Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.6 Exercises - Page 55: 1

Answer

see explanation

Work Step by Step

Goods Services 0.2 0.7 Goods 0.8 0.3 Services Each column refers to the fraction of output from one branch to the other $(\mathrm{s}) .$ e.g the second column says that Services sell 0.7 of its output to Goods and retains the rest 0.3 of its output to itself. Let $P_{G}$ be a price for goods be and let $P_{S}$ be a price for services. Goods spends $0.2 P_{G}+0.7 P_{S}$ and Services spends $0.8 P_{G}+0.3 P_{S} .$ To balance the income and outcome we need $0.2 P_{G}+0.7 P_{S}=P_{G}$ $0.8 P_{G}+0.3 P_{S}=P_{S}$ Here we simply equate outcome price to income price. We rewrite these equations as follows: $0.8 P_{G}-0.7 P_{S}=0$ $0.8 P_{G}-0.7 P_{S}=0$ It is actually the same equation with the solution $P_{G}=\frac{7}{8} P_{S}$ It follows that the prices can be chosen,e.g, as follows: $P_{S}=8, P_{G}=7$ or $P_{S}=80, P_{G}=70$ or $P_{S}=40, P_{G}=35$ The only condition is that the ratio of the prices is $P_{G}: P_{S}=7: 8$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.