The volume of the box as a function of the cut size (x) is
... v(x) = x(21 -2x)(48 -2x)
... = 4x³ -138x² +1008x
Setting the dirivative to zero lets us find the value of x that maximizes the volume.
... v'(x) = 12x² -276x +1008 = 0
Dividing by 12 gives ...
... x² -23x +84 = 0
and the quadratic formula tells us
... x = (23 -√(23²-4·84))/2 = 11.5 - √48.25 ≈ 4.55378
The volume for this cut dimension is
.. v(4.55378 in) ≈ 2106 in³
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Such a problem is easily solved by a graphing calculator that can tell you the extreme value of the volume function and where it is located.
