How to make contextual framed letters get proportional size automatically?

Question

I am creating a question sheet using a method (I don't know the exact name, maybe bubble sheet format) as follows:

My question is how to improve the appearance such that the framed letters get proportional size based on the context. If the frame letter must work as an exponent, it must be smaller. If it works as a lower index in integral or sum, it also must be smaller.

Here is my code snippet after "normalization" :-)

\documentclass[dvipsnames,dvips,cmyk]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{xcolor}
\usepackage{amsmath}
\usepackage{bera}


\fboxsep=2pt
\fboxrule=0.8pt

\newcommand{\boxy}[1]{\ensuremath{\,\fbox{\;\,{\color{red}\bf#1}\,\;}\,}}

\newcommand{\ltr}[1]{{\color{red}\bf#1}}

\begin{document}
\noindent%
Each letter \ltr{A}, \ltr{B}, \ltr{C}, etc in the equations represents a numeral (from 0 to 9) or the minus sign ($-$).


\[
\int_{\boxy{A}}^{\boxy{B}}f(x)\,\textrm{d}x=\frac{1-\boxy{P}}{\boxy{Q}+\boxy{R}}
\]

\[
-2 x^{\boxy{XYZ}}-\boxy{W}\sqrt{\boxy{V}+1}
\]

\[
\sum_{i=\boxy{A}}^{\boxy{BC}}=\tan\left(\boxy{D}-\boxy{E}\right)
\]
\end{document}

Answer 1

use the \text command, it takes the correct font setting:

\documentclass{article}
\usepackage{xcolor}
\usepackage{amsmath}
\fboxsep=2pt
\fboxrule=0.8pt

\newcommand\boxy[1]{\text{\,\fbox{\;\,{\color{red}$\mathbf{#1}$}\,\;}\,}}

\newcommand{\ltr}[1]{{\color{red}\bf#1}}

\begin{document}
\noindent%
Each letter \ltr{A}, \ltr{B}, \ltr{C}, etc in the equations 
represents a numeral (from 0 to 9) or the minus sign ($-$).
%
\[
\int\limits_{\boxy{A}}^{\boxy{B}}f(x)\,\textrm{d}x=\frac{1-\boxy{P}}{\boxy{Q}+\boxy{R}}
\]
%
\[
-2 x^{\boxy{XYZ}}-\boxy{W}\sqrt{\boxy{V}+1}
\]
%
\[
\sum_{i=\boxy{A}}^{\boxy{BC}}=\tan\left(\boxy{D}-\boxy{E}\right)
\]
\end{document}

Answer 2

You'll want to use \mathpalette: Replace your definition of \boxy with

\makeatletter
\newcommand{\boxy}[1]{\ensuremath{\mathpalette\boxy@helper{#1}}}
\newcommand{\boxy@helper}[2]{\,\fbox{\;\,{\color{red}$\m@th#1\mathbf{#2}$}\,\;}\,}
\makeatother

Note that I made another change to your macro: I use $s and \mathbf in the \fbox, and I added \m@th to ensure that you don't get additional horizontal space.