第 12 类 RD Sharma 解决方案 - 第 30 章线性规划 - 练习 30.3

问题 1. 两种食物 F ₁和 F ₂的饮食含有营养硫胺素、磷和铁。下表列出了每种食物中每种营养素的含量（以毫克/25 克计）。

Nutrients/Food	F1	F2
Thiamine	0.25	0.10
Phosphorus	0.75	1.50
Iron	1.60	0.80

饮食中营养素的最低要求为 1.00 毫克硫胺素、7.50 毫克磷和 10.00 毫克铁。 F ₁的成本是每 25 克 20 派萨，而 F ₂的成本是每 25 克 15 派萨。找出饮食的最低成本。

解决方案：

Let 25x grams of food F₁ and 25y grams of food F₂ be used to fulfill the minimum requirement of thiamine, phosphorus and iron.

Given        

The minimum requirement of the nutrients in the diet are 1.00 mg of thiamine,7.50 mg of phosphorus and 10.00 mg of iron. Therefore,

0.25x+0.10y ≥ 1

0.75x+1.50y ≥7.5

1.6x+0.8y ≥ 10

Since the quantity cannot be negative 

⇒ x,y ≥ 0

The cost of F₁ is 20 paise per 25 grams and the cost of F₂ is 15 paise per 25 grams. 

Therefore the cost of 25x grams of food F₁ and 25y grams of food F₂ is (0.20x+0.15y) rupees.

Let us minimize z=0.20x+0.15y

subject to

0.25x+0.10y ≥ 1

0.75x+1.50y ≥7.5

1.6x+0.8y ≥ 10

x, y ≥ 0

First we will convert the given inequations to equations. we get,

0.25x+0.10y = 1

0.75x+1.50y =7.5

1.6x+0.8y = 10

x=0 and y=0

Region represented by 0.25x+0.10y ≥ 1:

The line 0.25x+0.10y =1 meets the coordinate axis at A(4, 0) and B(0, 10).

Clearly (0, 0) do not satisfy the inequation 0.25x+0.10y≥1.So the region in XY plane that does not contain the origin represents the solution set of given inequation.

Region represented by 0.75x+1.50y ≥7.5:

The line 0.75x+1.50y =7.5 meets the coordinate axis at C(10, 0) and D(0, 5).

Clearly (0, 0) do not satisfy the inequation 0.75x+1.50y≥7.5.So the region in XY plane that doe snot contain the origin represents the solution set of this inequation.

Region represented by 1.6x+0.8y ≥ 10:

The line 1.6x+0.8y = 10 meets the coordinate axis at E(25/4, 0) and F(0, 25/2).

Clearly (0, 0) do not satisfy the inequation 1.6x+0.8y ≥ 10.So the region in XY plane that does not contain the origin represents the solution set of given inequation.

Region represented by x,y ≥ 0:

In the region represented by x≥0 and y≥0, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

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

The corner points of a Feasible region are F(0, 12.5), G(5, 2.5), C(10, 0)

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

Thus, the minimum cost is at G which is 1.375 rupees.

问题 2. 病人的饮食必须包含至少 4000 单位的维生素、50 单位的矿物质和 1400 单位的卡路里。两种食物 A 和 B 的价格分别为每单位 4 卢比和 3 卢比。如果一单位的 A 含有 200 单位的维生素、1 单位的矿物质和 40 卡路里，而 1 单位的食物 B 含有 100 单位的维生素、2 单位的矿物质和 40 卡路里，请找出应该使用哪种食物组合来最少成本？

解决方案：

Let the sick person takes x units and y units of food A and B respectively.

1 unit of food A costs Rs 4 and that of food B costs Rs 3.

Therefore, x units of food A costs Rs 4x and y units of food B costs Rs 3y.

Total cost =Rs (4x+3y) 

Let z=4x+3y

If one unit of A contains 200 units of vitamin and one unit of food B contains 100 units of vitamin.

Then, x units of food A and y units of food B contains (200x+100y) units of vitamin.

But a diet for a sick person must contain at least 4000 units of vitamin.

Thus 200x+100y≥4000

If one unit of A contains 1 unit of mineral and one unit of food B contains 2 units of mineral.

Then, x units of food A and y units of food B contains x+2y units of mineral.

But a diet for a sick person must contain at least 50 units of vitamin.

Thus, x+2y≥50

If one unit of A contains 40 calories and one unit of food B contains 40 calories.

Then x units of food A and y units of food B contains 40x+40y units of calories.

But a diet for a sick person must contain at least 1400 units of calories.

