问题 2:一家公司正在生产两种产品 A 和 B。生产一单位产品 A 和 B 的成本分别为 60 卢比和 80 卢比。根据协议,公司必须向其常客供应至少 200 件 B 产品。一个单位的产品 A 需要一个机器工时,而产品 B 在公司内有大量可用的机器工时。产品 A 可用的总机器时间为 400 小时。产品 A 和 B 各一个单位需要一个工时,总共有 500 个工时可用。公司希望通过满足给定的要求来最小化生产成本。将此问题表述为 LPP。
回答:
The given information can be written in tabular form as follows :
Minimum supply of product B is 200 units.
Let the production of product A be x units and production of product B be y units.
Profit on one unit of product A = Rs 60
Profit on x units of product A = Rs 60y
Profit on one unit of product B = Rs 80
Profit on y units of product B = Rs 80y
Let Z denote the total profit. So,
Z = 60x + 80y
Given, minimum supply of product B is 200.
So, y >= 200 (First constraint)
Given that the production of one unit of product A requires 1 hour of machine hours, so x units of product A requires x hours , but given total machine hours available for product A is 400 hours. So,
x <= 400 (Second constraint)
Given, each unit of product A and B requires one hour of labour hour, so x units of product A require x hour and y unit of product B require y hour of labour hours, but total labour hours available are 500. So,
x + y <= 500 (Third constraint)
Hence, mathematical formulation of LPP is,
Find x and y which minimize Z = 60x + 80y
Subject to constraints,
y >= 200,
x <= 400,
x + y <= 500
x, y >=0 (Since production cannot be less than 0.)
问题 3:一家公司生产 3 种产品 A、B 和 c。利润分别为 3 卢比、2 卢比和 4 卢比。该公司有 2 台机器,以下是每种产品每台机器所需的处理时间(以分钟为单位):
Gadget | Foundry | Machine-shop | Profit |
A | 10 | 5 | Rs. 30 |
B | 6 | 4 | Rs. 20 |
Firm’s capacity per week | 1000 | 600 |
机器 M1 和 M2 分别有 2000 和 2500 机器分钟。公司必须生产 100 个 A、200 个 B 和 50 个 C,但不得超过 150 个 A。建立 LPP 以最大化利润。
回答:
The given information can be written in tabular form as follows :
Let required production of product A, B and C be x, y and z units respectively.
Given, profit on one unit of product A, B and C are Rs 3, Rs. 2 and Rs 4 respectively
So, profit on x unit of A, y unit of B and z unit of C are given by Rs 3x, Rs. 2y and Rs 4z respectively.
Let U be the total profit, so
U = 3x + 2y + 4z
Given, one unit of product A, B and C requires 4, 3 and 5 minutes on machine M1. So, x units of product A, y units of product B and z units of product C requires 4x, 3y and 5z minutes on machine M1. Therefore,
4x + 3y + 5z <= 2000 (First constraint)
Given, one unit of product A, B and C requires 2, 2 and 4 minutes on machine M1. So, x units of product A, y units of product B and z units of product C requires 2x, 2y and 4z minutes on machine M2. Therefore,
2x + 2y + 4z <= 2500 (Second constraint)
Also, given that the firm must manufacture 100 A’s, 200 B’s and 50 C’s but not more than 150 A’s. So,
100 <= x <= 150
y >=200
z >=50
Hence the mathematical formulation of LPP is:
Find x, y and z which minimize U = 3x + 2y + 4z
Subject to constraints,
4x + 3y + 5z <= 2000
2x + 2y + 4z <= 2500
100 <= x <= 150
y >=200
z >=50
x, y, z >=0
问题 4:一家公司生产两种产品 A 和 B,并以 A 型 2 卢比和 B 型 3 卢比的利润出售它们。每种产品在两台机器 M1 和 M2 上加工。 A 类需要在 M1 上处理一分钟,在 M2 上处理两分钟; B 型需要在 M1 上一分钟,在 M2 上一分钟。 M1 机器的可用时间不超过 6 小时 40 分钟,而 M2 机器在任何工作日的可用时间为 10 小时。将此问题表述为 LPP。
回答:
The given information can be written in tabular form as follows :Product M1 M2 Profit A 1 2 2 B 1 1 3 Capacity 6 hours 40 min. = 400 min 10 hours = 600 min.
Let the required production of product A be x units and product B be y units.
Profit on one unit of product A = Rs 2
Profit on x units of product A = Rs 2x
Profit on one unit of product B = Rs 3
Profit on y units of product A = Rs 3y
Let the total profit be Z, so
Z = 2x + 3y
On machine M1,
Production of one unit of product A requires 1 minute.
Production of x units of product A requires x minute.
Production of one unit of product B requires 1 minute.
Production of y units of product B requires y minute.
But total time available on machine M1 is 600 minutes.
So, x + y <=400 (First constraint)
On machine M2,
Production of one unit of product A requires 2 minute.
Production of x units of product A requires 2x minute.
Production of one unit of product B requires 1 minute.
Production of y units of product B requires y minute.
But total time available on machine M2 is 600 minutes.
So, 2x + y <=600 (Second constraint)
Hence the mathematical formulation of LPP is:
Find x, y and z which maximize Z = 2x + 3y
Subject to constraints,
x + y <=400
2x + y <=600
x, y >=0 (Production cannot be less than zero.)
问题 5:一家橡胶公司生产 A、B、C 三种轮胎,每种类型需要加工两个工厂,工厂 1 和工厂 2。这两个工厂的产能,以每天的轮胎数量计,分别为如下:
Product | Machine hours | Labour hours | Profit |
A | 1 | 1 | Rs 60 |
B | – | 1 | Rs 80 |
Total Capacity | 400 for A | 500 |
轮胎A、B、C的月需求量分别为2500、3000、7000。如果工厂 1 每天的成本为 2500 卢比,而工厂 2 的运营成本为每天 3500 卢比,那么每个月应该运行多少天才能在满足需求的同时最大限度地降低成本?将此问题表述为 LPP。
回答:
The given information can be written in tabular form as follows :Plant A B C Cost 1 50 100 100 2500 2 60 60 200 3500 Monthly demand 2500 3000 7000
Let plant 1 requires x days and plant 2 requires y days per month to minimize cost.
Given, plant 1 and plant 2 requires Rs 2500 per day and Rs 3500 per day respectively.
So, cost to run plant 1 and 2 is Rs 2500x and Rs 3500y per month.
Let Z be the total cost per month, so
Z = 2500x + 3500y
Given production of tyre A from plant 1 and 2 per day is 50 and 60 respectively. So, production of tyre A from plant 1 and 2 per month will be 50x and 60y respectively. But the maximum demand of tyre A is 2500 per month. So,
50x + 60y >= 2500 (First constraint)
Given production of tyre B from plant 1 and 2 per day is 100 and 60 respectively. So, production of tyre B from plant 1 and 2 per month will be 100x and 60y respectively. But the maximum demand of tyre B is 3000 per month. So,
100x + 60y >= 3000 (Second constraint)
Given production of tyre C from plant 1 and 2 per day is 100 and 200 respectively. So, production of tyre C from plant 1 and 2 per month will be 100x and 200y respectively. But the maximum demand of tyre A is 7000 per month. So,
100x + 200y >= 7000 (Third constraint)
Hence the mathematical formulation of LPP is:
Find x and y which minimize Z = 2500x + 3500y
Subject to constraints,
50x + 60y >= 2500
100x + 60y >= 3000
100x + 200y >= 7000
x, y >= 0 (Since number of days cannot be less than zero)
问题 6:一家公司销售两种不同的产品 A 和 B。这两种产品是在一个共同的生产过程中生产的,并且在两个不同的市场上销售。生产过程总产能为45000工时。生产一单位A需要5小时,生产一单位B需要3小时。经过市场调查,公司负责人认为A的最大可以销售7000个,B的最大可以销售10000个。如果产品 A 的利润为每单位 60 卢比,产品 B 的利润为 40 卢比,每个产品应销售多少单位才能使利润最大化?将此问题表述为 LPP。
回答:
Product Man Hours Maximum demand Profit A 5 7000 60 B 3 10000 40 Total Capacity 45000
Let the required production of product A be x units and product B be y units.
Profit on one unit of product A = Rs 60
Profit on x units of product A = Rs 60x
Profit on one unit of product B = Rs 40
Profit on y units of product A = Rs 40y
Let the total profit be Z, so
Z = 60x + 40y
Production of one unit of product A requires 5 hours.
Production of x units of product A requires 5x hours.
Production of one unit of product B requires 3 hours.
Production of y units of product B requires 3y hours.
