Mr Smith decides to feed his pet Doberman pincher a combination of two dog foods. Each can of brand A contains 3 units of protein, 1 unit of carbohydrates, and 2 units of fat and costs 80 cents. Each can of brand B contains 1 unit of protein, 1 unit of carbohydrates, and 4 units of fat and costs 70 cents. Mr Smith feels that each day his dog should have at least 9 units of protein, 7 units of carbohydrates, and 24 units of fat. How many cans of each dog food should he give to his dog each day to provide the minimum requirements at the least cost?
| Brand | Protein | Carbohydrate | Fat | Cost per can (Cents) |
| A | 3 | 1 | 2 | 80 |
| B | 1 | 1 | 4 | 70 |
| Total | 9 | 7 | 24 |
Let x cans of brand A and y cans of brand B dog food should be given to the dog each day to provide the minimum requirements at the least cost.
Now, it is given that the cost of 1 can of brand A dog food is 80 cents. Therefore, the cost of x cans of brand A dog food will be obtained as:
Cost of x cans of brand A dog food
= Cost of 1 can of brand A dog food × Number of cans of brand A
= 80 × x
= 80x cents
It is also given that the cost of 1 can of brand B dog food is 70 cents. Therefore, the cost of y cans of brand B dog food will be obtained as:
Cost of y cans of brand B dog food
= Cost of 1 can of brand B dog food × Number of cans of brand B
= 70 × y
= 70y cents
The total cost of two dog foods
= Cost of x cans of brand A dog food + Cost of y cans of brand B dog food
= 80x + 70y
Now, the objective function will be as:
Minimise Z = 80x + 70y
Now, we will find the protein constraint. It is given that each can of brand A contains 3 unit of protein while each can of brand B contains 1 unit of protein. Therefore, x cans of brand A will have 3x unit of protein and y cans of brand B will have y units of protein.
Total quantity of protein
= Quantity of protein in x cans of brand A + Quantity of protein in y cans of brand B
= 3x + y
Now, Mr Smith feels that each day his dog should have at least 9 units of protein. Thus, we can write
Total quantity of protein ≥ 9
Or, 3x + y ≥ 9 (Protein Constraint)
Now, we will find the carbohydrate constraint. It is given that each can of brand A contains 1 unit of carbohydrate while each can of brand B contains 1 unit of carbohydrate. Therefore, x cans of brand A will have x unit of carbohydrate and y cans of brand B will have y units of carbohydrate.
Total quantity of carbohydrate
= Quantity of carbohydrate in x cans of brand A + Quantity of carbohydrate in y cans of brand B
= x + y
Now, Mr Smith feels that each day his dog should have at least 7 units of carbohydrate. Thus, we can write
Total quantity of carbohydrate ≥ 7
Or, x + y ≥ 7 (Carbohydrate Constraint)
Finally, we will find the fat constraint. It is given that each can of brand A contains 2 unit of fat while each can of brand B contains 4 unit of fat. Therefore, x cans of brand A will have 2x unit of fat and y cans of brand B will have 4y units of fat.
Total quantity of fat
= Quantity of fat in x cans of brand A + Quantity of fat in y cans of brand B
= 2x + 4y
Now, Mr Smith feels that each day his dog should have at least 24 units of fat. Thus, we can write
Total quantity of fat ≥ 24
Or, 2x + 4y ≥ 24 (Fat Constraint)
Thus, the formulation of the LPP is as:
Minimise Z = 80x + 70y
3x + y ≥ 9 (Protein Constraint)
x + y ≥ 7 (Carbohydrate Constraint)
2x + 4y ≥ 24 (Fat Constraint)
x ≥ 0; y ≥ 0 (Non-negative Constraint)
Now, to plot these inequations on the graph first we will have to convert these inequations into equations then we will find the points which lie on these equations.
First we will find the points which lie on the line 3x + y = 9. When we Put y = 0 in the equation, we get x = 3. So, the first point on this line is (3, 0). When we put x = 0 in the equation, we get y = 9. So, the second point on this line is (0, 9).
