Question 531437
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Once the box is folded, the length will be *[tex \Large 13\ -\ 2x], the width will be *[tex \Large 8\ -\ 2x], and the height will be *[tex \Large x].  Since the volume of a rectangular prism is given by length times width times height, we create the volume function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(x)\ =\ (13\ -\ 2x)(8\ -\ 2x)(x)\ =\ 4x^3\ -\ 42x^2\ +\ 104x]


The graph of which is NOT a parabola, although the portion of the graph of interest, that is where *[tex \Large x] is within an interval that allows for an actual physical box to be created, sorta-kinda <i>looks</i> like a parabola.


In order to actually create a box, *[tex \Large x] must be in the interval *[tex \Large \left(0,\,4)].  If *[tex \Large x] were negative you would be dealing with the absurdity of cutting a negative amount from the corners.  If *[tex \Large x\ =\ 0], then you aren't cutting anything away from the corners and all you have is a flat 13 by 8 piece of paper with zero volume.  If *[tex \Large x\ =\ 4] then *[tex \Large 8\ -\ 2x\ =\ 8\ -\ 2(4)\ =\ 0] and again, you have zero volume.


Somewhere in between is a maximum volume, and we know this because of the Mean Value Theorem.


If a function is continuous and twice differentiable over an interval and if the first derivitive of the function is equal to zero at some point in the interval, then there is a local extremum at that point.  If the second derivitive is negative at that point, then the extremum is a maximum.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dV}{dx}\ =\ 12x^2\ -\ 84x\ +\ 104]


Set the derivitive equal to zero:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12x^2\ -\ 84x\ +\ 104\ =\ 0]


Solve this quadratic and select the value that is within the previously discussed interval.  This is the *[tex \Large x] value that gives the extreme volume.  Evaluate the original volume function for this value to get the actual extreme volume.


Take the second derivitive:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2V}{dx^2}\ =\ 24x\ -\ 84]


and evaluate it at the value of *[tex \Large x] that gives the extreme volume.  The second derivitive must be negative if the extreme point is a maximum.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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