1. **Problem Statement:** We want to maximize the integral $$\int_0^{20} e^{\rho t} \log c(t) \, dt$$ subject to the constraint $$\dot{k}(t) = k^\alpha(t) - c(t) - \delta k(t)$$ with initial condition $$k(0) = 1$$ and free terminal capital $$k(20)$$. 2. **Set up the Hamiltonian:** The current-value Hamiltonian is $$ H = e^{\rho t} \log c(t) + \lambda(t) \left(k^\alpha(t) - c(t) - \delta k(t)\right) $$ where $$\lambda(t)$$ is the costate variable. 3. **First-order conditions:** - For consumption $$c(t)$$: $$ \frac{\partial H}{\partial c} = \frac{e^{\rho t}}{c(t)} - \lambda(t) = 0 \implies \lambda(t) = \frac{e^{\rho t}}{c(t)} $$ - For capital $$k(t)$$: $$ \dot{\lambda}(t) = -\frac{\partial H}{\partial k} = -\lambda(t) \left( \alpha k^{\alpha - 1}(t) - \delta \right) $$ 4. **Costate equation simplification:** $$ \dot{\lambda}(t) = -\lambda(t) \left( \alpha k^{\alpha - 1}(t) - \delta \right) $$ Dividing both sides by $$\lambda(t)$$: $$ \frac{\dot{\lambda}(t)}{\lambda(t)} = - \alpha k^{\alpha - 1}(t) + \delta $$ 5. **Substitute $$\lambda(t) = \frac{e^{\rho t}}{c(t)}$$ and differentiate:** $$ \dot{\lambda}(t) = \frac{d}{dt} \left( \frac{e^{\rho t}}{c(t)} \right) = \frac{\rho e^{\rho t} c(t) - e^{\rho t} \dot{c}(t)}{c^2(t)} = \lambda(t) \left( \rho - \frac{\dot{c}(t)}{c(t)} \right) $$ 6. **Equate the two expressions for $$\frac{\dot{\lambda}(t)}{\lambda(t)}$$:** $$ \rho - \frac{\dot{c}(t)}{c(t)} = - \alpha k^{\alpha - 1}(t) + \delta $$ Rearranged: $$ \frac{\dot{c}(t)}{c(t)} = \rho + \alpha k^{\alpha - 1}(t) - \delta $$ 7. **State equation:** $$ \dot{k}(t) = k^\alpha(t) - c(t) - \delta k(t) $$ 8. **Summary of system:** $$ \begin{cases} \dot{k}(t) = k^\alpha(t) - c(t) - \delta k(t) \\ \frac{\dot{c}(t)}{c(t)} = \rho + \alpha k^{\alpha - 1}(t) - \delta \end{cases} $$ 9. **Mangasarian's concavity principle:** The Hamiltonian is concave in $$c$$ and $$k$$ if $$0 < \alpha < 1$$ and $$\delta > 0$$, ensuring the optimal controls are a maximum. 10. **Optimal capital stock $$k^*(t)$$ and consumption $$c^*(t)$$:** They satisfy the above system with initial condition $$k(0) = 1$$ and free terminal $$k(20)$$. **Final answer:** The optimal paths $$k^*(t)$$ and $$c^*(t)$$ solve $$ \dot{k} = k^\alpha - c - \delta k, \quad \frac{\dot{c}}{c} = \rho + \alpha k^{\alpha - 1} - \delta, \quad k(0) = 1, \quad k(20) \text{ free} $$ with $$\lambda = \frac{e^{\rho t}}{c}$$ and the Hamiltonian concave under the given conditions.