Question 356869
An open box with a square base is required to have a volume of 10 cubic feet.
 Assume the box is to be made from a square piece of cardboard that is has
 original dimensions (x + 2h)-by-(x + 2h)
 by cutting out 4 h-by-h squares on the corners of the cardboard and folding
 up the sides where h is the height of the open box sides 
Base length x h by h square cut out to form open box
:
The removal of the h by h corners, reduces the dimension by 2h, therefore
The box dimensions will x by x by h, therefore:
Vol = x^2*h
Given the vol as 10 cu/ft:
x^2*h = 10 cu/ft
h = {{{10/x^2}}}
:
A] Express the surface area A(x) of the box as a function of the length of the base x.
Surface area of a 5 sided (open) box:
S.A. = x^2 + 4(x*h)
replace h with {{{10/x^2}}}
S.A. = x^2 + 4(x*{{{10/x^2}}})
Cancel x
S.A. = x^2 + 4({{{10/x}}})
S.A = x^2 + {{{40/x}}}; the surface area as a function of x
:
B] Use a graphing utility to graph A(x) and determine the dimensions of an open box with the smallest surface area possible.
In a TI83 or similar enter y= x^2+{{{40/x}}}, I used a scale: -2,+10; -10,+50
Looks like this:
{{{ graph( 300, 200, -2, 10, -10, 50, x^2+(40/x)) }}} 
Using the min feature on the calc, I got x=2.744, surface area of 22.1 is min
:
Using x=2.744,
h = {{{10/2.744^2}}}
h ~ 1.33 is the height
Find the Surface area from these values
S.A. = 2.744^2 + 4(2.744*1.334)
S.A. = 7.53 + 4(3.65)
S.A. = 7.53 + 14.6
S.A. = 22.13 which agrees with the Calc values
:
Did all this make sense to you?