But the total man hours available are 45000 hours, so
5x + 3y <= 450000 (First constraint)
Given, maximum demand for product A is 7000, so
x <= 7000 (Second constraint)
Given, maximum demand for product B is 10000, so
y <= 10000 (Third constraint)
Hence the mathematical formulation of LPP is:
Find x and y which minimize Z = 2500x + 3500y
Subject to constraints,
5x + 3y <= 450000
x <= 7000
y <= 10000
x, y >= 0 (Since production cannot be less than zero)
问题 7:为了保持健康,一个人必须满足几种营养素的某些最低每日需求量。假设营养素只有三种——钙、蛋白质和卡路里,而一个人的饮食只有两种食物,1和2,其价格和营养素含量如下表所示:
Machine | Products | ||
A | B | C | |
M1 | 4 | 3 | 5 |
M2 | 2 | 2 | 4 |
两种食物的哪种组合可以满足日常需求并花费最少?将此问题表述为 LPP。
回答:
Let x and y be the packets of 25 gm of Food 1 and Food 2 purchased. Let Z be the price paid. Obviously we have to minimize the price.
Take a mass balance on the nutrients from Food 1 and 2,
Calcium: 10x + 4y >= 20, i.e., 5x + 2y >= 10 (Equation 1)
Protein: 5x + 5y >= 20, i.e., x + y >= 20 (Equation 2)
Calories: 2x + 6y >=13 (Equation 3)
These become the constraints for the cost function, Z to be minimized , i.e, 0.6x + y = Z, given cost of Food 1 is Rs 0.6 and Rs 1 per lb.
From equation 1, 2 and 3 we get points on the X & Y- axis as (0,5) & (2,0); (0,4) & (4,0); (0,13/6) & (6.5, 0).
Plotting these,
The smallest value of Z is 2.9 at the point (2.75, 1.25). We cannot say that the minimum value of Z is 2.9 as the feasible region is unbounded.
Therefore, we have to draw the graph of the inequality 0.6x + y < 2.9.
Plotting this to see if the resulting line has any plot common with the feasible region. Since there are no common points, this is the minimum value of the function Z and the mix is
Food 1 = 2.75 lb; Food 2 = 1.25 lb; Price = Rs 2.9
When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities, this optimal value must occur at a corner point (vertex) of the feasible region.
Here the feasible region is the unbounded region A-B-C-D.
Computing the value of Z a t the corner points of the feasible region ABHGPoint Corner Point Value of Z = 0.6x + y A 2 ,5 6.2 B 0.67, 3.33 3.73 C 2.75, 1.25 2.9 D 6.5, 2.16 6.06
问题 8:制造商可以在给定的时间段内生产两种产品 A 和 B。这些产品中的每一种都需要四种不同的制造操作:磨削、车削、组装和测试。每单位产品 A 和 B 的制造要求如下所示:
Product | Machine(M1) | Machine(M2) | Profit |
A | 4 | 2 | 3 |
B | 3 | 2 | 2 |
C | 5 | 4 | 4 |
Maximum capacity | 2000 | 2500 |
在给定时间段内,这些操作的可用小时数为:研磨 30; 60岁;组装200个;测试 200。对利润的贡献是每单位 A 为 20 卢比,每单位 B 为 30 卢比。公司可以按现行市场价格出售其生产的所有产品。确定在给定时间段内生产的 A 和 B 的最佳数量。将此问题表述为 LPP。
回答:
Product Grinding Turning Assembling Testing Profit A 1 3 6 5 2 B 2 1 3 4 3 Maximum capacity 30 hours 60 hours 200 hours 200 hours
Let the required production of product A be x units and product B be y units.
Profit on one unit of product A = Rs 2
Profit on x units of product A = Rs 2x
Profit on one unit of product B = Rs 3
Profit on y units of product A = Rs 3y
Let the total profit be Z, so
Z = 2x + 3y
Production of one unit of product A requires 1 hours of grinding.
Production of x units of product A requires x hours of grinding.
Production of one unit of product B requires 2 hours of grinding.
Production of y units of product B requires 2y hours of grinding.
But the total time available for grinding is 30 hours, so
x + 2y <= 30 (First constraint)
Production of one unit of product A requires 3 hours of turning.
Production of x units of product A requires 3x hours of turning..
Production of one unit of product B requires 1 hours of turning..
Production of y units of product B requires y hours of turning..
But the total time available for turning.is 60 hours, so
3x + y <= 60 (Second constraint)
Production of one unit of product A requires 6 hours of assembling.
