dimensions of base : (6-2x) by (10-2x), where 0 < x < 3
volume = x(6-2x)(10-2x)
= 60x - 32x^2 + 4x^3
d(Volume)/dx = 60 - 64x + 12x^2
= 0 for a maximum volume
Using the quadratic formula I got
x = 4.12 or x = 1.214
but according to the above domain,
x = 1.214
A box with no top is to be built by taking a 6'' - by - 10'' sheet of cardboard and cutting x - in. squares out of each corner and folding up the sides. Find the value of x that maximizes the volume of the box.
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