Confused on when to take derivative

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    Confused Derivative
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The discussion revolves around the necessity of taking derivatives in the context of maximizing utility in a two-period consumption model. The problem involves individuals with a utility function defined as U = ln(C1) + ln(C2) and a lifetime budget constraint. To determine optimal consumption in each period, individuals must maximize their utility subject to this constraint. The key point highlighted is that when tasked with finding the maximum value of a function, it is essential to take the derivative and set it to zero. This indicates the critical points where the slope of the function is zero, which is a fundamental concept in calculus, even if not explicitly covered in the course material. The term "optimal" serves as a crucial identifier for recognizing the need to find the derivative to solve the problem effectively.
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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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