Thus, 40x+40y≥1400

Finally the quantities of food A and food B are nonnegative values.

So, x, y ≥ 0

Hence the required LPP is as follows:

Min z=4x+3y

subject to

200x+100y≥4000

x+2y≥50

40x+40y≥1400

 x, y ≥ 0

First we will convert the given inequations to equations. we get,

200x+100y=4000

x+2y=50

40x+40y=1400

x=0 and y=0

Region represented by 200x+100y≥4000:

The line 200x+100y=4000 meets the coordinate axis at A₁(20, 0) and B₁(0, 40) respectively.

Clearly (0, 0) does not satisfy inequation 200x+100y≥4000.

So the region in XY plane that does not contain the origin represents the solution set of this inequation.

Region represented by x+2y≥50:

The line x+2y=50 meets the coordinate axis at C₁(50, 0) and D₁(0, 25) respectively.

Clearly (0, 0) does not satisfy inequation x+2y≥50.

So the region in XY plane that does not contain the origin represents the solution set of this inequation.

Region represented by 40x+40y≥1400:

The line 40x+40y=1400 meets the coordinate axis at E₁(35, 0) and F₁(0, 35) respectively.

Clearly (0, 0) does not satisfy inequation 40x+40y≥1400.

So the region in XY plane that does not contain the origin represents the solution set of this inequation.

Region represented by x≥ 0 and y≥0:

Every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

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

The corner points of a Feasible region are B₁(0, 40), G₁(5, 30), C₁(50, 0), H₁(20, 15).

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

Thus the minimum cost is at G₁ which is 110 rupees.

问题 3. 为了保持健康，一个人必须满足以下三种营养素的最低每日需求量：钙、蛋白质和卡路里。饮食仅包括 I 和 II 项目，其价格和营养成分如下所示：

Nutrients/Food	F1	F2
Thiamine	0.25	0.10
Phosphorus	0.75	1.50
Iron	1.60	0.80

找出食物的组合，以使成本最小化。

解决方案：

Let the person takes x units and y units of food I and II respectively.

Since, per unit of food I costs Rs 0.60 and that of food II costs Rs.1.00.

Therefore, x units of food I costs Rs 0.60x and y units of food II costs Rs1.00y.

Total cost per day=0.60x+1.00y

Let z=0.60x+1.00y

Since, each unit of food I contains 10 units of calcium.

Therefore, x units of food I contains 10x units of calcium.

Each unit of food II contains 4 units of Calcium.

So, y units of food II contains 4y units of calcium.

Thus, x units of food I and y units of food II contains (10x+4y) units of calcium.

But the minimum requirement is 20 units of calcium.

Thus, 10x+4y≥20

Since, each unit of food I contains 5 units of protein.

Therefore, x units of food I contains 5x units of protein.

Each unit of food II contains 6 units of protein.

So, y units of food II contains 6y units of protein.

Thus, x units of food I and y units of food II contains (5x+6y) units of protein.

But the minimum requirement is 20 units of protein.

Thus, 5x+6y≥20

Since, each unit of food I contains 2 units of Calories.

Therefore, x units of food I contains 2x units of Calories.

Each unit of food II contains 6 units of Calories.

So, y units of food II contains 6y units of Calories.

Thus, x units of food I and y units of food II contains (2x+6y) units of Calories.

But the minimum requirement is 12 units of Calories.

Thus, 2x+6y≥12

Finally the quantities of food I and food II are nonnegative values.

So, x, y ≥ 0

Hence the required LPP is as follows:

Min z=0.60x+1.00y

subject to

10x+4y≥20

5x+6y≥20

2x+6y≥12

x, y ≥ 0

First we will convert the given inequations to equations. we get,

10x+4y=20

5x+6y=20

2x+6y=12

x=0 and y=0

Region represented by 10x+4y≥20:

The line 10x+4y=20 meets the coordinate axes at A(2, 0) and B(0, 5) respectively.

Clearly (0, 0) does not satisfy inequation 10x+4y≥20.

So the region in XY plane that does not contain the origin represents the solution set of this inequation.

Region represented by 5x+6y≥20:

The line 5x+6y=20 meets the coordinate axes at C(4, 0) and D(0, 10/3) respectively.

Clearly (0, 0) does not satisfy inequation 5x+6y≥20.

So the region in XY plane that does not contain the origin represents the solution set of this inequation

Region represented by 2x+6y≥12:

The line 2x+6y=12 meets the coordinate axes at E(6, 0) and F(0, 2) respectively.

