Let the following optimal control problem $(r>0)$:
$$\max \int_0^T\ln q(t)\exp(-rt)dt$$
$$x(0)=s \quad \text{and} \quad x'(t)=-q(t)$$
My idea : let $\lambda \in C^1[0,T]$
then $\mathcal{H}(t,x,q,\lambda)=\ln q\exp(-rt)+\lambda(-q)$ $$\frac{\partial \mathcal{H}}{\partial q}=0 \implies \exp(-rt)/q(t)=\lambda(t)$$ $$\frac{\partial \mathcal{H}}{\partial x}=-\lambda'(t) \quad \text{implies} \quad \lambda'(t)=0 \quad \text{implies} \quad \lambda(t)=0 \text{ (since }\lambda(T)=0) $$ Then we'll have : $\exp(-rt)/q(t)=0 \implies 1/q(t)=0 $
I'm stuck there !