<p>You try to solve
<span class="math-container">$$
\max_{x,y} \int_0^\infty e^{-\rho t}  A U(x,y) \textrm{d} t
$$</span>
where <span class="math-container">$U(x,y) = x^\beta y^{1-\beta}$</span> and <span class="math-container">$x = z(1-y)^\alpha - c$</span>.  (I think there should be a minus sign in front of <span class="math-container">$\rho$</span>).</p>
<p>There is no intertemporal constraint, so one can solve this model by simply maximizing each period instantaneous utility function:
<span class="math-container">$$
\max_{x,y} A x^\beta y^{1-\beta} \text{ s.t. } x = z(1-y)^\alpha - c
$$</span>
Substituting the constraint into the objective gives:
<span class="math-container">$$
\max_y A (z(1-y)^\alpha - c)^\beta y^{1 - \beta}.
$$</span>
The first order condition gives:
<span class="math-container">$$
-A \beta (z(1-y)^\alpha - c)^{\beta - 1} \alpha(1-y)^{\alpha-1}y^{1 - \beta} + A(z(1-y)^{\alpha} - c)^\beta (1-\beta) y^{-\beta} = 0.
$$</span>
This can be simplified to:
<span class="math-container">$$
\frac{(1-y)^{\alpha - 1}y}{z(1-y)^\alpha - c} = \frac{1-\beta}{\beta}.
$$</span>
I don't think this has a closed form solution.</p>
<p>Anyway, the level of <span class="math-container">$x$</span> and <span class="math-container">$y$</span> in very time period will be constant.</p>

CC BY-SA 4.0
2023-08-24T22:43:20.567Z
thank you, I understand we cannot get explicit solution here because the last equation is nonlinear, is there an implicit solution you could show? Also, if I would want to force it would I get the same result with Hamiltonian or in this case that method simply doesn't work?
WilliamT
CC BY-SA 4.0
2023-08-24T07:49:33.573Z
2023-08-24T07:49:33.573Z
tdm
<p>I am trying to solve problem that looks like this; there is utility function that takes <span class="math-container">$x$</span> and <span class="math-container">$y$</span> as inputs, <span class="math-container">$x$</span> is produced by production function that depends on labor <span class="math-container">$l+y=1$</span>. <span class="math-container">$x, y$</span> depend on <span class="math-container">$t$</span> but I dropped <span class="math-container">$(t)$</span> from <span class="math-container">$x(t)$</span> to make it more readable.</p>
<p><span class="math-container">$\max_{x,y}  V(0) = \int_0^{\infty} e^{\rho t}AU(x,y)dt \quad where \quad U(x,y)= x^{\beta}y^{1-\beta} \quad \text{s.t.} \quad  x=f(1-y)=z(1-y)^{\alpha} -c$</span></p>
<p>I know that on conceptual level I can use Hamiltonian. Using generic functions that gives me:</p>
<p><span class="math-container">$$  H(x,y,t,\mu) =  e^{\rho t}AU(x,y) + \mu f(1-y) $$</span></p>
<p>when I substitute functions I get:</p>
<p><span class="math-container">$$  H(x,y,t,\mu) =  e^{\rho t}Ax^{\beta}y^{1-\beta} + \mu \left(z(1-y)^{\alpha} -c \right)$$</span></p>
<p>If I understand my textbook correctly I need to find values such that <span class="math-container">$H'(x^*,y^*,t) = -\mu'$</span>, but I am not sure if I understood it correctly. There are also supposed to be transversality conditions in scenarios when <span class="math-container">$\mu \geq 0$</span> and <span class="math-container">$\mu =0$</span> but how can I check that in a problem without numbers?</p>
<p>Can someone please give me explanation how to solve the problem above?</p>

CC BY-SA 4.0
2023-08-18T08:51:51.650Z
Your problem seems a bit ill defined. Where does $k$ come from? Is it the stock of capital. If so, then you should add a law of motion that determines how capital changes over time. If not, it should be part of the decision variables. But then, there should be some budget constraint to limit the possible values of $y$ and $k$. What is $c$?
tdm
CC BY-SA 4.0
2023-08-18T19:14:21.243Z
@tdm you are right I made mistake because I accidentally combined 2 problems in one... sorry posted this problem late in evening, I edited it. There shouldn't be $k$ in this problem, there should only be labor (1-y) because y is free time and x is some good produced by labor. $c$ is just a constant because it is assumed that at least some minimum amount of good x has to be produced so people don't starve... please let me know if more clarification is needed
WilliamT
CC BY-SA 4.0
2023-08-21T07:06:12.410Z
As I see it, you do not really have a dynamic optimization problem, as you have no intertemporal constraint (law of motion). You can easily solve it by looking at the optimal amount of $x$ and $y$ at every moment in time: $\max AU(x,y)$ subject to $x = f(1-y)$.
tdm
CC BY-SA 4.0
2023-08-22T17:46:51.663Z
@tdm so I do not need hamiltonian? but y and x are function of t I just simplified the notation it is actually x(t) and y(t)
WilliamT
CC BY-SA 4.0
2023-08-23T07:15:20.613Z
Indeed. If your objective function is additively separable and if your constraints have no cross-time restrictions, then the best thing you can do is simply optimize at every period in time.
tdm
CC BY-SA 4.0
2023-08-23T17:38:25.050Z
@tdm but what about the $\int e^{tr} dt$ part? Do I just ignore it? I know how to handle the problem without that but I am new to dynamic problems, I know you are probably busy but would you mind attempting to answer my problem? I would like to see solution so I can better understand it.
WilliamT
<utility><self-study><dynamic-programming>
Applying dynamic programming to constrained utility
WilliamT
# Applying dynamic programming to constrained utility

