Problem: Let $\mathcal{B}$ be the set of rectangular boxes with surface area $54$ and volume $23$ . Let $r$ be the radius of the smallest sphere that can contain each of the rectangular boxes that are elements of $\mathcal{B}$ . The value of $r^2$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$

Solution: Denote by $x$ $y$ $z$ the length, width, and height of a rectangular box.
We have
\begin{align*}
xy + yz + zx & = \frac{54}{2}  \hspace{1cm} (1) \\
xyz & = 23 \hspace{1cm} (2)
\end{align*}
We have
\begin{align*}
4 r^2 & = x^2 + y^2 + z^2 \\
& = \left( x + y + z \right)^2 - 2 \cdot \left( xy + yz + zx \right) \\
& = \left( x + y + z \right)^2 - 54 .
\end{align*}
Therefore, we solve the following constrained optimization problem:
\begin{align*}
\max_{x,y,z} \ & x + y + z \\
\mbox{subject to } & (1), (2) 
\end{align*}
First, we prove that an optimal solution must have at least two out of $x$ $y$ $z$ that are the same.
Denote by $\lambda$ and $\eta$ lagrangian multipliers of constraints (1) and (2), respectively. 
Consider the following Lagrangian:
\begin{align*}
\max_{x,y,z, \lambda, \eta} & x + y + z + \lambda \left( xy + yz + zx - 27 \right)
+ \eta \left( xyz - 23 \right) .
\end{align*}
Taking first-order-condition with respect to $x$ $y$ $z$ , respectively, we get
\begin{align*}
1 + \lambda \left( y + z \right) + \eta yz & = 0 \hspace{1cm} (3) \\
1 + \lambda \left( z + x \right) + \eta zx & = 0 \hspace{1cm} (4) \\
1 + \lambda \left( x + y \right) + \eta xy & = 0 \hspace{1cm} (5)
\end{align*}
Suppose there is an optimal solution with $x$ $y$ $z$ that are all distinct.
Taking $(4)-(3)$ , we get \[ \left( x - y \right) \left( \lambda + \eta z \right) = 0 . \]
Because $x \neq y$ , we have \[ \lambda + \eta z = 0 \hspace{1cm} (6) \]
Analogously, we have
\begin{align*}
\lambda + \eta x & = 0 \hspace{1cm} (7) 
\end{align*}
Taking $(6) - (7)$ , we get $\eta \left( z - x \right) = 0$ .
Because $z \neq x$ , we have $\eta = 0$ . Plugging this into (6), we get $\lambda = 0$
However, the solution that $\lambda = \eta = 0$ is a contradiction with (3).
Therefore, in an optimal solution, we cannot have $x$ $y$ , and $z$ to be all distinct.
W.L.O.G, in our remaining analysis, we assume an optimal solution satisfies $y = z$
Therefore, we need to solve the following two-variable optimization problem:
\begin{align*}
\max_{x,y} \ & x + 2y \\
\mbox{subject to } & 2 xy + y^2 = 27 \\
& xy^2 = 23
\end{align*}
Replacing $x$ with $y$ by using the constraint $xy^2 = 23$ , we solve the following single-variable optimization problem:
\begin{align*}
\max_y \ &  \frac{23}{y^2} + 2y \hspace{1cm} (8) \\
\mbox{subject to } & \frac{46}{y} + y^2 = 27 \hspace{1cm} (9)
\end{align*}
By solving (9), we get $y = 2$ and $-1 + 2 \sqrt{6}$
Plugging $y = 2$ into (8), we get $\frac{23}{y^2} + 2y = \frac{39}{4}$
Plugging $y = -1 + 2 \sqrt{6}$ into (8), we get $\frac{23}{y^2} + 2y = \frac{96 \sqrt{6} - 21}{23}$
We have $\frac{96 \sqrt{6} - 21}{23} < \frac{39}{4}$ .
Therefore, the maximum value of $x + y + z$ is $\frac{39}{4}$
Therefore,
\begin{align*}
r^2 & = \frac{1}{4} \left( \left( x + y + z \right)^2 - 54 \right) \\
& = \frac{1}{4} \left( \left( \frac{39}{4} \right)^2 - 54 \right) \\
& = \frac{657}{64} .
\end{align*}
Therefore, the answer is $657 + 64 = \boxed{721}$

Answer:  721