Production of x units of product A requires 6x hours of assembling.
Production of one unit of product B requires 3 hours of assembling.
Production of y units of product B requires 3y hours of assembling.
But the total time available for assembling is 200 hours, so
6x + 3y <= 200 (Third constraint)
Production of one unit of product A requires 5 hours of testing.
Production of x units of product A requires 5x hours of testing.
Production of one unit of product B requires 4 hours of testing.
Production of y units of product B requires 4y hours of testing.
But the total time available for testing is 200 hours, so
5x + 4y <= 200 (Fourth constraint)
Hence the mathematical formulation of LPP is:
Find x and y which maximize Z = 2x + 3y
Subject to constraints,
x + 2y <= 30
3x + y <= 60
6x + 3y <= 200
5x + 4y <= 200
x, y >= 0 (Since production cannot be negative)
问题 9:维生素 A 和 B 存在于两种不同的食物 F1 和 F2 中。一单位食物 F1 含有 2 单位维生素 A 和 3 单位维生素 B。一单位食物 F2 含有 4 单位维生素 A 和 2 单位维生素 B。一单位食物 F1 和 F2 分别花费 50 卢比和 25 卢比。维生素 A 和 B 的最低每日需求量分别为 40 和 50 单位。假设超过维生素 A 和 B 的每日最低需求量的任何东西都无害,以最低成本找到食物 F1 和 F2 的最佳混合物,以满足维生素 A 和 B 的每日最低需求量。将此问题表述为 LPP。
回答:
Given information can be tabulated as below:Foods Vitamin A Vitamin B Cost F1 2 3 5 F2 4 2 2.5 Minimum daily requirement 40 50
Let the required quantity of Food F1 be x units and quantity of food F2 be y units.
Given, cost of one unit of food F1 and F2 are Rs 5 and Rs 2.5 respectively. So, cost of x units of Food F1 and y units of food F2 are Rs 5x and 2.5y respectively.
Let Z be the total cost, so
Z = 5x + 2.5y
Given, one unit of food F1 and F2 contain 2 and 4 units of Vitamin A respectively, so x unit of food F1 and y units of food F2 contain 2x and 4y units of Vitamin A respectively, but minimum requirement of Vitamin A is 40 unit, so
2x + 4y >= 40 (First constraint)
Given, one unit of food F1 and F2 contain 3 and 2 units of Vitamin B respectively, so x unit of food F1 and y units of food F2 contain 3x and 2y units of Vitamin B respectively, but minimum requirement of Vitamin A is 50 unit, so
3x + 2y >= 50 (Second constraint)
Hence the mathematical formulation of LPP is:
Find x and y which maximize Z = 5x + 2.5y
Subject to constraints,
2x + 4y >= 40
3x + 2y >= 50
x, y>= 0 (Since requirement of food F1 and F2 cannot be less than zero)
问题 10:一家汽车制造商在分为两个车间的工厂生产汽车和卡车。进行基本装配操作的车间 A 必须在每辆卡车上工作 5 个工作日,但在每辆汽车上只能工作 2 个工作日。 B 车间进行精加工作业,其生产的每辆汽车或卡车必须工作 3 个工作日。由于人员和机器的限制,A 店每周有 180 个工作日,而 B 店每周有 135 个工作日。如果制造商每辆卡车的利润为 30000 卢比,每辆汽车的利润为 2000 卢比,他应该分别生产多少才能使利润最大化?进行基本装配操作的车间 A 必须在每辆卡车上工作 5 个工作日,但在每辆汽车上只能工作 2 个工作日。
回答:
Let the number of automobiles produced be x and let the number of trucks produced be y.
Let Z be the profit function to be maximized.
Z = 2000x + 30000y
The constraints are on the man hours worked
Shop A : 2x + 5y <= 180 (Equation 1)
Shop B : 3x + 3y <= 135 ((Equation 2)
Corner points can be obtained from
2x + 5y = 180 , i.e, x= 0; y = 36 & x = 90; y = 0
3x + 3y = 135 , i.e. , x =0; y = 45 & x = 45; y = 0
Solving Equation 1 and Equation 2 gives x = 15 & y = 30Corner point Value of Z = 2000x + 30000y 0, 0 0 0, 36 1080000 15, 30 930000 45, 0 90000
0 automobiles and 36 trucks will give maximum profit of 1080000 Rs.