I found a neat example. It answers the question:
Given constraints on its expected values, which distribution has maximum entropy?
The answer can be shown to be the exponential family, using the variational calculus (see here).
The exponential family is the set of distributions with density:
$$
f_X(x|\lambda) = h(x)g(\lambda)\exp\left(\vec{\eta}(\lambda)\cdot\vec{T}(x)\right)
$$
for parameter vector $\lambda$.
Then, the following is (a slightly weaker version of) a theorem due to Boltzmann:
Suppose we consider $n$ measurable functions $\{f_j\}_{j=1}^n$ and numbers $a_j\in\mathbb{R}\;\forall\;j=1,\ldots,n$.
Then among all probability distributions with support on $S$ that satisfy the constraints
$$
\mathbb{E}[f_j(x)]=a_j\;\forall\;j=1,\ldots,n
$$
the one with probability density
$$
p(x) = c \exp\left( \sum_j \lambda_j f_j(x) \right)
$$
has maximum entropy.
The proof works as follows. Define the variational functional:
$$
J[p]=\int_S p(x)\ln p(x)dx - \lambda_0 \left[ \int_S p(x)dx-1 \right]
- \sum_j\lambda_j\left[ \int_S f_j(x)p(x)dx - a_j \right]
$$
with Lagrange multipliers $\lambda_i$. Notice term 2 forces $p$ to integrate to 1 and term 3 forces the moments to satisfy the constraints, while term 1 maximizes the negative entropy.
Then, the vanishing of the first variation
$$
\frac{\delta J}{\delta p} = 0
$$
can be shown to imply
$$
p(x) = c \exp\left( \sum_j \lambda_j f_j(x) \right)
$$
which is an exponential family member.
