Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 4 / 28J (Optimization and Control)

The requested properties are not properties of the optimum under the printed assumption $\alpha +\beta >1$. We first derive their origin, and then solve the actual constrained problem.

Put $s=1-\ell$, so $0\le s\le 1$, and write the Hamiltonian of an optimal-control problemFor an interior stationary control, $H_c=H_s=0$ givesThis proves the printed identity only as a necessary stationarity condition for an interior candidate.

It fails the required maximization condition. For fixed positive $p$ and fixed $s$, maximization over $c$ givesSince $\beta /(1-\alpha )>1$, this is strictly convex in $s$. Its maximum on $[0,1]$ is at an endpoint, not at its interior stationary point. Thus the actual optimum uses $\ell =1,c=0$, or $\ell =0,c>0$. On any consumption interval the printed relation would require $\beta c=\alpha$, although the actual optimal consumption varies with time.

This proves a genuine failure of the printed premise, not just failure to check a boundary condition. If the parameter inequality were reversed and an interior solution stayed within the control bounds, these first-order equations would instead arise from a concave objective.

The adjoint equation is $\dot{p}=-rp$, so $p(t)=p_0{e}^{-rt}$. For the interior stationary candidate from part (i), writing $s=\beta c/\alpha$ gives $p\propto c^{\alpha +\beta -1}$. Therefore it formally haswhich derives the printed formula as a stationary, generally nonoptimal candidate.

For the work-then-consume optimum for a nonconcave consumption-leisure utility, the endpoint maximization in part (i) giveswhenever consumption is chosen, and $\ell =1,c=0$ otherwise. Compare the two endpoint values: leisure and consumption are preferred whenBecause the unconstrained consumption is increasing in time, there is at most one switch: work first, then cease work and consume at an increasing rate.

Here is a global verification, not just a Pontryagin maximum principle condition. The terminal constraint is equivalent toFor any $p_0>0$, pointwise maximization of $c^{\alpha }{s}^{\beta }-p_0{e}^{-rt}(c+s-1)$ gives an upper bound on every feasible objective, after adding $p_0{x}_0$. A control attaining this pointwise maximum and using the full budget attains the bound, so it is globally optimal.

Let $a=r\alpha /(1-\alpha )$. Ifthe optimal switch time is $t_{\ast }=0$ and $C=C_0$. Otherwise choose the unique $t_{\ast }\in (0,T)$ such thatThe left side of the second equality after substitution decreases strictly from $c_{\ast }(e^{aT}-1)/a$ to zero, while the right side increases from $x_0$ to a positive number. Thus the switch exists and is unique. The control is $(\ell ,c)=(1,0)$ before $t_{\ast }$ and $(0,C{e}^{rt/(1-\alpha )})$ after it. It saturates the budget and attains the dual upper bound, completing the proof of optimality for the actual printed parameter range.

For the interior stationary candidate, let $\gamma =1/(\alpha +\beta -1)$. Its formal relation $\ell =1-\beta c/\alpha$ givesIntegration therefore gives $x=A{e}^{rt}+B{e}^{-\gamma rt}-1/r$. Imposing $x(0)=x_0$ and $x(T)=0$ yieldsThis recovers the printed expression algebraically, but does not turn its stationary control into an optimum; it may also violate $\ell \ge 0$.

The actual globally optimal wealth from part (ii) isThe budget equation makes the two branches agree at the switch. Direct differentiation gives $\dot{x}=rx+\ell -c$ on both branches, and $x(T)=0$. This solution remains nonnegative throughout, even though only the terminal nonnegativity constraint was imposed. It provides the complete corrected solution for $\alpha +\beta >1$.