Question 968969
Let {{{X}}} , {{{Y}}} , and {{{Z}}} be the dimensions (in feet) of the rectangular prism.
Let's say {{{X<=Y<=Z}}} .
{{{drawing(550,400,-1,10,-1,7,
line(0,0,6,0),line(0,2,6,2),
line(0,0,0,2),line(6,0,6,2),
line(4,6,9,6),line(0,2,4,6),
line(6,2,9,6),line(6,0,9,4.5),
line(9,4.5,9,6),locate(0.1,1.2,X),
locate(6.1,1.2,X),locate(9.1,5.3,X),
locate(2.9,0,Y),locate(2.9,2,Y),
locate(6.4,6,Y),locate(2,4,Z),
locate(7.5,4,Z),locate(7.7,2.5,Z)
)}}}
The surface area of the base of that {box) is {{{YZ}}} square feet.
The surface area of the top of that {box) is {{{YZ}}} square feet.
The surface area of the sides of that {box) is {{{2XY+2XZ}}} square feet.
The total surface area of the base of that {box) in square feet is
is {{{2XY+2XZ+2YZ=2(XY+XZ+YZ)}}} .
We are told that the total surface area is {{{160}}} square feet, so we know that
{{{2(XY+XZ+YZ)=160}}}<--->{{{XY+XZ+YZ=160/2}}}<--->{{{XY+XZ+YZ=80}}} .
We can give values to {{{X}}} and {{{Y}}} and solve for {{{Z}}} .
Then we can calculate the volume of that box (in cubic feet) as
{{{Volume=X*Y*Z}}} , by multiplying together {{{X}}} , {{{Y}}} , and {{{Z}}} .
 
Let's say that {{{system(X=1,"and",Y=2)}}} .
Then, substituting those values, we get
{{{1*2+1*Z+2*Z=80}}}-->{{{2+Z+2Z=80}}}-->{{{2+3Z=80}}}-->{{{3Z=80-2}}}-->{{{3Z=78}}}-->{{{Z=78/3}}}-->{{{Z=26}}} .
{{{Volume=1*2*26=52}}} .
 
Let's say that {{{system(X=1,"and",Y=8)}}} .
Then, substituting those values, we get
{{{1*8+1*Z+8*Z=80}}}-->{{{8+9Z=80}}}-->{{{9Z=80-8}}}-->{{{9Z=72}}}-->{{{Z=72/8}}}-->{{{Z=8}}} .
{{{Volume=1*8*8=64}}} .
 
Let's say that {{{system(X=2,"and",Y=2)}}} .
Then, substituting those values, we get
{{{2*2+2*Z+2*Z=80}}}-->{{{4+4Z=80}}}-->{{{4Z=80-4}}}-->{{{4Z=76}}}-->{{{Z=76/4}}}-->{{{Z=19}}} .
{{{Volume=2*2*19=76}}} .
 
If we start with {{{system(X=2,"and",Y=4)}}} , we find
{{{system(Z=12,"and",Volume=96)}}} .
 
If we start with {{{system(X=2,"and",Y=5)}}} , we find
{{{system(Z=10,"and",Volume=100)}}} .
 
If we start with {{{system(X=4,"and",Y=4)}}} , we find
{{{system(Z=8,"and",Volume=128)}}} .
 
Other choices yield measurements that are not whole numbers.
For example, if we start with {{{system(X=5,"and",Y=5)}}} , we find
{{{system(Z=5.5,"and",Volume=137.5)}}} ,
and if we wanted a cube, with {{{X=Y=Z}}}, we would have
{{{X*Y+X*Z+Y*Z=80}}}--->{{{X*X+X*X+X*X=80}}}--->{{{X^2+X^2+X^2=80}}}--->{{{3X^2=80}}}--->{{{X^2=80/3}}}--->{{{Z=Y=X=sqrt(80/3)=4sqrt(5/3)=4sqrt(15)/3=about5.164}}}
and {{{Volume=X*Y*Z=X*X*X=X^2*X=(80/3)*(4sqrt(15)/3)=320sqrt(15)/9=about137.706}}}
 
NOTE:
The way to answer the question depends on the math level of the class where this question was asked.
Was it asked in elementary school? In a college advanced calculus class? somewhere in between.
I do not believe this is a question from a multivariate calculus class, but intuition would tell us that the rectangular prism with greatest volume will be a cube.