Question 964177
This is the base of the box:
{{{drawing(350,275,-1,13,-1,10,rectangle(0,0,12,9),
green(line(0,0,12,9)),locate(6,4.5,green(d)),
locate(5.5,0,12),locate(12.1,5,9),
locate(-.25,0,A),locate(12,0,B),locate(12,9.5,C),
rectangle(11.5,0.5,12,0)
)}}} The length of the diagonal AC (in inches) is
{{{green(d)=sqrt(12^2+9^2)=sqrt(144+81)=sqrt(225)=15}}} (so says Pythagoras).
It is clear that you cannot lay down the bat on tha base of the box,
along that diagonal, but there may be a way.
However, you could rest one end of the bat on one corner of the base of the box,
and see if the other end can fit above the opposite corner of the base,
but still within the box.
Let's see what that box would look in almost 3-D:
{{{drawing(400,320,-1,19,-1,15,
line(0,0,12,0),line(0,0,0,8),
line(0,8,12,8),line(12,8,12,0),
line(6,14,18,14),line(18,14,18,6),
line(12,0,18,6),line(12,8,18,14),line(0,8,6,14),
locate(-.25,0,A),locate(12,0,B),locate(18,6,C),
locate(18.1,14.5,D),locate(-0.35,8.5,E)
)}}} It does not look realistic, because I did not draw true perspective.
For true perspective I should draw shorter the lines are supposed to be further back,
for example making CD shorter than AE.
Instead, I made them the same length.
Maybe you could rest one end of the bat on one corner of the base of the box (A),
and see if the other end can fit above the opposite corner of the base (C),
but still within the box (maybe at D, or a little below D along line CD).
We just need to measure the distance AD to see if the bat will fit.
Let me draw the other edges and diagonals, as if the box were see-through
(I will just make them different colors, instead of black):
{{{drawing(400,320,-1,19,-1,15,
line(0,0,12,0),line(0,0,0,8),
line(0,8,12,8),line(12,8,12,0),
line(6,14,18,14),line(18,14,18,6),
line(12,0,18,6),line(12,8,18,14),line(0,8,6,14),
locate(-.25,0,A),locate(12,0,B),locate(18,6,C),
locate(18.1,14.5,D),locate(-0.35,8.5,E),
locate(5.5,0,12),locate(15,3,9),
locate(18.1,10.5,8),locate(9,3,green(15)),
blue(line(0,0,6,6)),blue(line(18,6,6,6)),
blue(line(6,14,6,6)),green(line(0,0,18,6)),
red(line(0,0,18,14))
)}}} Since AC is horizontal and CD is vertical, angle ACD is a right angle, and triangle ACD is a right triangle.
{{{drawing(425,250,-1,16,-1,9,
rectangle(15,0,14.5,0.5),line(15,0,15,8),
green(line(0,0,15,0)),red(line(0,0,15,8)),
locate(-.25,0,A),locate(15,0,C),locate(15.1,8.5,D),
locate(7,0,green(15)),locate(15.1,4.5,8)
)}}}
So, according to the Pythagorean theorem
{{{AD=sqrt(15^2+8^2)=sqrt(225+64)=sqrt(289)=17}}} .
So the box is big enough, and the bat fits.
 
Now, what about a square box, with side length {{{x}}} inches?
The length (in inches) of a diagonal of the base (like AC) is
{{{sqrt(x^2+x^2)=sqrt(2x^2)}}} .
That diagonal and a vertical edge of length {{{x}}} inches form a right angle,
and with a line connecting opposite corners of the box (like AD) they form a right triangle.
The distance between opposite corners of the box is
{{{sqrt((sqrt(2x^2))^2+x^2)=sqrt(2x^2+x^2)=sqrt(3x^2)=sqrt(3)x}}} .
Since {{{sqrt(3)=about 1.73}}} ,
for {{{x=9}}} , {{{9sqrt(3)=about15.6}}} ,
so a cube-shaped box of edge length {{{9}}} inches is too small,
but for {{{x=10}}} , {{{10sqrt(3)=about17.3}}} ,
so a cube-shaped box of edge length {{{highlight(10)}}} inches is large enough to fit the bat.