= Solution
Write $T=t_0$, and specify the physical <Black-Scholes model> as $dS_t/S_t=\mu\,dt+\sigma\,dW_t$, with constant $\sigma>0$ and bank rate $\rho$. The physical drift $\mu$ is needed for an <expected utility> calculation; a pricing model specified only under a <risk-neutral measure> does not determine the investor's preferences over risky positions. Put
$$
\vartheta=\frac{\mu-\rho}{\sigma},\quad Z_t=e^{-\vartheta W_t-\vartheta^2t/2},\quad \xi_t=e^{-\rho t}Z_t.
$$
Here $\vartheta$ is the <market price of risk>, $Z_T$ is the density of the <risk-neutral measure> $Q$, and $\xi$ is the <state-price density>. The <Girsanov theorem> makes $W_t^Q=W_t+\vartheta t$ a <Brownian motion> under $Q$.
For a dollar holding $\pi_t$ in the <stock> and <consumption> rate $c_t$, the <portfolio wealth> equation is
$$
dV_t=\{\rho V_t+(\mu-\rho)\pi_t-c_t\}dt+\sigma\pi_t\,dW_t,\qquad V_0=w_0.
$$
Since $d\xi_t=-\rho\xi_tdt-\vartheta\xi_tdW_t$, integration by parts gives the useful <deflated wealth equation with consumption>
$$
d(\xi_tV_t)+\xi_tc_tdt=\xi_t(\sigma\pi_t-\vartheta V_t)\,dW_t.
$$
For strategies with a true deflated gain <martingale>, terminal wealth $X=V_T$ and <consumption> obey the <state-price budget constraint>
$$
\mathbb E_P\left[\xi_TX+\int_0^T\xi_tc_tdt\right]=w_0.
$$
For nonnegative wealth and consumption, the admissibility condition can instead give a budget inequality; increasing utilities use the full budget at an optimum. Conversely, suitable integrable payoff and consumption plans with this budget are financeable in the Brownian <complete market>. To construct the strategy, set
$$
M_t=\mathbb E_P\left[\xi_TX+\int_0^T\xi_sc_sds\,\middle|\,\mathcal F_t\right],\qquad
V_t=\xi_t^{-1}\left(M_t-\int_0^t\xi_sc_sds\right).
$$
By the <Brownian martingale representation theorem>, write $dM_t=H_t\,dW_t$. The required dollar stock holding is $\pi_t=(H_t/\xi_t+\vartheta V_t)/\sigma$; substitution in the deflated gain equation verifies financing. Remaining wealth goes into the <bank account>. For square-integrable data the representation holds directly, and localization handles the customary integrable admissible data.
First take no <consumption>. For an increasing strictly <concave> differentiable <utility function> whose marginal derivative can be inverted, let $I_u=(u')^{-1}$. Choose $\lambda>0$ so that
$$
\boxed{X^*=I_u(\lambda\xi_T),\qquad\mathbb E_P[\xi_TX^*]=w_0.}
$$
This candidate is not merely a first-order guess. The <concave supporting-tangent inequality> gives, pointwise,
$$
u(X)\leq u(X^*)+u'(X^*)(X-X^*)=u(X^*)+\lambda\xi_T(X-X^*).
$$
Taking <expectations> and using the budget proves $\mathbb E u(X)\leq\mathbb E u(X^*)$ for every feasible competitor. Finance the optimizer by its <risk-neutral pricing> value $V_t=e^{-\rho(T-t)}\mathbb E_Q[X^*\mid\mathcal F_t]$, using the preceding representation to find the holdings.
For a merely <concave> differentiable <utility function>, the general formulation is to select $X^*(\omega)$ from the maximizers of $u(x)-\lambda\xi_T(\omega)x$ over its wealth domain, and choose the multiplier to satisfy the budget. If increasing utility is not assumed, the equality-budget multiplier may have either sign; the positive-multiplier inverse formula above is for increasing utility. Flat portions allow multiple maximizers; boundaries require one-sided inequalities rather than equality of marginal derivatives. Existence of a multiplier, integrability and an admissible maximizer must be checked. Concavity alone does not guarantee a well-posed problem: with unrestricted dollar positions, linear utility $u(x)=x$ and $\mu>\rho$ permits unbounded expected terminal wealth. The explicit utilities below satisfy the needed conditions in their stated domains.
