# If a Feasible Region is Unbounded — LP keywords: cost, vitamin, diet

📌 Question 1. Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs \$60/kg and Food Q costs \$80/kg. Food P contains 3 units /kg of vitamin A and 5 units /kg of vitamin B while food Q contains 4 units /kg of vitamin A and 2 units /kg of vitamin B. Determine the minimum cost of the mixture?

Let the mixture contain x kg of food P and y kg of food Q. Therefore,

x≥0 and y≥0.

The given information can be compiled in a table as follows.

 Vitamin A(units/kg) Vitamin B(units/kg) cost Food P 3 5 60 Food Q 4 2 80 Requirement (units/kg) 8 11

The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B. Therefore, the constraints are

3x+4y≥8
5x+2y≥11

Total cost, Z, of purchasing food is, Z=60x+80y.
The mathematical formulation of the given problem is
Minimize Z=60x+80y … (1)
subject to the constraints,
3x+4y≥8 … (2)
5x+2y≥11 … (3)
x,y≥0 … (4)

The feasible region determined by the system of constraints is as follows. It can be seen that the feasible region is unbounded.
The corner points of the feasible region are A(8/3, 0), B(2, ½), and C(0, 11/2).
The values of Z at these corner points are as follows.

 Corner point Z=60x+80y A(8/3, 0) 160 }Minimum B(2, ½) 160 C(0, 11/2) 440

As the feasible region is unbounded, therefore, 160 may or may not be the minimum value of Z.
For this, we graph the inequality, 60x+80y<160 or 3x+4y<8, and check whether the resulting half plane has points in common with the feasible region or not. It can be seen that the feasible region has no common point with 3x+4y<8. Therefore, the minimum cost of the mixture will be \$160 at the line segment joining the points (8/3, 0) and (2, ½). 📌 Q2. A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs \$4 per unit food and F2 costs \$6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements?

Let the diet contain x units of food F1 and y units of food F2. Therefore, x≥0 and y≥0.
The given information can be complied in a table as follows.

 Vitamin A (units) Mineral (units) Cost per unit (\$) Food F1 (x) 3 4 4 Food F2 (y) 6 3 6 Requirement 80 100

The cost of food F1 is \$4 per unit and of Food F2 is \$6 per unit. Therefore, the constraints are

3x+6y≥80
4x+3y≥100
x,y≥0

Total cost of the diet, Z=4x+6y.
The mathematical formulation of the given problem is minimizing Z=4x+6y … (1)
subject to the constraints,
3x+6y≥80…(2)
4x+3y≥100…(3)
x,y≥0…(4)

The feasible region determined by the constraints is as follows. It can be seen that the feasible region is unbounded.
The corner points of the feasible region are A(8/3,0), B(2, ½), and C(0, 11/2).
The corner points are A(80/3,0), B(24, 4/3), and C(0, 100/3).
The values of Z at these corner points are as follows.

 Corner point Z=4x+6y A(80/3, 0) 320/3=106.67 B(24, 4/3) 104 → Minimum C(0, 100/3) 200

As the feasible region is unbounded, therefore, 104 may or may not be the minimum value of Z.
For this, we draw a graph of the inequality, 4x+6y<104 or 2x+3y<52, and check whether the resulting half plane has points in common with the feasible region or not. It can be seen that the feasible region has no common point with 2x+3y<52. Therefore, the minimum cost of the mixture will be \$104. 📌 Q3. A farmer mixes two brands P and Q of cattle feed. Brand P, costing \$250 per bag contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing \$200 per bag contains 1.5 units of nutritional elements A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag? ✍ Let the farmer mix x bags of brand P and y bags of brand Q.
The given information can be compiled in a table as follows.

 Vitamin A (units/kg) Vitamin B (units/kg) Cost (\$/kg) Food P 3 5 60 Food Q 4 2 80 Requirement (units/kg) 8 11

The given problem can be formulated as follows.
Minimize z=250x+200y…(1)
subject to the constraints,

3x+1.5y≥18…(2)
2.5x+11.25y≥45…(3)
2x+3y≥24…(4)
x,y≥0…(5)

The feasible region determined by the system of constraints is as follows. The corner points of the feasible region are A(18, 0), B(9, 2), C(3, 6), and D(0, 12). The values of z at these corner points are as follows.

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

As the feasible region is unbounded, therefore, 1950 may or may not be the minimum value of z.

For this, we draw a graph of the inequality, 250x+200y<1950 or 5x+4y<39, and check whether the resulting half plane has points in common with the feasible region or not. It can be seen that the feasible region has no common point with 5x+4y<39. Therefore, the minimum value of z is 2000 at (3, 6).
Thus, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimize the cost to \$1950.

📌 Q4. A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin content of one kg food is given below:

 Food vitamin A vitamin B vitamin C X 1 2 3 Y 2 2 1

One kg of food X costs \$16 and one kg of food Y costs \$20. Find the least cost of the mixture which will produce the required diet?

Let the mixture contain x kg of food X and y kg of food Y.
The mathematical formulation of the given problem is as follows.
Minimize z=16x+20y…(1)
subject to the constraints,

x+2y≥10…(2)
x+y≥6…(3)
3x+y≥8…(4)
x,y≥0…(5)

The feasible region determined by the system of constraints is as follows. The corner points of the feasible region are A(10, 0), B(2, 4), C(1, 5), and D(0, 8). The values of z at these corner points are as follows.

 Corner point z=16x+20y A(10, 0) 160 B(2, 4) 112 → Minimum C(1, 5) 116 D(0, 8) 160

As the feasible region is unbounded, therefore, 112 may or may not be the minimum value of z.
For this, we draw a graph of the inequality, 16x+20y<112 or 4x+5y<28, and check whether the resulting half plane has points in common with the feasible region or not. It can be seen that the feasible region has no common point with 4x+5y<28. Therefore, the minimum value ofz is 112 at (2, 4). Thus, the mixture should contain 2 kg of food X and 4 kg of food Y. The minimum cost of the mixture is \$112.

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