My Math Forum Macro-economics

 Economics Economics Forum - Financial Mathematics, Econometrics, Operations Research, Mathematical Finance, Computational Finance

 April 9th, 2018, 02:28 AM #1 Member   Joined: Mar 2017 From: Israel Posts: 70 Thanks: 2 Macro-economics Hello Can you help me please about the next question: For a certain family, a private consumption curve/function that has a marginal tendency to consume is constant at any available income (the income after deduction because of debts, tax and more) level which equal to 0.75. It is known that when the available income level is zero, the planned consumption of the family is positive and equal to 120. Find the private consumption and available income of the family if in the current available income level, the average tendency to consume is equal to 0.81. I'm trying to solve this exercise but I don't even know what is the current available income level or how to find it. Thanks! And of course if you didn't understand something, I will explain again.
 April 9th, 2018, 04:16 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,038 Thanks: 423 I THINK that this is what you are looking for, but the question does not seem well enough worded for me to be at all sure. $c(a) = \text { PLANNED consumption.}$ $a = \text { available income.}$ $x(a) = \text { average tendency (???) to consume.}$ $c'(a) = 0.75 \text { given } a \ge 0.$ $x(a) = \dfrac{c(a)}{a} \text { given } a > 0.$ The above is what I think the question means in math. If I am correct in what I think that problem is trying to say, the solution is mathematically simple. $\displaystyle c'(a) = 0.75 \implies c(a) = \int 0.75\ da = 0.75a + K.$ $c(0) = 120 = 0.75(0) + K \implies K = 120 \implies c(a) = 0.75a + 120.$ $x(a) = 0.81 = \dfrac{0.75a + 120}{a} \implies 0.06a = 120 \implies$ $a = 2000 \implies c(a) = 0.75 * 2000 + 120 = 1620.$ Thanks from IlanSherer
April 9th, 2018, 10:27 AM   #3
Member

Joined: Mar 2017
From: Israel

Posts: 70
Thanks: 2

Quote:
 Originally Posted by JeffM1 I THINK that this is what you are looking for, but the question does not seem well enough worded for me to be at all sure. $c(a) = \text { PLANNED consumption.}$ $a = \text { available income.}$ $x(a) = \text { average tendency (???) to consume.}$ $c'(a) = 0.75 \text { given } a \ge 0.$ $x(a) = \dfrac{c(a)}{a} \text { given } a > 0.$ The above is what I think the question means in math. If I am correct in what I think that problem is trying to say, the solution is mathematically simple. $\displaystyle c'(a) = 0.75 \implies c(a) = \int 0.75\ da = 0.75a + K.$ $c(0) = 120 = 0.75(0) + K \implies K = 120 \implies c(a) = 0.75a + 120.$ $x(a) = 0.81 = \dfrac{0.75a + 120}{a} \implies 0.06a = 120 \implies$ $a = 2000 \implies c(a) = 0.75 * 2000 + 120 = 1620.$
I found it, the average tendency to consume - it's APC (Average propensity to consume).
If there are another words which you didn't understand, just say and I will try to find a way to explain you

April 10th, 2018, 03:47 AM   #4
Member

Joined: Mar 2017
From: Israel

Posts: 70
Thanks: 2

Quote:
 Originally Posted by JeffM1 I THINK that this is what you are looking for, but the question does not seem well enough worded for me to be at all sure. $c(a) = \text { PLANNED consumption.}$ $a = \text { available income.}$ $x(a) = \text { average tendency (???) to consume.}$ $c'(a) = 0.75 \text { given } a \ge 0.$ $x(a) = \dfrac{c(a)}{a} \text { given } a > 0.$ The above is what I think the question means in math. If I am correct in what I think that problem is trying to say, the solution is mathematically simple. $\displaystyle c'(a) = 0.75 \implies c(a) = \int 0.75\ da = 0.75a + K.$ $c(0) = 120 = 0.75(0) + K \implies K = 120 \implies c(a) = 0.75a + 120.$ $x(a) = 0.81 = \dfrac{0.75a + 120}{a} \implies 0.06a = 120 \implies$ $a = 2000 \implies c(a) = 0.75 * 2000 + 120 = 1620.$
Yes, you understood the question correctly, thanks a lot!

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Simonsky Elementary Math 2 January 28th, 2017 12:50 PM jiasyuen Calculus 0 June 5th, 2015 12:47 AM mathormyth Economics 2 February 11th, 2015 10:02 AM miscloto Linear Algebra 2 February 7th, 2015 03:30 PM JJAtlanta New Users 4 August 8th, 2011 11:01 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top