Question 287659
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Assume you have a square of paper/cardboard/whatever of dimension *[tex \Large a] by *[tex \Large a].  If you cut squares of dimension *[tex \Large x] by *[tex \Large x] from each of the four corners, then the square base of your box will have dimension *[tex \Large a\ -\ 2x], and the height of the box will be simply *[tex \Large x].


Therefore the volume of the box will be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(x)\ =\ x(a\ -\ 2x)^2 = 4x^3\ -\ 4ax^2\ +\ a^2x]


Take the first derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dV}{dx}\ =\ 12x^2\ -\ 8ax\ +\ a^2]


Which we need to set equal to zero and solve to find a local extremum:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12x^2\ -\ 8ax\ +\ a^2\ =\ 0]


Since for this problem we know that *[tex \Large a] is a multiple of 12 we can simplify further by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ \frac{8a}{12}x\ +\ \frac{a^2}{12}\ =\ 0]


Now substitute the given 84" for *[tex \Large a]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 56x\ +\ 588\ =\ 0]


Which factors to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 42)(x\ -\ 14)\ =\ 0]


Hence the roots and potential extrema are at *[tex \Large x\ =\ 42] and *[tex \Large x\ =\ 14].  Quite obviously we can exclude *[tex \Large x\ =\ 42] in looking for maximum volume because if you cut 42" squares from the corners of an 84" by 84" sheet of paper, you won't have anything left for the bottom of the box.


Take the second derivative


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2V}{dx^2}\ =\ 24x\ -\ 8a\ =\ 24x\ -\ 672]


Now evaluate the second derivative at the extreme point:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V''(14)\ =\ 24(14)\ -\ 672\ <\ 0]


A negative second derivative at a local extremum indicates a local maximum.


Therefore the cutout for the maximum volume is 14".


The volume of the box when *[tex \Large x\ =\ 14] is given by the volume function evaluated at 14:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(14)\ = 4(14)^3\ -\ 4(84)(14)^2\ +\ (84)^2(14)] 


You can do your own arithmetic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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