Clearly (0, 0) does not satisfy inequation 2x+6y≥12.

So the region in XY plane that does not contain the origin represents the solution set of this inequation.

Region represented by x, y ≥ 0:

Every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

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

The corner points of a Feasible region are B(0, 5), G(1, 5/2), E(6, 0), H(8/3, 10/9).

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

Points	Value of z
B	0.6(0)+5=5
G	0.6(1)+5/2=3.1
H	0.6(8/3)+(10/9)=2.711
E	0.6(6)+0=3.6

From the table, we can see that minimum cost is at H(8/3, 10/9) and that value is Rs 2.71

问题 4. 医院营养师希望找到 A 和 B 两种食物的最便宜组合，其中至少含有 0.5 毫克硫胺素和至少 600 卡路里的热量。每单位 A 含有 0.12 毫克硫胺素和 100 卡路里，而每单位 B 含有 0.10 毫克硫胺素和 150 卡路里。如果每种食物每单位成本为 10 派萨，那么每种食物应以最低成本组合多少单位？

解决方案：

Let the dietician wishes to mix x units of food A and y units of food B

Therefore x,y≥0

The given information can be tabulated as follows

	Thiamine(mg)	Calories(mg)
Food A	0.12	100
Food B	0.1	150
Minimum requirement	0.5	600

According to question,

The constraints are

0.2x+0.1y≥0.5

100x+150y≥600

It is given that each food costs 10 paise per unit

Let the total cost be z

 z=10x+10y

Now the mathematical formulation of given linear programming problem is

0.2x+0.1y≥0.5

100x+150y≥600

Region represented by 0.12x+0.1y≥0.5:

The line 0.12x+0.1y=0.5 meets the coordinate axes at A₁(25/6, 0) and B₁(0, 5) respectively.

Clearly (0, 0) does not satisfy inequation 0.12x+0.1y≥0.5.

So the region in XY plane that does not contain the origin represents the solution set of this inequation.

Region represented by 100x+150y≥600:

The line 100x+150y=600 meets the coordinate axes at C₁(6, 0) and D₁(0, 4) respectively.

Clearly (0, 0) does not satisfy inequation 100x+150y≥600.

So the region in XY plane that does not contain the origin represents the solution set of this inequation.

Region represented by x,y≥0:

Every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

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

The corner points of a Feasible region are B₁(0, 5), E₁(15/8, 11/4), C₁(6, 0).

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

Corner point	z=10x+10y
B1	50
E1	46.2
C1	60

The minimum of z is at E₁(15/8, 11/4) and the value is Rs 46.2.

Hence the cheapest combination of foods will be 1.875 units of food A and 2.75 units of food B.

问题 5. 营养师将两种食物混合在一起，使混合物中至少含有 6 个单位的维生素 A、7 个单位的维生素 B、11 个单位的维生素 C 和 9 个单位的维生素 D。 1 公斤食物 X 和 1 公斤食物 Y 如下：

Points	Value of z
F	0.20(0)+0.15(12.5)=1.875
G	0.20(5)+0.15(2.5)=1.375
C	0.20(10)+00.15(0)=200

一公斤食物 X 的价格为 5 卢比，而一公斤的食物 Y 的价格为 8 卢比。找出能够产生所需饮食的混合物的最低成本。

解决方案：

Let the dietician wishes to mix x kg of food X and y kg of food Y.

Therefore, x,y≥0

Given

	Vitamin A	Vitamin B	Vitamin C	Vitamin D
Food X	1	1	1	2
Food Y	2	1	3	1

It is given that the mixture should contain at least 6 units of vitamin A,7 units of vitamin B ,11 units of vitamin C and 9 units of vitamin D.

Therefore, the constraints are

x+2y≥6

x+y≥7

x+3y≥11

2x+y≥9

It is given that cost of food X is Rs 5 per kg and cost of food Y is Rs 8 per kg.

Let the total cost be z.

Thus, z=5x+8y

Now the mathematical formulation of given linear programming problem is

Minimize Z=5x+8y

subject to

x+2y≥6

x+y≥7

x+3y≥11

2x+y≥9

First we will convert the given inequations to equations. we get

x+2y=6

x + y=7

x+3y=11

2x+y=9

x=0 and y=0

Region represented by x+2y≥6:

The line x+2y=6 meets the coordinate axes at A₁(6, 0) and B₁(0, 3) respectively.

Clearly (0, 0) does not satisfy inequation x+2y≥6.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by x+y≥7:

The line x + y=7 meets the coordinate axes at C₁(7, 0) and D₁(0, 7) respectively.