For <exponential utility> $u(x)=(1-e^{-ax})/a$ on $\mathbb R$, $u'(x)=e^{-ax}$ and $I_u(y)=-\log y/a$. Therefore
$$
X^*=-\frac1a\log(\lambda\xi_T)
=-\frac{\log\lambda}{a}+\frac{\rho T}{a}+\frac{\vartheta W_T}{a}+\frac{\vartheta^2T}{2a}.
$$
The budget is equivalently $\mathbb E_QX^*=w_0e^{\rho T}$, since $\xi_T=e^{-\rho T}dQ/dP$. Under $Q$, $\mathbb E_QW_T=-\vartheta T$, giving $\log\lambda=\rho T-\vartheta^2T/2-aw_0e^{\rho T}$. Hence
$$
X^*=w_0e^{\rho T}+\frac{\vartheta}{a}W_T^Q,\qquad
V_t=e^{-\rho(T-t)}\left(w_0e^{\rho T}+\frac{\vartheta}{a}W_t^Q\right).
$$
Matching the Brownian coefficient of $dV$ with $\sigma\pi_t$ gives the <finite-horizon exponential-utility portfolio>
$$
\boxed{\pi_t^*=\frac{\mu-\rho}{a\sigma^2}e^{-\rho(T-t)}.}
$$
Hold $\pi_t^*/S_t$ shares and invest $V_t-\pi_t^*$ in the <bank account>. This is a dollar amount, not a fixed wealth fraction. The Gaussian exponential formula also gives the optimal <expected utility>
$$
\boxed{\mathbb E_Pu(X^*)=\frac{1-\exp(-aw_0e^{\rho T}-\vartheta^2T/2)}{a}.}
$$
This calculation uses unrestricted wealth and an admissibility class permitting the displayed Gaussian terminal wealth, with integrable deflated gains and the relevant exponential moment. A nonnegative-wealth constraint would change the optimizer; no such constraint is imposed on this exponential-utility example.
For both terminal and running <utility function>, choose $X$ and a nonnegative adapted <consumption> rate $c$ to maximize $\mathbb E_P[u(X)+\int_0^Tv(c_t)dt]$ subject to the combined budget. Pointwise maximization of $u(x)-\lambda\xi_Tx$ and $v(c)-\lambda\xi_tc$ gives, for interior strictly <concave> cases,
$$
X^*=I_u(\lambda\xi_T),\qquad c_t^*=I_v(\lambda\xi_t),\qquad
\mathbb E_P\left[\xi_TX^*+\int_0^T\xi_tc_t^*dt\right]=w_0.
$$
Apply the <concave supporting-tangent inequality> to $u$ and to $v$ at every time, then integrate and take <expectations>. The difference in objectives is at most $\lambda$ times the difference in budget expenditures, which is nonpositive. This proves optimality, not just the marginal conditions. The financing value is
$$
V_t=\xi_t^{-1}\mathbb E_P\left[\xi_TX^*+\int_t^T\xi_sc_s^*ds\,\middle|\,\mathcal F_t\right].
$$
The question's consumption rate $R_t$ is denoted $c_t$ here to keep it distinct from interest-rate notation.
For <logarithmic utility> with $u(x)=a\log x$, $v(x)=b\log x$, assume $a,b,w_0>0$ and positive wealth and <consumption>. The marginal conditions and budget give
$$
\lambda=\frac{a+bT}{w_0},\qquad X^*=\frac{a}{\lambda\xi_T},\qquad c_t^*=\frac{b}{\lambda\xi_t}.
$$
Each state-price-weighted payment is deterministic, so conditional valuation immediately gives $V_t=[a+b(T-t)]/(\lambda\xi_t)$. Put $A_t=a+b(T-t)$. Since the <Itô formula> gives $d(1/\xi_t)=(1/\xi_t)[(\rho+\vartheta^2)dt+\vartheta dW_t]$, this value satisfies
$$
dV_t=V_t\left(\rho+\vartheta^2-\frac{b}{A_t}\right)dt+\vartheta V_t\,dW_t.
$$
Comparison with the wealth equation verifies the complete <finite-horizon logarithmic investment and consumption> strategy:
$$
\boxed{V_t=\frac{A_t}{\lambda\xi_t},\qquad R_t^*=c_t^*=\frac{bV_t}{A_t},\qquad\pi_t^*=\frac{\mu-\rho}{\sigma^2}V_t.}
$$
Thus the optimal share holding is $\pi_t^*/S_t$, and the remaining value $V_t-\pi_t^*$ is in the <bank account>; borrowing or shorting is allowed if that value or the stock holding is negative. Terminal wealth is positive and consumption is positive throughout. Finally $\mathbb E\log\xi_t=-(\rho+\vartheta^2/2)t$, so the optimal objective, if desired explicitly, is
$$
\boxed{a\log\frac{aw_0}{a+bT}+bT\log\frac{bw_0}{a+bT}+\left(\rho+\frac{\vartheta^2}{2}\right)\left(aT+\frac{bT^2}{2}\right).}
$$
Finite-horizon lognormal moments ensure the budget and expected utilities are finite, completing the feasibility and optimality checks.
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