Question 1040831
Let {{{ W }}} = the width
{{{ 2W }}} = the height
Let {{{ L }}} = the length
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given:
{{{ W*2W*L = 18 }}} ft3
{{{ 2L*W^2 = 18 }}}
{{{ L*W^2 = 9 }}}
{{{ L = 9 / (W^2) }}}
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Let {{{ S }}} = the inside surface area
{{{ S = 2*W*2W + 2*L*2W }}}
{{{ S = 4W^2 + L*4W }}}
{{{ S = 4W^2 + ( 9/(W^2))*4W }}}
{{{ S = 4W^2 + 36/W }}}
Is this in a calculus course? If so take the 
derivative of {{{ S }}} with respect to {{{ W }}}
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{{{ s[1] = 8W - 72/(W^2) }}}
{{{ 8W - 72/(W^2) = 0 }}}
{{{ 8W = 72/(W^2) }}}
{{{ W^3 = 9 }}}
{{{ W = root( 3, 9 ) }}}
{{{ W = 2.08008 }}}
height = {{{ 2W = 4.16016 }}}
length =  {{{ L = 9 / (W^2) }}}
length = {{{ 9 / 4.32673 }}}
length = {{{ 2.08008 }}}
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check:
{{{ 2.08008*2.08008*4.16016 = 18 }}}
{{{ 17.9999 = 18 }}}
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Here's a plot of the surface area function,
{{{ S }}} and {{{ W }}}
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{{{ graph( 400, 400, -10, 10, -100, 100, 4x^2 + 36/x ) }}}
Hope I got it -definitely get another opinion on this!!!