Clearly (0, 0) does not satisfy inequation x+y≥7.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by x+3y≥11:

The line x+3y=11 meets the coordinate axes at E₁(11, 0) and F₁(0, 11/3) respectively.

Clearly (0, 0) does not satisfy inequation x+3y≥11.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by 2x+y≥9:

The line 2x+y=9 meets the coordinate axes at G₁(9/2, 0) and H₁(0, 9) respectively.

Clearly (0, 0) does not satisfy inequation 2x+y≥9.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by x,y≥0:

Every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

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

The corner points of a Feasible region are H₁(0, 9), I₁(2, 5), E₁(11, 0), J₁(5, 2).

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

Corner Point	z=5x+8y
H1	72
I1	50
J1	41
E1	55

The minimum value of z is at J₁(5, 2) which is Rs 41.

Hence cheapest combination of foods will be 5 units of food X and 2 units of food Y.

问题 6. 饮食应包含至少 80 单位的维生素 A 和 100 单位的矿物质。有两种食物 F ₁和 F ₂可供选择。食品 F ₁的成本为每单位 4 卢比，而 F ₂的成本为每单位 6 卢比。一个单位的食物 F ₁含有 3 个单位的维生素 A 和 4 个单位的矿物质。一个单位的食物 F ₂含有 6 个单位的维生素 A 和 3 个单位的矿物质。将此公式化为一个线性规划问题，并以图形方式找到由这些食物的混合物组成的同时满足矿物质营养要求的最低饮食成本。

解决方案：

Let the dietician wishes to mix x units of food F₁ and y units of food F₂.

Clearly x,y≥0

The given information can be tabulated as follows

	Vitamin A	Vitamin B
Food F1	3	4
Food F2	6	3
Minimum  requirement	80	100

The constraints are

3x+6y≥80

4x+3y≥100

It is given that cost of food F₁ is Rs 4 per unit and cost of food F₂ is Rs 6 per unit.

Therefore cost of x units of food F₁ and y units of food F₂ is Rs 4x and Rs 6y respectively.

Let the total cost be z.

z=4x+6y

Now the mathematical formulation of given linear programming problem is

Minimize Z=4x+6y

subject to

3x+6y≥80

4x+3y≥100

 x,y≥0

First we will convert the given inequations to equations. we get

3x+6y=80

4x+3y=100

x=0 and y=0

Region represented by 3x+6y≥80:

The line 3x+6y=80 meets the coordinate axes at A(80/3, 0) and B(0, 40/3) respectively.

Clearly (0, 0) does not satisfy inequation 3x+6y≥80.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by 4x+3y≥100:

The line 4x+3y=100 meets the coordinate axes at C(25, 0) and D(0, 100/3) respectively.

Clearly (0, 0) does not satisfy inequation 4x+3y≥100.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by x,y≥0:

Every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

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

The corner points of a Feasible region are D(0, 100/3), E(24, 4/3), A(80/3, 0).

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

Corner Point	z=4x+6y
D(0, 100/3)	200
E(24, 4/3)	104
A(80/3, 0)	320/3

The minimum value of z is at E(24, 4/3) and the value is Rs 104.

问题 7. Kellogg 是一种由麸皮和大米混合而成的新谷物，含有至少 88 克蛋白质和至少 36 毫克铁。知道麸皮每公斤含有 80 克蛋白质和 40 毫克铁，大米每公斤含有 100 克蛋白质和 30 毫克铁，如果麸皮成本为每公斤 5 卢比，大米成本为每公斤4卢比。

解决方案：

Let the cereal contain x kg of bran and y kg of rice.

Therefore x,y≥0

The given information can be tabulated as follows

	Protein(gms)	Iron(milligrams)
Bran	80	40
Rice	100	30
Minimum requirement	88	36

Bran and Rice contains at least 88 grams of protein and at least 36 milligrams of iron.

Thus the constraints are

80x+100y≥88

40x+30y≥36

It is given that bran costs Rs 5 per kg and rice costs Rs 4 per kg. Therefore cost of x kg of bran and y kg of rice is Rs 5x and Rs 4y respectively.

Hence, total profit is Rs(5x+4y).

Let the total cost be z

Thus, z=5x+4y

Now the mathematical formulation of given linear programming problem is

Minimize Z=5x+4y

subject to

80x+100y≥88

40x+30y≥36

x,y≥0

First we will convert the given inequations to equations. we get

80x+100y=88

40x+30y=36

x=0 and y=0

Region represented by 80x+100y≥88:

The line 80x+100y=88 meets the coordinate axes at A(1.1, 0) and B(0, 0.88) respectively.

