Systems of linear equations

Question

Rabbits in a lab are to be kept on a strict daily diet that includes 25 g of protein, 22 g of fat, and 40 g of carbohydrates. The scientist has only three food mixes available with the following grams of nutrients per unit.

Mix A = 3, 4, 5 (protein/fat/carbohydrates)

Mix B = 4, 2, 6

Mix C = 5, 2, 8

Find how many units of each mix are needed daily to meet each rabbits dietary need.

Answer 1

To find how many units of each type of food are needed, an equation needs to be created for type of nutrient, protein, fat, and carbohydrates.

Each unit of Mix A will have 3g protein. In 'a' units the amount of protein is 3a.

Each unit of Mix B will have 4g protein. In 'b' units the amount of protein is 4b.

Each unit of Mix C will have 5g protein. In 'c' units the amount of protein is 5c.

If each of these amounts are added they need to equal 25. This can be used to create the equation:

3a + 4b + 5c = 25

The amounts in each unit and the totals for fat and carbohydrates can be used to create 2 more equation:

4a + 2b + 2c = 22

5a + 6b + 8c = 40

We will examine the first 2 equations:

3a + 4b + 5c = 25; 4a + 2b + 2c = 22

Our goal is to eliminate one of the variables. To do this we multiply the second equation by -2. This gives:

3a + 4b + 5c = 25 for the 1st and

-8a - 4b - 4c = -44 for the second. We then add these equations. This results in:

-5a + c = -19

We will now examine the 2nd and 3rd equations:

4a + 2b + 2c = 22; 5a + 6b + 8c = 40 We will multiply the 2nd equation by -3 This gives:

-12a - 6b - 6c = -66 for the 2nd and

5a + 6b + 8c = 40 for the 3rd. These will be added to give:

-7a +2c = -26

We will use the two new equations that were created

-5a + c = -19 and -7a + 2c = -26 We will isolate the variable c in -5a + c = -19.

5a will be added to each side:

c = 5a - 19 This value of c will be substituted in to -7a + 2c = -26 This results in:

-7a + 2(5a - 19) = -26 Then distribute

-7a +10a - 38 = -26 Combine like terms

3a -38 = -26 Add 38 to both sides

3a = 12 Divide both sides by 3

a = 4 This value of a can be substituted in to the equation that was solved for c, c = 5a - 19. This gives:

c = 5(4) - 19 Multiply

c= 20 -19 Subtract

c = 1 The found values of a and c will be substituted in to the first original equation, 3a + 4b + 5c = 25:

3(4) +4b +5(1) = 25 Multiply

12 + 4b + 5 = 25 Combine like terms

4b + 17 = 25 Subtract 17 from both sides

4b = 8 Divide both sides by 4

b = 2

We need to use 4 units of A, 2 units of B, and 1 unit of C.

Answer 2

Let

a = units from Mixture A,

b = units from Mixture B,

c = units from Mixture C,

Then our system of equations to solve are as follows:

3a + 4b + 5c = 25

4a + 2b + 6c = 22

5a + 2b + 8c = 40

This problem requires the use of matrices and the calculation of a matrix inverses. I prefer to use a free advanced math interpreter like Octave (i.e., https://octave-online.net/#)

Let

M = Matrix of coefficients - given

d = vector of desired outcomes- given

s = vector of amounts from the various Mixes - unknowns

Then, we are to solve the following matrix equation:

M * s = d

What follows is the dialog from Octave:

>> M = [3,4,5; 4,2,6;5,2,8]'

M =

3 4 5

4 2 2

5 6 8

>> d = [25;22;40]

d =

25

22

40

>> s = inv(A) * d

s =

4.0000

2.0000

1.0000

>> A * s

ans =

25.000

22.000

40.000

Thus, we need

4 units from Mixture A

2 units from Mixture B

1 unit from Mixture C

Answer 3

Using EXCEL Solver, it appears that you use 4 units of A, 2 units of B, & 1 unit of C to get the required nutrients with minimum number of mixes. (P = Protein, F = Fat, & C = Carbs)