Q. 57 - Precalculus Enhanced with Graphing Utilities

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 raisin and $1$ dozen chocolate chip; the medium box has $2$ dozen oatmeal raisin, $1$ dozen chocolate chip, and $1$ dozen shortbread; the large box contains $2$ 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, how many of each size box should you buy?

Step-by-Step Solution

Answer

The number of small boxes should be bought is $1$, that of medium box is $5$ and the large box is $2$

Step 1. Given

The small box contains $1$ dozen oatmeal and $1$ dozen chocolate chip.

The medium box contains $2$ dozen oatmeal raisin, $1$ dozen chocolate chip and $1$ dozen shortbread.

The large contains $2$ dozen oatmeal raisin, $2$ dozen chocolate chip, and $3$ dozen shortbread.

Step 2. Form equations.

Let $x,y,z$ denote the small, medium and large boxes respectively.

So the total number of raisins required is given by, 

$x+2y+2z=15$

The number of chocolate chip required is given by,

$x+y+2z=10$

The number of short bread required is given by,

$y+3z=11$

Step 3. Write the system of equations

The system of equation will be,

$\left\{\begin{array}{l}x+2y+2z=15\\ x+y+2z=10\\ y+3z=11\end{array}\right.$

Step 4. Solve the system of equations.

Subtract second equation from first equation, we get,

$y=5$

Step 5. Find z

Substitute $y=5$ in the third equation, we get,

$$\begin{gathered} y+3z=11 \\ \left(5\right)+3z=11 \\ 3z=11-5 \\ 3z=6 \\ z=2 \end{gathered}$$

Step 6. Find x

Substitute $y=5; z=2$ in the first equation, we get,

$$\begin{gathered} x+2y+2z=15 \\ x+2\left(5\right)+2\left(2\right)=15 \\ x+14=15 \\ x=1 \end{gathered}$$

The number of small box required is $1$, the medium box is $5$ and the large box is $2$