Clearly (0, 0) does not satisfy inequation 80x+100y≥88.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by 40x+30y≥36:

The line 40x+30y=36 meets the coordinate axes at C(0.9, 0) and D(0, 1.2) respectively.

Clearly (0, 0) does not satisfy inequation 40x+30y≥36.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by x,y≥0:

Every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

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

The corner points of a Feasible region are D(0, 1.2), E(0.6, 0.4), A(1.1, 0).

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

Corner Point	z=5x+4y
D(0, 1.2)	4.8
E(0.6, 0.4)	4.6
A(1.1, 0)	5.5

The minimum value of z is Rs 4.6 at E(0.6, 0.4)

问题 8. 批发商经营 A 和 B 两种坚果混合物。每公斤混合物A含有60克杏仁、30克腰果和30克榛子。每公斤混合物B含有30克杏仁、60克腰果和180克榛子。两种混合物的其余部分是每个坚果。经销商正在考虑使用混合物 A 和 B 制作一个包含至少 240 克杏仁、300 克腰果和 540 克榛子的袋子。混合物 A 的价格为每公斤 8 卢比，混合物 B 的价格为每公斤 12 卢比。假设混合物 A 和 B 是均匀的，使用图解法来确定他应该使用的每种混合物的公斤数，以尽量减少袋子的成本。

解决方案：

Let x kg of kind A and y kg of kind B were used.

Since quantity cannot be negative.

Therefore x,y≥0

The given information can be tabulated as follows

Nut	Almonds(grams)	Cashew nuts(grams)
A(x)	60	30
B(x)	30	60
Availability	240	300

Thus the constraints are

60x+30y≥240

30x+60y≥300

30x+180y≥540

Mixture A costs Rs 8 per kg and mixture B costs Rs 12 per kg

Total cost= 8x+12y

Let the total cost be z

Thus, z=8x+12y

Now the mathematical formulation of given linear programming problem is

Minimize Z=8x+12y

subject to

2x+y≥8

x+2y≥10

x+6y≥18

x,y≥0

First we will convert the given inequations to equations. we get

2x+y=8

x+2y=10

x+6y=18

x=0 and y=0

Region represented by 2x+y≥8:

The line 2x+y=8 meets the coordinate axes at A₁(4, 0) and B₁(0, 8) respectively.

Clearly (0, 0) does not satisfy inequation 2x+y≥8.

So the region in XY plane that does not contain the origin represents the solution set of this equation

Region represented by x+2y≥10:

The line x+2y=10 meets the coordinate axes at C₁(10, 0) and D₁(0, 5) respectively.

Clearly (0, 0) does not satisfy inequation x+2y≥10.

So the region in XY plane that does not contain the origin represents the solution set of this equation.

Region represented by x+6y≥18:

The line x+6y=18 meets the coordinate axes at E₁(18, 0) and F₁(0, 3) respectively.

Clearly (0, 0) does not satisfy inequation x+6y≥18.

So the region in XY plane that does not contains the origin represents the solution set of this equation

Region represented by x,y≥0:

Since every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

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

The corner points of a Feasible region are B₁(0, 8), G₁(2, 4), H₁(6, 2) and E₁(18, 0).

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

Corner Point	z=8x+12y
B1	96
G1	64
H1	72
E1	144

The minimum value of z is at G₁(2, 4) and the value is Rs 64.

Thus the minimum cost is Rs 64 obtained when 2 units of kind A and 4 units of kind B nuts were used.

问题 9. 一种蛋糕需要 300 克面粉和 15 克脂肪，另一种蛋糕需要 150 克面粉和 30 克脂肪。假设不缺少用于制作蛋糕的其他成分，请找出可以由 7.5 公斤面粉和 600 克脂肪制成的蛋糕的最大数量。将其作为 LPP 并以图形方式解决。

解决方案：

Let there be x cakes of first kind and y cakes of second kind.

Therefore x,y≥0

The given information can be tabulated as follows

	Flour(gm)	Fat(gm)
Cakes of first kind	300	15
Cakes of second kind	150	30
Availability	7500	600

Thus the constraints are

300x+150y≤7500

15x+30y≤600

Total number of cakes that can be made =x + y

Let the total number of cakes be z.

