when-money-is-spent-on-goods-and-services-those-who-receive-the-money-also-spend-some-of-it-the-peop

Question

When money is spent on goods and services, those who receive the money also spend some of it. The people receiving some of the twice-spent money will spend some of that, and so on. Economists call this chain reaction the multiplier effect. In a hypothetical isolated community, the local government begins the process by spending $ D $ dollars. Suppose that each recipient of spent money spends $ 100c% $ and saves $ 100s% $ of the money that he or she receives. The values $ c $ and $ s $ are called the marginal propensity to consume and the marginal propensity to save and, of course, $ c + s = 1. $
(a) Let $ S_n $ be the total spending that has been generated after $ n $ transactions. Find an equation for $ S_n. $
(b) Show that $ \lim_{n 	o \infty} S_n = kD, $ where $ k = 1/s. $ The number $ k $ is called the multiplier. What is the multiplier if the marginal propensity to consume is $ 80%? $
Note: The federal government uses this principle to justify lending a large percentage of the money that they receive in deposits.

J Hardin · Accepted Answer

So, for part A, the government starts off by giving away D dollars to the public and then among those that received The DE dollars they spend See Time&#x27;s that and then among those who received the CD dollars they spend C Square D and so on. So going on to the next page after in transactions that&#x27;s horrible spending is given by Listen to note this by S n So the sum So the total spending in the end, it&#x27;s for the transactions. So we have the from the government that d. C from the first transaction after that and so on. And since we started off with with no sea, we go all the way up so and minus one. But this is just a geometric series. So we can just we have a formula for this? No. So they&#x27;re the formula. This is the type of some is a fine, eh? She a mattress Siri&#x27;s. So all we did was factor out the D and then didn&#x27;t want minus C, And then we increase their exponents by one. So we go from and minus wantto and then divided by one minus the common ratio. See? So this will be the answer for part a And then let&#x27;s go on to the next page from our peak. Okay, Here. So let me first recall part, eh? So as n from party, eh? So let me put a party there to let us know where we got it. D one minus either the end over one minus c. Let&#x27;s take the limit. Herbal sides. So let&#x27;s say thus. Limit ascend and goes to infinity is limit and goes to infinity d one minus seat of the end over one minus c, however, sees a number between zero and one. Therefore the limit of CN is just equal to zero. Any number between negative one and one, Not including the end points. When you multiply it by itself, the number is the whole good smaller and then the limited ghost zero. So using that fast, this just becomes D over one minus c. We were given, however, that C plus s equals one So we can write This is s equals one minus C. And so what we wanted was D over s. That was what the requirement wass and this is. Listen, Katie. So let me go on to the next page and we could plug into given value of see because we&#x27;re given C equals eighty percent. So that&#x27;s point a now going on to the next page. We have us equals one minus C. So it&#x27;s one minus point eight point two. It&#x27;s o K equals one over us equals one over points. Who which equals five. This was the value that we wanted. This was the multiplier, and that&#x27;s our final answer.

Problem

A certain ball has the property that each time it…

Problem 73 Hard Difficulty

Answer

(A). $S_{n}=\frac{D\left(1-c^{n}\right)}{1-c}$
(B). Multiplier is $k=\frac{1}{s}=\frac{1}{0.2}=5$

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 2 / BC Educators

Catherine R.

Anna Marie V.

Heather Z.

Samuel H.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 11

Video Transcript

Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Heather Z.

Samuel H.

Integration Review - Intro

Definite vs Indefinite

(A). $S_{n}=\frac{D\left(1-c^{n}\right)}{1-c}$(B). Multiplier is $k=\frac{1}{s}=\frac{1}{0.2}=5$

Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Heather Z.

Samuel H.

Integration Review - Intro

Definite vs Indefinite

Catherine R.

Anna Marie V.

Heather Z.

Samuel H.

(A). $S_{n}=\frac{D\left(1-c^{n}\right)}{1-c}$
(B). Multiplier is $k=\frac{1}{s}=\frac{1}{0.2}=5$