Maximise $p_1p_2p_3p_4p_5$ subject to constraints 
Given $x_5 \geq x_4 \geq x_3 \geq x_2 \geq x_1 \geq 0$, solve the following optimization problem in $p_1, p_2,\dots, p_5$.
$$\max p_1p_2p_3p_4p_5$$
subject to:
$$p_1 x_1 + p_2 (x_2 - x_1) + p_3(x_3-x_2) + p_4(x_4-x_3)+p_5(x_5-x_4) = 1 $$
$$0\leq p_5 \leq p_4 \leq p_3 \leq p_2 \leq p_1$$

Is there a closed form solution to this problem?  I am not sure how to go about approaching it.
 A: Well you could try with Am-Gm inequality:
$$ 5\sqrt[5]{p_1p_2...p_5x_1(x_2-x_1)...(x_5-x_4)}\leq 1$$
So $$p_1p_2...p_5 \leq {1\over 5^5x_1(x_2-x_1)...(x_5-x_4)}$$
and this value is achivable if $$ p_1 ={1\over 5x_1}$$
$$ p_2 ={1\over 5(x_2-x_1)}$$
$$ p_3 ={1\over 5(x_3-x_2)}$$
$$...$$
A: This answer addresses the case when
$$
\frac{5}{x_5}\geq\max_{i=1,\dots,4}\big\{\frac{i}{x_i}\big\}.
$$
In this case, the solution using the AM-GM inequality fails to satisfy the ordering constraint. There are still other cases to be considered, for example, $x=(1,\;3,\;4.75,\;8.75,\;13.75)$ which produces the solution $p_1=1/5$, $p_2=p_3=8/75$, $p_4=1/20$ and $p_5=1/25$ (computed this solution numerically).

While the other answers are clever, I don't believe that they respect the ordering constraint $0\leq{p_5}\leq\dots\leq{p_1}$. My proposed solution is found using the KKT conditions. (Note: I assume $x_i>0$ for all $i$--the case when $x_i=0$ should following from a similar technique to what I use below).
Answer: $p_1=p_2=p_3=p_4=p_5=1/x_5$.
Solution: By taking the logarithm of the objective function, the given optimization problem can be formulated as a convex optimization problem in standard form:
$$
\begin{array}{rl}
\min & -\log(p_1)-\log(p_2)-\cdots-\log(p_5)\\
\text{s.t.} & p_1x_1+p_2(x_2-x_1)+p_3(x_3-x_2)+p_4(x_4-x_3)+p_5(x_5-x_4)-1=0\\
&p_2-p_1\leq0\\
&p_3-p_2\leq0\\
&p_4-p_3\leq0\\
&p_5-p_4\leq0\\
&\hspace{12pt}-p_5\leq0
\end{array}
$$
Associate the multiplier $\lambda\in\mathbb{R}$ with the equality constraint, and $\mu_i\geq0$ with the remaining constraints $i=1,\dots,5$. The stationarity conditions for this optimization problem are
\begin{align*}
\frac{1}{p_1}&=\lambda{x_1}-\mu_1\\
\frac{1}{p_2}&=\lambda(x_2-x_1)+(\mu_1-\mu_2)\\
\frac{1}{p_3}&=\lambda(x_3-x_2)+(\mu_2-\mu_3)\\
\frac{1}{p_4}&=\lambda(x_4-x_3)+(\mu_3-\mu_4)\\
\frac{1}{p_5}&=\lambda(x_5-x_4)+(\mu_4-\mu_5)\\
\end{align*}
A solution to the above system is 
\begin{align*}
p_i&=\frac{1}{x_5}&&\text{for }i=1,\dots,5\\
\mu_i&=5x_i-i\cdot{x_5}&&\text{for }i=1,\dots,5\\
\lambda&=5
\end{align*}
It's easy to verify that the other KKT conditions are satisfied (complementary slackness, etc.), and to check that the problem satisfies the Slater condition. Hence, we conclude that the proposed point is indeed globally optimal for this problem.
A: Letting
$y_1 = x_1$
and
$y_i 
=x_i-x_{i-1}$ 
for $i>1$,
this is
$\sum_{i=1}^n p_iy_i
=1
$
with $n=5, y_i \ge 0$.
As Maria Mazur suggested,
using AM-GM gives
$\prod_{i=1}^n p_iy_i
\le (\frac1{n}\sum_{i=1}^n p_iy_i)^n
=\frac1{n^n}
$
with equality iff
$p_iy_i =\frac1{n}
$
or
$p_i = \frac1{ny_i}
$.
Note that there may be a problem
if some $y_i = 0$;
in this case,
$p_i$ can be any value.
Perhaps the conditions should be
$y_i > 0$.
(Added after a comment)
Regarding the requirement
that
$p_{i+1} \le p_i$:
This will hold if
$y_{i+1} \ge y_i$.
If not,
perhaps use the same idea
of defining
$q_i = p_i-p_{i+1}
$,
rewriting in terms of the $q_i$,
and requiring $q_i \ge 0$.