Similarly, we will find the points which lie on the line x + y = 7. When we Put y = 0 in the equation, we get x = 7. So, the first point on this line is (7, 0). When we put x = 0 in the equation, we get y = 7. So, the second point on this line is (0, 7).
Finally, we will find the points which lie on the line 2x + 4y = 24. When we Put y = 0 in the equation, we get x = 12. So, the first point on this line is (12, 0). When we put x = 0 in the equation, we get y = 6. So, the second point on this line is (0, 6).
Now, we will solve the equations 3x + y = 9 and x + y = 7 to find their point of intersection. Subtracting equation x + y = 7 from equation 3x + y = 9, we get
(3x + y) – (x + y) = 9 – 7
Or, 3x + y – x – y = 2
Or, 3x – x = 2
Or, 2x = 2
Or, x = 2/2
Or, x = 1
Putting the value of x in x + y = 7, we get
x + y = 7
Or, 1 + y = 7
Or, y = 7 – 1
Or, y = 6
So, the point of intersection of the equations 3x + y = 9 and x + y = 7 is (1, 6).
Now, we will solve the equations 2x + 4y = 24 and x + y = 7 to find their point of intersection. Multiplying the equation x + y = 7 by 4 then Subtracting it from equation 2x + 4y = 24, we get
(2x + 4y) – (4x + 4y) = 24 – 28
Or, 2x + 4y – 4x – 4y = – 4
Or, 2x – 4x = – 4
Or, – 2x = – 4
Or, 2x = 4
Or, x = 4/2
Or, x = 2
Putting the value of x in x + y = 7, we get
x + y = 7
Or, 2 + y = 7
Or, y = 7 – 2
Or, y = 5
So, the point of intersection of the equations 2x + 4y = 24 and x + y = 7 is (2, 5).
Finally, we will solve the equations 2x + 4y = 24 and 3x + y = 9 to find their point of intersection. Multiplying the equation 3x + y = 9 by 4 then Subtracting it from equation 2x + 4y = 24, we get
(2x + 4y) – (12x + 4y) = 24 – 36
Or, 2x + 4y – 12x – 4y = – 12
Or, 2x – 12x = – 12
Or, – 10x = – 12
Or, 10x = 12
Or, x = 12/10
Or, x = 1.2
Putting the value of x in 3x + y = 9, we get
3x + y = 9
Or, 3×1.2 + y = 9
Or, 3.6 + y = 9
Or, y = 9 – 3.6
Or, y = 5.4
So, the point of intersection of the equations 2x + 4y = 24 and 3x + y = 9 is (1.2, 5.4).
Now, we will plot all the points on the graph to find the number of cans.
We can observe from the above graph that the feasible region is unbounded because each inequation has the sign of greater then equal to and the corner points are as: A (0, 9), B (1, 6), C (2, 5), and D (12, 0)
Now, we will find the value of Z at the corner points A, B, C, and D.
The value of Z at corner point A (0, 9) is obtained as:
Z = 80x + 70y
Or, Z = 80×0 + 70×9
Or, Z = 0 + 630
Or, Z = 630 cent
The value of Z at corner point B (1, 6) is obtained as:
Z = 80x + 70y
Or, Z = 80×1 + 70×6
Or, Z = 80 + 420
Or, Z = 500 cent
The value of Z at corner point C (2, 5) is obtained as:
Z = 80x + 70y
Or, Z = 80×2 + 70×5
Or, Z = 160 + 350
Or, Z = 510 cent
The value of Z at corner point D (12, 0) is obtained as:
Z = 80x + 70y
Or, Z = 80×12 + 70×0
Or, Z = 960 + 0
Or, Z = 960 cent
Now, we can see that the minimum value of Z is 500 cent and it is obtained at the corner point B (1, 6).
Hence, we can write Mr Smith should give his dog 1 can of brand A and 6 cans of Brand B to provide the minimum requirement at the least cost.