Thus z=x + y

Now the mathematical formulation of given linear programming problem is

Maximize z=x + y

subject to

300x+150y≤7500

15x+30y≤600

x,y≥0

First we will convert the given inequations to equations. we get

300x+150y=7500

15x+30y=600

x=0 and y=0

Region represented by 300x+150y≤7500:

The line 300x+150y=7500 meets the coordinate axes at A(25, 0) and B(0, 50) respectively.

Clearly (0, 0) satisfies the equation 300x+150y=7500.

So the region which contains origin that represents the solution set of this inequation.

Region represented by 15x+30y≤600:

The line 15x+30y=600 meets the coordinate axes at C(40, 0) and D(0, 20) respectively.

Clearly (0, 0) satisfies the equation 15x+30y=600.

So the region in XY plane that contains the origin represents the solution set of this inequation.

Region represented by x,y≥0:

Since every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by these inequations.

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

The corner points of a Feasible region are O(0, 0), A(25, 0), D(0, 20) and E(20,10).

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

Corner Points	z=x+y
O(0, 0)	0
A(25, 0)	25
E(20, 10)	30
D(0, 20)	20

Thus maximum number of cakes is 30 and that can be made from 20 of the first kind and 10 of other kind.

问题 10. Reshma 希望混合两种食物 P 和 Q，使混合物中的维生素含量至少包含 8 单位的维生素 A 和 11 单位的维生素 B。食物 P 的价格为每公斤 60 卢比，食物 Q每公斤售价80卢比。食物 P 每公斤含有 3 单位维生素 A 和每公斤维生素 B 5 单位，而食物 Q 每公斤含有 4 单位维生素 A 和每公斤维生素 B 2 单位。确定混合物的最低成本。

解决方案：

Let x units of food P and y units of food Q are mixed together to make the mixture.

The cost of food P is Rs 60 per kg and that of food Q is Rs 80 per kg.

So, x kg of food P and y kg of food Q will cost Rs(60x+80y).

Since one kg of food P contains 3 units of vitamin A and one kg of food Q contains 4 units of vitamin A, therefore x kg of food P and y kg of food Q will contain (3x+4y) units of vitamin A. But the mixture should contain at least 8 units of vitamin A.

Thus, 3x+4y≥8

Similarly x kg of food P and y kg of food Q will contain 5x+2y units of vitamin B. But the mixture should contain at least 11 units of vitamin B.

Thus, 5x+2y≥11

Now the mathematical formulation of given linear programming problem is

Minimize z=60x+80y

subject to

3x+4y≥8

5x+2y≥11

x,y≥0

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

The corner points of a Feasible region are A(8/3, 0), B(2, 1/2), C(0, 11/2).

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

Corner point	z=60x+80y
A	160(Minimum)
B	160(Minimum)
C	440

The smallest value of z is 160 at A(8/3, 0) and B(2, 1/2).

It can be verified that the open half-plane represented by 60x+80y<160 has no common points with the feasible region.

So, the minimum cost of mixture is Rs 160 since the minimum value of z is 160.

问题 11. 一种蛋糕需要 200 克面粉和 25 克脂肪，另一种蛋糕需要 100 克面粉和 50 克脂肪。找出用 5 公斤面粉和 1 公斤脂肪可以制作的蛋糕的最大数量。假设没有储存用于制作蛋糕的其他成分。

解决方案：

Let the number of cakes of one kind and another kind be x and y respectively.

Therefore, total number of cakes produced are (x+y).

One kind of cake requires 200g of flour and another kind of cake requires 100 gms of flour. 

So, x cakes of one kind and y cakes of another kind requires (200x+100y) grams of flour. But cakes should contain at most 5 kg of flour.

Thus, 200x+100y≤5000

⇒  2x+y≤50

One kind of cake requires 25g of fat and another kind of cake requires 50 gms of fat.

So, x cakes of one kind and y cakes of another kind requires (25x+50y) grams of fat. But cakes should contain at most 1 kg of fat.

Thus, 25x+50y≤1000

⇒ x+2y≤40

Now the mathematical formulation of given linear programming problem is

Minimize z=x + y

subject to

2x+y≤50

x+2y≤40

x,y≥0

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

The corner points of a Feasible region are O(0, 0),  A(25, 0), B(20, 10), C(0, 20).

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

Corner point	z=x+y
O	0+0=0
A	25+0=25
B	20+10=30(Maximum)
C	0+20=20

Thus the maximum value of z is 30 at B(20,10).

Hence the maximum number of cakes which can be made are 30.

