Question 1092961
The more math you know, the more difficult it gets.
 
THE MIDDLE-SCHOOLER SOLUTION:
{{{361=19^2}}}
If the width and lenth are {{{w=l=19inch}}} ,
the box volume would be
{{{(19inch)*(19inch)*(1inch)=361}}}{{{cubic}}}{{{inches}}} .
The shape of such a box could work for a small pizza,
but the "shortest distance around the box" would be
{{{19inches+1inch+19inches+1inch=40inches}}} .
We can also get a volume of {{{361}}}{{{cubic}}}{{{inches}}}
if we make {{{w=l=19/2}}}{{{inch=9.5inch}}} and {{{height=4inch}}} .
Then, the volume in cubic inches is
{{{(19/2)*(19/2)*4=19^2=361}}} .
In that case, the shortest way around the box is
{{{9.5inch+4inch+19inch+9.5inche+4inch=19inch+8inch=27inch}}} ,
and {{{27inch+1inch=28inch}}} .
So, the solution is
{{{highlight(system(length=width=9.5inches,height=4inches))}}} .
 
THE HIGH-SCHOOLER (OR COLLEGE STUDENT) SOLUTION:
We are expected to find all possible solutions,
and "show our work" with equations,
because we already studied algebra, 
and maybe even calculus.
We were given {{{w}}} and {{{l}}}
(for the width and length measuerments of the base/bottom/lid of the box).
We will say that those dimensions are are measured in inches,
and we must define another variable:
{{{h}}}= height of the box in inches.
{{{Volume=w*l*h=hw^2=361}}}
We must write our equations as
{{{hw^2=361}}} for the volume, and
{{{system(4w+1=28,"or",2(w+h)+1=28)}}} for the length of the tape closing the box.
That may give us more than one solution.
 
One of the {{{or}}} choices gives us
{{{system(4w+1=28,hw^2=361)}}} --> {{{system(4w+1=28,h=361/w^2)}}} --> {{{...}}} --> {{{highlight(system(length=width=9.5inches,height=4inches))}}} .
 
That solution above is one solution,
but we have to look for any other possible solution.
The other {{{(or)}}} choice gives us
{{{system(2(w+h)+1=28,hw^2=361)}}} --> {{{system(w+h=27/2,h=361/w^2)}}} --> 
{{{system(w+361/w^2=27/2,h=361/w^2)}}} --> {{{system(2w^3+722=27w^2,h=361/w^2)}}} --> {{{"?!?!"}}}
That {{{2w^3+722=27w^2}}}<-->{{{2w^3-27w^2+722=0}}} is a cubic equation.
I studied polynomial function, so I know that
it must have at least positive real solution for {{{w}}} .
I may want to use my graphing calculator.
Could there be a rational root?
If there is, it would be of the form {{{p/q}}} ,
where {{{p}}}= is a factor of {{{722=2*19^2}}} ,
and {{{q}}} is a factor of {{{2}}} .
That includes the previously found {{{w=19/2}}} .
Dividing {{{2w^3-27w^2+722}}} by {{{w-19/2}}} , I get
{{{2w^2-8w-76}}} , so the solutions to {{{2w^3-27w^2+722=0}}}
that work for this problem
are {{{highlight(w=19/2=9.5)}}}
and any positive solution to
{{{2w^2-8w-76=0}}}<-->{{{w^2-4w-38=0}}} .
I cannot solve that by factoring,
but "completing the square" or using the quadratic formula I find
{{{w=2 +- sqrt(42)}}} , one positive and one negative solution.
I could have found the approximate value of {{{2 + sqrt(42)}}} as {{{8.40874}}} ,
but the exact {{{highlight(w=2 + sqrt(42)=approximately8.40874)}}}
was probably required.
Then {{{h=361/w^2="?!?!"}}} .
{{{highlight(h=approximately5.01926)}}}
Am I expected to express that as the exact solution too?
{{{(2+sqrt(42))^2=4+42+4sqrt(42)=46+4sqrt(42)}}}
{{{h=361/(46+4sqrt(42))=361(46-4sqrt(42))/((46+4sqrt(42))(46-4sqrt(42)))=(16606-1444sqrt(42))/(2116-16*42)=(16606-1444sqrt(42))/(2116-672)=(16606-1444sqrt(42))/1444=11.5-sqrt(42)}}}
So,
{{{highlight(system(h=11.5-sqrt(42),l=w=2 + sqrt(42)))}}}
or
{{{highlight(system(h=approximately5.01926,l=w=approximately8.40874))}}}
is the other solution.