Confused on when to take derivative

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    Confused Derivative
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Discussion Overview

The discussion centers around understanding when to take derivatives in the context of maximizing utility in an economic model. Participants explore the identification of critical points and the necessity of derivatives in optimization problems, particularly in relation to a specific problem involving consumption over two periods.

Discussion Character

  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the reasoning behind taking the derivative in the given problem, seeking clarity on what indicators suggest the need for this step.
  • Another participant explains that critical points occur where the slope of a function is zero, which is determined by the derivative.
  • A further response emphasizes that the problem's request for the maximum value of a function implies the need to find and set the derivative to zero, suggesting that a foundational understanding of calculus is assumed.
  • Participants note that the term "optimal" in the context of the utility function indicates the necessity of finding a maximum, which involves taking the derivative.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the explicit indicators for taking derivatives in optimization problems. There is no consensus on the clarity of these indicators based on the course material.

Contextual Notes

Some participants highlight that the course may not cover all necessary calculus tools, implying a reliance on prior knowledge to recognize when to apply derivatives in optimization scenarios.

surferbarney0729
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Here is a question we did in class...Why in the answer do we take the derivatve? How do we know to take the derivative? What in the problem would have set us off to look for the derivative and set to 0?

Consider an economy that is composed of identical individuals who live for two periods. These individuals have preferences over consumption in periods 1 and 2 given
by U = ln(C1) + ln(C2). They receive an income of 100 in period 1 and an income of 50
in period 2. They can save as much of their income as they like in bank accounts,
earning an interest rate of 10% per period. They do not care about their children, so
they spend all their money before the end of period 2. Each individual’s lifetime budget constraint is given by C1+ C2/(1 + r) = Y1+ Y2/(1+ r). Individuals choose consumption in each period by maximizing lifetime utility subject to this lifetime budget constraint.

here is part a. and the answer

a. What is the individual’s optimal consumption in each period? How much saving
does he or she do in the first period?
Individuals solve
max U = ln(C1) + ln(C2) subject to C1+ C2/(1.1) = 100 + 50/(1.1).
Rearrange the budget constraint C2= 110 + 50 – 1.1C1 and plug into the maximand: max U = ln(C1) + ln(160 – 1.1C1).
Then take the derivative and set it equal to zero:1/C1= 1.1/(160 – 1.1C1), or 2.2C1
= 160.

So C1
≈ 72.7, and savings 100 – C1≈ 27.3. The optimal consumption in the second period is then 50 + 1.1(100 – C1) = 80
 
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The critical points of a function are where the slope of the function is zero.
The maximum is a critical point.
The slope is given by the derivative.
 
What was the key identifier in that problem that said...you must find the derivative? There is no mention in the chapter of derivatives at all, so I am just trying to appreciate what signifies the derivative needing to be set to zero to solve.
 
swoodward said:
What was the key identifier in that problem that said...you must find the derivative? There is no mention in the chapter of derivatives at all, so I am just trying to appreciate what signifies the derivative needing to be set to zero to solve.
When it asked you to find the maximum value of a function.
A particular course may not explicitly teach all the tools needed for the course - you have been presumed to have a minimum calculus knowledge such as how to find and characterize the critical points of a function. The rest is reading/comprehension: do you recognize when you have a function and when you have to find a max or a min.
Here the key word is "optimal".

You have a function U = ln(C1) + ln(160 – 1.1C1) and the optimum U is when this function is a maximum (hence the "maximand" wording).
U is a maximum for C1: dU/dC1 = 0.
 
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