问题 12. 营养师必须使用两种食物 P 和 Q 制定特殊饮食。每包（含 30 克）食物 P 含有 12 个单位的钙、4 个单位的铁、6 个单位的胆固醇和 6 个单位的维生素 A。每包相同数量的食物Q含有3个单位的钙、20个单位的铁、4个单位的胆固醇和3个单位的维生素A。饮食需要至少240个单位的钙、至少460个单位的铁和最多300个单位的胆固醇。每种食物应使用多少包以尽量减少饮食中维生素 A 的含量？维生素A的最低含量是多少？

解决方案：

Let x packets of food P and y packets of food Q be used to make the diet.

Each packet of food P contains 6 units of vitamin A and each packet of food Q contains 3 units of vitamin A.

Therefore, x packets of food P and y packets of food Q contains (6x+3y) units of vitamin A.

Since each packet of food P contains 12 units of calcium and each packet of food Q contains 3 units of calcium.

Therefore, x packets of food P and y packets of food Q will contain (12x+3y) units of calcium. But the diet should contain at least 240 units of calcium.

Thus 12x+3y≥240

⇒ 4x+y≥80

Similarly,  x packets of food P and y packets of food Q contains (4x+20y) units of iron.

But the diet should contain at least 460 units of iron.

Thus 4x+20y≥460

⇒ x+5y≥115

Also, x packets of food P and y packets of food Q contains (6x+4y) units of cholesterol.

But the diet should contain atmost 300 units of cholesterol.

Thus 6x+4y≤300

⇒ 3x+2y≤150

Now the mathematical formulation of given linear programming problem is

Minimize z=6x+3y

subject to

4x+y≥80

x+5y≥115

3x+2y≤150

x,y≥0

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

The corner points of a Feasible region are A(2, 72), B(15, 20), C(40, 15).

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

Corner Point	z=6x+3y
A	6(2)+3(72)=228
B	6(15)+3(20)=150
C	6(40)+3(15)=285

The smallest value of z is 150 which is obtained at point B(15, 20).

Thus,15 packets of food P and 20 packets of food Q should be used to minimize the amount of vitamin A on the diet.

Hence, the minimum amount of vitamin A is 150 units.

问题 13. 一位农民混合了 P 和 Q 两种品牌的牛饲料。品牌 P，每袋售价 250 卢比，含有 3 单位营养元素 A、2.5 单位元素 B 和 2 单位元素 C。品牌 Q 每袋售价 200 卢比，含有 1.5 单位营养元素 A、11.25 单位元素 B 和3个单位的元素C。营养素A、B和C的最低要求分别为18个单位、45个单位和24个单位。确定应混合的每个品牌的袋子数量，以产生每袋成本的混合物？每袋混合物的最低成本是多少？

解决方案：

Let x bags of brand P and y bags of brand Q should be mixed to produce the mixture.

Each bag of brand P costs Rs 250 per bag and each bag of brand Q costs Rs 200 per bag.

Therefore, x bags of brand P and y bags of brand Q costs Rs(250x+200y).

Since each bag of brand P contains 3 units of nutritional element A and each bag of brand Q contains 1.5 units of nutritional element A.

Therefore, x bags of brand P and y bags of brand Q will contain (3x+1.5y) units of nutritional element A.

But the minimum requirement of nutrients A is 18 units.

Thus, 3x+1.5y≥18

⇒  2x+y≥12

Similarly, x bags of brand P and y bags of brand Q will contain (2.5x+11.25y) units of nutritional element B.

But the minimum requirement of nutrients B is 45 units.

Thus, 2.5x+11.25y≥45

⇒  2x+9y≥36

Also, x bags of brand P and y bags of brand Q will contain (2x+3y) units of nutritional element C.

But the minimum requirement of nutrients C is 24 units.

Thus, 2x+3y≥24

Now the mathematical formulation of given linear programming problem is

Minimize z=250x+200y

subject to

 2x+y≥12

2x+9y≥36

2x+3y≥24

x,y≥0

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

The corner points of a Feasible region are A(18, 0), B(9, 2), C(3, 6), D(0, 12).

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

Corner point	z=250x+200y
A	250(18)+200(0)=4500
B	250(9)+200(2)=2650
C	250(3)+200(6)=1950
D	250(0)+200(12)=2400

The smallest value of z is 1950 obtained at C(3, 6).

So, 3 bags of brand P and 6 bags of brand Q should be used to minimize the cost.

Hence, the minimum cost of mixture per bag is Rs 1950.

