Confused on when to take derivative

[surferbarney0729]

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

 

[Simon Bridge]

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.

 

[surferbarney0729]

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.

 

[Simon Bridge]

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.

 

[LudusRex]

.

I would like to clarify that taking the derivative in this problem is necessary in order to find the optimal consumption in each period. In economics, we use calculus to find the maximum or minimum value of a function, which in this case is the individual's utility function. By taking the derivative and setting it equal to zero, we are finding the critical point where the function reaches its maximum value. This is known as the first-order condition in optimization problems.

In this problem, we know to take the derivative because we are trying to maximize the individual's utility function subject to their budget constraint. The budget constraint sets a limit on the individual's total consumption, and by maximizing their utility, we can find the optimal allocation of consumption between the two periods.

In general, in optimization problems like this, we look for clues in the problem that indicate we need to take the derivative. In this case, the problem explicitly states that the individual is trying to maximize their utility, which is a clear indication that we need to use calculus. Additionally, the budget constraint and the fact that the individual has a limited income also point towards using calculus to find the optimal solution.

In summary, taking the derivative is a necessary step in finding the optimal solution in this problem because we are using calculus to maximize the individual's utility function subject to their budget constraint. By setting the derivative equal to zero, we can find the critical point and determine the optimal consumption in each period.