# Gaus Elimination Help

• Sep 22nd 2008, 05:48 PM
Kdocimo90
Gaus Elimination Help
Ok im totally stumped because the variables got switched up on me... here is the problem

A furniture copmany makes loungers, chairs, and footstools made out of wood, fabric and stuffing. The number of units of each of these materials needed for each of the products is given in the table below. How many of each product can be made if tehre are 54 units of wood, 63 units of fabric, and 43 units of stuffing availabe?

__________Wood__Fabric__Stuffing
Lounger_____1_____2_______2
Chair_______2_____2_______1
Footstool___3_____1_______1
• Sep 22nd 2008, 08:27 PM
Soroban
Hello, Kdocimo90!

Quote:

A furniture company makes loungers, chairs, and footstools made out of wood,
fabric and stuffing. The number of units of each of these materials needed
for each of the products is given in the table below.

How many of each product can be made if there are 54 units of wood,
63 units of fabric, and 43 units of stuffing availabe?

$\begin{array}{|c||c|c|c|}
& \text{Wood} & \text{Fabric} & \text{Stuffing} \\ \hline \hline
\text{Lounger }(x) & 1 & 2 & 2 \\ \hline
\text{Chair }(y) & 2 & 2 & 1 \\ \hline
\text{Stool }(z) & 3 & 1 & 1 \\ \hline \hline
\text{Total} & 54 & 63 & 43 \\ \hline \end{array}$

The equations are: . $\begin{array}{ccc}x + 2y + 3z &=& 54 \\ 2x + 2y + z &=& 63 \\ 2x + y + z &=& 43 \end{array}$

We have: . $\begin{array}{|ccc|c|}
1 & 2 & 3 & 54 \\ 2 & 2 & 1 & 63 \\ 2 & 1 & 1 & 43 \end{array}$

$\begin{array}{c}\\ R_2-2R_1 \\ R_3-R_2\end{array} \begin{array}{|ccc|c|}
1 & 2 & 3 & 54 \\ 0 & \text{-}2 & \text{-}5 & \text{-}45 \\ 0 & \text{-}1 & 0 & \text{-}20 \end{array}$

$\begin{array}{c}R_1+R_2 \\ \text{-}1\cdot R_2 \\ \text{-}1\cdot R_3 \end{array} \begin{array}{|ccc|c|} 1& 0 & \text{-}2 & 9 \\ 0 & 2 & 5 & 45 \\ 0 & 1 & 0 & 20 \end{array}$

$\begin{array}{c} \\ R_2-R_3 \\ \\ \end{array} \begin{array}{|ccc|c|}
1 & 0 & -2 & 9 \\ 0 & 1 & 5 & 25 \\ 0 & 1& 0 & 20 \end{array}$

$\begin{array}{c}\\ \\ R_3-R_2 \end{array} \begin{array}{|ccc|c|}
1 & 0 & \text{-}2 & 9 \\ 0 & 1 & 5 & 25 \\ 0 & 0 & \text{-}5 & \text{-}5 \end{array}$

$\begin{array}{c} \\ \\ \text{-}\frac{1}{5}\cdot R_3\end{array} \begin{array}{|ccc|c|}
1 & 0 & \text{-}2 & 9 \\ 0 & 1 & 5 & 25 \\ 0 & 0 & 1 & 1 \end{array}$

$\begin{array}{c}R_1+2R_3 \\ R_2 - 5R_3 \\ \\ \end{array} \begin{array}{|ccc|c|} 1&0 & 0 & 11 \\ 0 & 1 & 0 & 20 \\ 0 & 0 & 1 & 1 \end{array}$

Therefore: . $x = 11,\;y = 20,\;z = 1$