Answered: Assume that the production of 1 unit of…

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter2: Systems Of Linear Equations

Section2.4: Applications

Problem 27EQ

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Assume that the production of 1 unit of magnesium requires 0.5 units of magnesium and 0.6 units of aluminum, and that the production of 1 unit of aluminum requires 0.5 units of magnesium and 0.2 units of aluminum.

Find the production schedule that satisfies an external demand for 60 units of magnesium and 20 units of aluminum.

Expert Solution

Step 1

Given that the production of 1 unit of magnesium requires 0.5 units of magnesium and 0.6 units of aluminum, and that the production of 1 unit of aluminum requires 0.5 units of magnesium and 0.2 units of aluminum.

Thus, the contribution of magnesium to 1 unit of production of magnesium is 0.5 units and the contribution of magnesium to 1 unit of production of aluminium is 0.5 units.

Similarly, the contribution of aluminium to 1 unit of production of magnesium is 0.6 units and the contribution of aluminium to 1 unit of production of aluminium is 0.2 units.

Let 1 represent magnesium and 2 represent aluminium.

Use $a_{i j}$ to denote the contribution of item i to item j.

Then, $a_{11} = 0.5, a_{12} = 0.5, a_{21} = 0.6, a_{22} = 0.2$ .

Obtain the technology matrix A as follows.

$A = [a_{i j}] = [\begin{matrix} 0.5 & 0.5 \\ 0.6 & 0.2 \end{matrix}]$

Step 2

The external demand for magnesium is 60 units and that for aluminium is 20 units.

Hence, the demand vector $D = [\begin{matrix} 60 \\ 20 \end{matrix}]$ .

We have to find the production schedule $X = [\begin{matrix} x \\ y \end{matrix}]$ .

The production schedule is given by the equation $X = {(I - A)}^{- 1} D$ where $I$ is the identity matrix.

Hence,

$X = {(I - A)}^{- 1} D \Rightarrow [\begin{matrix} x \\ y \end{matrix}] = {([\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - [\begin{matrix} 0.5 & 0.5 \\ 0.6 & 0.2 \end{matrix}])}^{- 1} [\begin{matrix} 60 \\ 20 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \end{matrix}] = {[\begin{matrix} 0.5 & - 0.5 \\ - 0.6 & 0.8 \end{matrix}]}^{- 1} [\begin{matrix} 60 \\ 20 \end{matrix}]$

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