Question

A cookie company makes three kinds of cookies (oatmeal raisin, chocolate chip, and shortbread) packaged in small, medium, and large boxes. The small box contains 1 dozen oatmeal raisinand 1 dozen chocolate chip; the medium box has 2 dozenoatmeal raisin, 1 dozen chocolatechip, and 1 dozenshortbread; the large boxcontains 2 dozen oatmeal raisin, 2 dozen chocolate chip, and 3 dozen shortbread. If you require exactly 15 dozen oatmeal raisin, 10 dozen chocolate chip, and 11 dozen shortbread, ow many of each size box should you buy

Answer 1

1 small box

5 medium box

2 large box

Explanation:

Let x be the small box, y be the medium box and z be the large box.

So, we have:

`\begin{array}{cccc}{} & {Oatmeal} & {Chocolate} & {Shortbread} & {x} & {1} & {1} & {0} & {y} & {2} & {1} & {1}& {z} & {2} & {2} & {3} & {Total} & {15} & {10} & {11} \ \end{array}`

From the above table, we have the following equations:

Oatmeal:

`x + 2y + 2z = 15` --- (1)

Chocolate Chip

`x + y + 2z = 10` --- (2)

Shortbread

`y + 3z = 11` --- (3)

Make y the subject in (3)

`y =11 - 3z`

Substitute
`y =11 - 3z` in (1) and (2)

`x + 2y + 2z = 15` --- (1)

`x + 2(11-3z) + 2z = 15`

`x + 22 - 6z + 2z = 15`

`x -6z +2z = 15 - 22\\`

`x-4z = -7`

`x = 4z - 7` --- (4)

`x + y + 2z = 10` --- (2)

`x + 11 - 3z + 2z = 10`

`x - 3z + 2z = 10 - 11`

`x -z = -1`

`x = z - 1` --- (5)

Equate (4) and (5)

`4z - 7 = z - 1`

`4z - z = 7 - 1`

`3z = 6`

`z = 2`

Substitute
`z = 2` in (5)

`x = z - 1` --- (5)

`x =2-1`

`x =1`

Substitute
`z = 2` in
`y =11 - 3z`

`y = 11 - 3 * 2`

`y = 11 - 6`

`y = 5`

So, the solution is:

`x =1`
`y = 5`
`z = 2`