Applying dynamic programming to constrained utility

Question

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?

Answer 1

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.