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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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marked as duplicate by Ross Millikan, LVK, tomasz, Chris Eagle, J. M. Aug 19 '12 at 3:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
"...since it's volume a triple integral is involved..." - for a simple box, such complicated machinery isn't needed. –  J. M. Aug 15 '12 at 12:18
    
the answer is $a^3$ it may help you formulate why.. –  picakhu Aug 15 '12 at 12:20
    
This was asked before with (before the edits) very similar wording. –  André Nicolas Aug 15 '12 at 12:36

2 Answers 2

If the dimensions of the box are $x,y,z$, then the surface are is $2(xy + yz + zx)$ while the volume is $xyz$. The given constraint is therefore $2(xy + yz + zx) = 6a^2 = \text{constant}$. Now you can go about optimizing this in whatever you like, such as with the arithmetic-geometric mean inequality:

$$ x^2 y^2 z^2 = (xy)(yz)(zx) \leq \left(\frac{xy + yz + zx}{3}\right)^3 = a^6, $$

so the volume is at most $a^3$, with equality iff $x=y=z$ (i.e., a cube).

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If the only variable in the formula for the rectangular solid is a it sounds like you have a cube. If the sum of the area for the six sides is 6 a$^2$ then each side has area a$^2$. The volume would be a$^3$. But the constraint does not limit the volume. So a can be as large as you like and there is no finite maximum.

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