= Solution
Ignoring terms independent of $\theta$, the <log-likelihood> is
$$
\ell(\theta)
=\sum_i\bigl(Y_i\log\theta-\theta x_i\bigr)+\text{constant}.
$$
Its <score function> is
$$
\ell'(\theta)=\frac{\sum_iY_i}{\theta}-\sum_i x_i.
$$
Since $\ell''(\theta)=-(\sum_iY_i)/\theta^2\leq0$, the interior critical point is the maximum, giving
$$
\boxed{\widehat\theta_{MLE}
=\frac{\sum_iY_i}{\sum_i x_i}}.
$$
The sum of independent <Poisson distributions> is Poisson with mean $\theta\sum_i x_i$, so
$$
\mathbb E\widehat\theta_{MLE}=\theta.
$$
\b[Thus the <maximum-likelihood estimator> is also unbiased.]
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