# Macroeconomics/Optimal Growth

## Optimal Growth

### Assumptions and Story

The topic of Optimal Growth is often introduced with mention of the economy of a single person dealing with a single commodity. Robinson Crusoe, for instance, might be stuck on an Island with only one commodity, coconuts perhaps, denoted by ${\displaystyle c}$. In this case, just for fun, we may want Crusoe to live forever, but for now we may let him have a finite lifetime. This assumption changes how we will deal with the problem, with a finite lifetime lending itself to the Calculus of Variations or the Hamiltonian methods of analysis, and an infinite lifetime often requiring recursive methods. Also our one good is consumed, and is used for "once" production each day.

### Model

In this model, we have on good, ${\displaystyle c\in C\subseteq \mathbb {R} }$.

Our agent has an increasing (or at least non-decreasing) utility function, ${\displaystyle u:C\to \mathbb {R_{+}} }$.

The production function in the economy is nondecreasing ${\displaystyle f:C\to \mathbb {R_{+}} }$. For some reason we use ${\displaystyle k}$ as the argument in this function. We'll see the relationship between c and k when we talk about our constraint in just one minute.

Crusoe wants to maximize his utility over the period he lives, but perhaps he discounts the future by a factor of ${\displaystyle \beta }$, so we can write our objective function ${\displaystyle \sum _{t=0}^{T}\beta ^{t}u(c_{t})}$ with ${\displaystyle c_{t}}$ the amount consumed at time ${\displaystyle t}$ (for now we are considering a discrete time model).

We want our solution to be feasible, so our sequence ${\displaystyle \{c_{0},c_{1},\ldots \}}$ must be "possible". For this to happen, we need the sequence to satisfy the constraint that ${\displaystyle c}$ consumed plus ${\displaystyle k}$ used tomorrow in production must be less than or equal to the amount produced. We will take as given an initial endowment of coconuts, ${\displaystyle k_{0}}$. So our problem is:

${\displaystyle \max \sum _{t=0}^{T}\beta ^{t}u(c_{t})}$

such that ${\displaystyle 0\leq c_{t}+k_{t+1}\leq f(k_{t})}$

### Solution for finite T

since our functions f and u are nondecreasing, we can rewrite our problem as:

${\displaystyle \max \sum _{t=0}^{T}\beta ^{t}u(f(k_{t})-k_{t+1})}$

such that ${\displaystyle k_{t+1}\leq f(k_{t})}$ for ${\displaystyle t=0,1,\ldots ,T-1}$

We can solve this using Calculus of Variations (Euler's Equation) or by first order conditions of the Legrangian with respect to ${\displaystyle k_{t+1}}$, so we get:

${\displaystyle u^{\prime }\left(f(k_{t})-k_{t+1}\right)=\beta u^{\prime }\left(f(k_{t+1})-k_{t+2}\right)f^{\prime }(k_{t+1})}$ for ${\displaystyle t=0,1,\ldots ,T-1}$.

Given ${\displaystyle k_{0}}$, Weierstrass' Theorem guarantees that this problem has a solution since the objective function is continuous and the constraint set is closed and bounded (hence compact). If T goes to infinity, however, we cannot guarantee a solution this way. A discussion of this will come later.

### Solution for finite T over continuous time

Now instead of a summation, we could use an integral to get a similar problem over continuous time.

${\displaystyle \max \int _{t=0}^{T}\beta ^{t}u(f(k_{t})-k_{t+1})}$

such that ${\displaystyle k_{t+1}\leq f(k_{t})}$ for ${\displaystyle t=0,1,\ldots ,T-1}$

We can solve this using Calculus of Variations (Euler's Equation), or use Hamiltonians if necessary, which we remember is very similar in form to first order conditions of the Legrangian with respect to ${\displaystyle k_{t+1}}$. Using Euler's equation, we get: