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Maximizing volume of a rectangular solid, given surface area

Maximize the volume of a rectangular solid, given that the sum of the areas of the six faces is 6a^2 for a constant 'a'.

So I know if its a rectangle, I have 6 sides... 2 sides are 'a x a' and 4 sides should be '2a x a'... but that gives me 2a^2 + 8a^2 = 10a^2 which is more than the given 6a^2... im not sure what im doing wrong or how to go about this question? Im pretty sure since its volume a triple integral is involved...

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    "...since it's volume a triple integral is involved..." - for a simple box, such complicated machinery isn't needed.2012-08-15
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    the answer is $a^3$ it may help you formulate why..2012-08-15
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    This was [asked before](http://math.stackexchange.com/questions/179689/maximizing-volume-of-a-rectangular-solid-given-surface-area) with (before the edits) very similar wording.2012-08-15

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