A researcher wants to provide a rabbit exactly 162 units ofprotein, 72 units of carbohydrates, and 30 units of vitamin A. The rabbit is fed three types of food. Each gram of Food A has 5 units of protein, 2 units of carbohydrates, and 1unit of vitamin A. Each gram of Food B contains 11 units of protein, 5 units ofcarbohydrates, and 2 units of vitamin A. Each gram of Food C contains 23 units of protein, 11 units of carbohydrates, and 4 units of vitamin A. How many grams of each food should the rabbit be fed?

There are 2 units of carbohydrates in each gram of food A, 5 units of carbohydrates in each gram of food B and 11 units of carbohydrates in each gram of food C. So:

[tex]2x + 5y + 11z = 72[/tex]

A researcher wants to provide a rabbit exactly 30 units of Vitamin A:

There is 1 unit of Vitamin A in each gram of food A, 2 units of Vitamin A in each gram of food B and 4 units of Vitamin A in each gram of food C. So:

[tex]x + 2y + 4z = 30[/tex].

We have to solve the following system of equations:

[tex]5x + 11y + 23z = 162[/tex]

[tex]2x + 5y + 11z = 72[/tex]

[tex]x + 2y + 4z = 30[/tex].

I think that the easier way to solve this is reducing the augmented matrix of this system.

This system has the following augmented matrix:

[tex]\left[\begin{array}{cccc}5&11&23&162\\2&5&11&72\\1&2&4&30\end{array}\right][/tex]

To help reduce this matrix, i am going to swap the first line with the third

[tex]L_{1} <-> L_{3}[/tex]

Now we have the following matrix:

[tex]\left[\begin{array}{cccc}1&2&4&30\\2&5&11&72\\5&11&23&162\end{array}\right][/tex]

Now i am going to do these following operations, to reduce the first row:

[tex]L_{2} = L_{2} - 2L_{1}[/tex]

[tex]L_{3} = L_{3} - 5L_{1}[/tex]

Now we have

[tex]\left[\begin{array}{cccc}1&2&4&30\\0&1&3&12\\0&1&3&12\end{array}\right][/tex]

Now, to reduce the second row, i do:

[tex]L_{3} = L_{3} - L_{2}[/tex]

The matrix is:

[tex]\left[\begin{array}{cccc}1&2&4&30\\0&1&3&12\\0&0&0&0\end{array}\right][/tex]

This means that z is a free variable, so we are going to write y and x as functions of z.

From the second line, we have

[tex]y + 3z = 12[/tex]

[tex]y = 12 - 3z[/tex]

From the first line, we have

[tex]x + 2y + 4z = 30[/tex]

[tex]x + 2(12 - 3z) + 4z = 30[/tex]

[tex]x + 24 - 6z + 4z = 30[/tex]

[tex]x = 6 + 2z[/tex]

Our solution is: [tex]x = 6 + 2z, y = 12 - 3z, z = z[/tex].

However, we can not give a negative number of grams of a food. So

[tex]y \geq 0[/tex]

[tex]12 - 3z \geq 0[/tex]

[tex]-3z \geq -12 *(-1)[/tex]

[tex]3z \leq 12[/tex]

[tex]z \leq 4[/tex]

The rabbit should be fed:

[tex]6 + 2z[/tex] grams of food A

[tex]12 - 3z[/tex] grams of food B

[tex]z[/tex] grams of food C

For [tex]z \leq 4[/tex].