Question 443207
On thinking this over last night, I discovered a big error in this, let me correct it here
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a company wishes to manufacture a box with a volume of 36 cubic feet that is open on top and is twice as long as it is wide.
 find the dimensions of the box produced from the minimum amount of material.
Let x = the width 
Let 2x = the length
Let h = the height 
then
vol = x*2x*h
so we have
2x^2*h = 36
h = {{{36/(2x^2)}}}
h = {{{18/x^2}}}

Surface area: two ends + 1 bottom + 2 sides (no top)
S.A. = 2(x*h) + 1(2x*x) + 2(2x*h)
:
I am truly sorry for this error.  Carl
:
S.A. = 2xh + 2x^2 + 4xh
S.A. = 2x^2 + 6xh
Replace h with {{{18/x^2}}}
S.A = 2x^2 + 6x({{{18/x^2}}})
S.A = 2x^2 + 6({{{18/x}}})
S.A = 2x^2 + ({{{108/x}}})
Graph this equation to find the value of x for minimum material
{{{ graph( 300, 200, -1, 5, -20, 150, 2x^2+(108/x)) }}}
Min surface area when x = 3.0 is the width
then
2(3) = 6 is the length
;
Find the height:
h = {{{18/3^2}}}
h = 2
:
Box dimensions for min surface area: 3 by 6 by 2; much better numbers
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Check the vol of these dimensions: 3*6*2 ~ 36