问题 14. 一位营养师希望将两种食物 X 和 Y 混合在一起，使混合物至少含有 10 个单位的维生素 A、12 个单位的维生素 B 和 8 个单位的维生素 C。公斤食物如下：

Points	Value of z
B1	4(0)+3(40)=120
G1	4(5)+3(30)=110
H1	4(20)+3(15)=125
C1	4(50)+3(0)=200

一公斤食物 X 的价格为 16 卢比，一公斤食物 Y 的价格为 20 卢比。找出能产生所需饮食的混合物的最低成本？

解决方案：

Let x kg of food X and y kg of food Y are mixed together to make the mixture.

The cost of food X is Rs 16 per kg and that of food Y is Rs 20 per kg.

So, x kg of food X and y kg of food Y will cost Rs (16x+20y).

Since one kg of food X contains 1 unit of vitamin A and one kg of food Y contains 2 units of vitamin A. Therefore, x kg of food X and y kg of food Y will contain (x+2y) units of vitamin A.

But the mixture should contain at least 10 units of vitamin A.

Thus, x+2y≥10

Similarly x kg of food X and y kg of food Y will contain (2x+2y) units of vitamin B. But the mixture should contain at least 12 units of vitamin B.

Thus, 2x+2y≥12

⇒ x+y≥6

Also, x kg of food X and y kg of food Y will contain (3x+y) units of vitamin C. But the mixture should contain at least 8 units of vitamin C.

Thus, 3x+y≥8

Now the mathematical formulation of given linear programming problem is

Minimize z=16x+20y

subject to

x+2y≥10

x+y≥6

3x+y≥8

x,y≥0

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

The corner points of a Feasible region are A(10, 0), H(2, 4), G(1, 5), F(0,8).

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

Corner Point	z=16x+20y
A	16(10)+20(0)=160
H	16(2)+20(4)=112(Minimum)
G	16(1)+20(5)=116
F	16(10)+20(8)=160

The smallest value of z is 112 obtained at H(2, 4).

Hence the least cost of mixture which will produce the required diet is Rs 112.

问题 15. 一个果农可以在他的花园里使用两种肥料，P 牌和 Q 牌。每个牌号的袋子中的氮、磷酸、钾和氯的含量见下表。测试表明，花园需要至少 240 公斤的磷酸、至少 270 公斤的钾肥和近 310 公斤的氯。

	Food I	Food II	Minimum daily  requirement
Calcium	10	4	20
Protein	5	6	20
Calories	2	6	12
Price	Rs 0.60 per unit	Rs 1.00 per unit

如果种植者想尽量减少向花园添加的氮量，每个品牌应该使用多少袋？花园中添加的最低氮量是多少？

解决方案：

Let x bags of fertilizer of brand P and y bags of fertilizer of brand Q are used in the garden.

One bag of brand P contains 3 kg of nitrogen and one bag of brand Q contains 3.5 kg of nitrogen, therefore, x bags of brand P and y bags of brand Q will contain (3x+3.5y) kg of nitrogen.

Since, One bag of brand P contains 1 kg of phosphoric acid and one bag of brand Q contains 2 kg of phosphoric acid, therefore, x bags of brand P and y bags of brand Q will contain (x+2y) kg of phosphoric acid.

But the garden needs at least 240 kg of phosphoric acid.

Thus, x+2y≥240

Similarly, One bag of brand P contains 3 kg of potash and one bag of brand Q contains 1.5 kg of potash, therefore, x bags of brand P and y bags of brand Q will contain (3x+1.5y) kg of potash.

But the garden needs at least 270 kg of  potash.

Thus, 3x+1.5y≥270

⇒  2x+y≥180

Also, One bag of brand P contains 1.5 kg of Chlorine and one bag of brand Q contains 2 kg of Chlorine, therefore, x bags of brand P and y bags of brand Q will contain (1.5x+2y) kg of Chlorine.

But the garden needs atleast 310 kg of  Chlorine.

Thus, 1.5x+2y≤310

Now the mathematical formulation of given linear programming problem is

Minimize z=3x+3.5y

subject to

x+2y≥240

2x+y≥180

1.5x+2y≤310

x,y≥0

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

The corner points of a Feasible region are A(40, 100), B(140, 50), C(20, 140).

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

Corner point	z=3x+3.5y
A(40, 100)	3(40)+3.5(100)=470
B(140, 50)	3(140)+3.5(50)=595
C(20, 140)	3(20)+3.5(140)=550

The smallest value of z is 470 obtained at A(40, 100).

Hence 40 bags of fertilizer of brand P and 100 bags of fertilizer of brand Q are used in the garden. The minimum amount of nitrogen added in the garden is 470 kg.