Question 1128295
It is good to figure out by ourselves if we have the correct solution.
First, think if the answer seems possible.
You should not get a negative value for {{{x}}} ,
because it is the length of the side of a box,
so {{{x=-5.07}}} is obviously a wrong result.
Maybe you could have found {{{x=5.07}}} , but not {{{x=-5.07}}} .
  
From {{{A=4xh +x^2}}} and {{{h=32/x^2)}}}
you do get
{{{A=4x(32/x^2)+x^2,h=32/x^2}}}
{{{A(x)= 128/x + x^2}}} , not  {{{A(x)= 4(32/ x^2) + x^2}}}
 
How do we check if a formula like that is right?
If for each {{{x}}} you calculate {{{h}}},
and then use the two formulas,
A=4xh +x^2 and A(x)= 128/x + x^2
do you get the same results?
I did:
{{{matrix(9,4,
x,h,A,A(x),
1,32,129,129,
2,8,68,68,
3,3&5/9,51&2/3,51&2/3,
4,2,48,48,
5,1.28,50.6,50.6,
6,8/9,57&1/3,57&1/3,
7,32/49,67&2/7,67&2/7,
8,0.5,80,80)}}}
 
If I were to graph A(x),
I would see that the minimum is at {{{x=4}}} with {{{h=2}}} .
I suppose you are expected to interpret "use a calculator to answer the question"
to mean use a certain graphing calculator to graph A(x).
You might have to use your expensive graphing calculator for the SAT.
I guess that is why they teach you to use it.
I have never been asked to use a graphing calculator at work,
but most of us spend a lot of time in front of computers with Microsoft Excel on them,
and we use it a lot for graphing.
It would be nice if schools taught you to use Excel.