% Answers for galley proofs:

\ansno15.1:
 The last three page-break calculations would have been
\begintt
% t=503.0 g=528.0 b=3049 p=150 c=3199#
% t=514.0 g=528.0 b=533 p=-100 c=433#
% t=542.0 g=528.0 b=* p=0 c=*
\endtt
so the break would have occurred at the same place. The badness would have
been@533, but the page would still have looked tolerable. \ (On the other
hand if that paragraph had been two lines shorter instead of one,
the first two lines of the next ``dangerous bend'' paragraph
would have appeared on that page; the value $t=531\pt$ would have
been able to shrink to $g=528\pt$ because the three ``medskips'' on
the page would have had a total shrinkability of $6\pt$. This would certainly
have been preferable to a stretched-out page whose badness was@3049; but the
author might have seen it and written another sentence or two, so that
the paragraph would not have been broken up. After all, this manual is supposed
to be an example of good practice.)

\ansno15.2:
 The next legal break after the beginning of a dangerous bend
paragraph occurs $28\pt$ later, because there is $6\pt$ additional space for
a |\medskip|, followed by two lines of $11\pt$ each. \TeX\ does not
allow breaking between those two lines; the ↑{*clubpenalty} is set briefly
to 10000 in Appendix@E, since the dangerous bend symbol is two lines tall.

\ansno15.3:
 A page always contains at least one box, if there are no
insertions, since the legal breakpoints are discarded otherwise.
Statement@(a) fails if the height of the topmost box exceeds $10\pt$.
Statement@(b) fails if the depth of the bottommost box exceeds $2\pt$, or
if some glue or kern comes between the bottommost box and the page break
(unless that glue or kern exactly cancels the depth of the box).

\ansno15.4:
 |\topinsert\vskip2in\rjustline{\vbox{\hsize|\wedge|...|\wedge
|artwork.}}\endinsert|
does the job; but ↑{*leftskip}|3.5in| would be slightly more efficient than
↑{:rjustline}, since \TeX\ would not have to read the text of the caption
twice. To compute the amount of |\leftskip|, one could of course say
|\setdimen0=\the\hsize \advdimen0 by-3in \leftskip=\dimen0|.

\ansno15.5:
 It would appear on page@25, since it does fit there. A |\midinsert|
will jump ahead of other insertions only if it is not carried over to another
page; for example, if the second 3-inch insertion were a |\midinsert|, it would
not appear on page@26, because it is converted to a |\topinsert| as soon as the
|\midinsert| macro notices that the insertion is too big for page@25.

\ansno15.6:
 Set |\count1| to 50,
then |\dimen2| to@$50\pt$,
then |\count1| to@6,
then |\skip2| to@$-10\pt$ plus@$6{\rm fil}$ minus@$50\pt$,
then |\skip2| to@$60\pt$ plus@$-36{\rm fil}$ minus@$-300\pt$,
then |\skip2| to@$1{\rm sp}$ minus@$-6{\rm sp}$,
then |\count6| to@1,
then |\skip1| to@$25\pt$ plus@$1{\rm sp}$ minus@$1{\rm fill}$,
then |\skip2| to@$25\pt$ minus@$-150\pt$,
then |\skip1| to@$0\pt$ plus@$1{\rm sp}$ minus@$1{\rm fill}$.

\ansno15.7:
 If |\skip4| has infinite stretchability, |\skip5| will be zero;
otherwise it will be $0\pt$ plus@$1\pt$.

\ansno15.8:
 |\advdimen2 by.5dm3 \divdimen2 by\dimen3 \multdimen2 by\dimen3|.

\ansno15.9:
 |\count1| takes the values 5, 2 (save@5), 4 (make this global),
8 (save@4); finally the value@4 is restored, and this is the answer.

\ansno15.10:
 |\hbox{\hbox{A}A}|. After `|\unbox5|', |\box5| is absent; |\copy5|
yields nothing.

\ansno15.11:
 |\hbox{A}|. But after `|{\global\setbox3=\hbox{A}\setbox3=\box3}|',
|\box3| will be absent.

\ansno15.12:
 |\newcount\notenumber|\parbreak
|\def\clearnotenumber{\setcount\notenumber=0}|\parbreak
|\def\note{\advcount\notenumber by1|\parbreak
|  \footnote{$↑{\the\count\notenumber}$}}|

\ansno15.13:
 Yes, in severe circumstances.
(1)@If there is no other legal breakpoint, \TeX\ will take a
break whose cost is@$\infty$. (2)@If |\vadjust{\eject}| occurs on the
same line as a footnote, before that footnote, the reference will be
forcibly detached. (3)@Other |\vadjust| commands on that line could
also interpose breakpoints before the insertion.