(Asked by: username_0 on 2023-08-17)

I am trying to solve problem that looks like this; there is utility function that takes $x$ and $y$ as inputs, $x$ is produced by production function that depends on labor $l+y=1$. $x, y$ depend on $t$ but I dropped $(t)$ from $x(t)$ to make it more readable.

$\max\_{x,y} V(0) = \int\_0^{\infty} e^{\rho t}AU(x,y)dt \quad where \quad U(x,y)= x^{\beta}y^{1-\beta} \quad \text{s.t.} \quad x=f(1-y)=z(1-y)^{\alpha} -c$

I know that on conceptual level I can use Hamiltonian. Using generic functions that gives me:

$$ H(x,y,t,\mu) = e^{\rho t}AU(x,y) + \mu f(1-y) $$

when I substitute functions I get:

$$ H(x,y,t,\mu) = e^{\rho t}Ax^{\beta}y^{1-\beta} + \mu \left(z(1-y)^{\alpha} -c \right)$$

If I understand my textbook correctly I need to find values such that $H'(x^\*,y^\*,t) = -\mu'$, but I am not sure if I understood it correctly. There are also supposed to be transversality conditions in scenarios when $\mu \geq 0$ and $\mu =0$ but how can I check that in a problem without numbers?

Can someone please give me explanation how to solve the problem above?

**Tags:** utility, self-study, dynamic-programming
> **Comments:**
> **username_1**: Your problem seems a bit ill defined. Where does $k$ come from? Is it the stock of capital. If so, then you should add a law of motion that determines how capital changes over time. If not, it should be part of the decision variables. But then, there should be some budget constraint to limit the possible values of $y$ and $k$. What is $c$?  
> **username_0**: @tdm you are right I made mistake because I accidentally combined 2 problems in one... sorry posted this problem late in evening, I edited it. There shouldn't be $k$ in this problem, there should only be labor (1-y) because y is free time and x is some good produced by labor. $c$ is just a constant because it is assumed that at least some minimum amount of good x has to be produced so people don't starve... please let me know if more clarification is needed  
> **username_1**: As I see it, you do not really have a dynamic optimization problem, as you have no intertemporal constraint (law of motion). You can easily solve it by looking at the optimal amount of $x$ and $y$ at every moment in time: $\max AU(x,y)$ subject to $x = f(1-y)$.  
> **username_0**: @tdm so I do not need hamiltonian? but y and x are function of t I just simplified the notation it is actually x(t) and y(t)  
> **username_1**: Indeed. If your objective function is additively separable and if your constraints have no cross-time restrictions, then the best thing you can do is simply optimize at every period in time.  
> **username_0**: @tdm but what about the $\int e^{tr} dt$ part? Do I just ignore it? I know how to handle the problem without that but I am new to dynamic problems, I know you are probably busy but would you mind attempting to answer my problem? I would like to see solution so I can better understand it.  

---

## 1 answers

### Answer by username_1 on 2023-08-24. Score: 2 | Accepted answer

You try to solve
$$
\max\_{x,y} \int\_0^\infty e^{-\rho t} A U(x,y) \textrm{d} t
$$
where $U(x,y) = x^\beta y^{1-\beta}$ and $x = z(1-y)^\alpha - c$. (I think there should be a minus sign in front of $\rho$).

There is no intertemporal constraint, so one can solve this model by simply maximizing each period instantaneous utility function:
$$
\max\_{x,y} A x^\beta y^{1-\beta} \text{ s.t. } x = z(1-y)^\alpha - c
$$
Substituting the constraint into the objective gives:
$$
\max\_y A (z(1-y)^\alpha - c)^\beta y^{1 - \beta}.
$$
The first order condition gives:
$$
-A \beta (z(1-y)^\alpha - c)^{\beta - 1} \alpha(1-y)^{\alpha-1}y^{1 - \beta} + A(z(1-y)^{\alpha} - c)^\beta (1-\beta) y^{-\beta} = 0.
$$
This can be simplified to:
$$
\frac{(1-y)^{\alpha - 1}y}{z(1-y)^\alpha - c} = \frac{1-\beta}{\beta}.
$$
I don't think this has a closed form solution.

Anyway, the level of $x$ and $y$ in very time period will be constant.

> **Comments:**
> **username_0**: thank you, I understand we cannot get explicit solution here because the last equation is nonlinear, is there an implicit solution you could show? Also, if I would want to force it would I get the same result with Hamiltonian or in this case that method simply doesn't work?  

---

