Question

A furniture factory uses three types of wood namely, teakwood, rosewood and satinwood for manufacturing three types of furniture, that are, table, chair and cot.

The wood requirements (in tonnes) for each type of furniture are given below:

	Table	Chair	Cot
Teakwood	2	3	4
Rosewood	1	1	2
Satinwood	3	2	1

It is found that 29 tonnes of teakwood, 13 tonnes of rosewood and 16 tonnes of satinwood are available to make all three types of furniture.

Using the above information, answer the following questions:

1. Express the data given in the table above in the form of a set of simultaneous equations.
2. Solve the set of simultaneous equations formed in subpart (i) by matrix method.
3. Hence, find the number of table(s), chair(s) and cot(s) produced.

Answer 1

Let the number of tables, chairs and cots produced be x, y and z.

i. Then, the system of simultaneous equations produced is:

2x + 3y + 4z = 29

x + y + 2z = 13

3x + 2y + z = 16

ii. Part (i) equations are in matrix form as follows:

${\displaystyle \left[\begin{array}{ccc}2& 3& 4\\ 1& 1& 2\\ 3& 2& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}29\\ 13\\ 16\end{array}\right]}$

i.e., AX = B

${\displaystyle \Rightarrow }$ X = A^–1B

I A I = 2(1 – 4) – 3(1 – 6) + 4(2 – 3)

= – 6 + 15 – 4

= 15 – 10

= 5 ≠ 0

As a result, the inverse exists.

Then, a₁₁ = (–1)¹⁺¹(1 – 4) = – 3

a₁₂ = (–1)¹⁺²(1 – 6) = 5

a₁₃ = (–1)¹⁺³(2 – 3) = – 1

a₂₁ = (–1)²⁺¹(3 – 8) = 5

a₂₂ = (–1)²⁺²(2 – 12) = – 10

a₂₃ = (–1)²⁺³(4 – 9) = 5

a₃₁ = (–1)³⁺¹(6 – 4) = 2

a₃₂ = (–1)³⁺²(4 – 4) = 0

a₃₃ = (–1)³⁺³(2 – 3) = – 1

adj A = ${\displaystyle {\left[\begin{array}{ccc}-3& 5& -1\\ 5& -10& 5\\ 2& 0& -1\end{array}\right]}^1=\left[\begin{array}{ccc}-3& 5& 2\\ 5& -10& 0\\ -1& 5& -1\end{array}\right]}$

∴ A^–1 = ${\displaystyle \frac{\left[adjA\right]}{\left|A\right|}=\frac{1}{5}\left[\begin{array}{ccc}-3& 5& 2\\ 5& -10& 0\\ -1& 5& -1\end{array}\right]}$

∴ X = A^–1B

∴ ${\displaystyle \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{5}\left[\begin{array}{ccc}-3& 5& 2\\ 5& -10& 0\\ -1& 5& -1\end{array}\right]\left[\begin{array}{c}29\\ 13\\ 16\end{array}\right]}$

= ${\displaystyle \frac{1}{5}\left[\begin{array}{c}-87+65+32\\ 145-130+0\\ -29+65-16\end{array}\right]}$

= ${\displaystyle \frac{1}{5}\left[\begin{array}{c}10\\ 15\\ 20\end{array}\right]=\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]}$

Hence, x = 2, y = 3, z = 4

iii. ∴ Number of table(s) produced = 2

Number of chair(s) produced = 3

Number of cot(s) produced = 4