0.66 inches of material is needed to be cut off to make the volume maximum.
maximum and minimum points test
When the second derivative of a function is negative, the function has a maximum point and if the second derivative is positive, the function has a minimum point.
Analysis:
After cut and folded, length = 8-2x
Width = 3-2x
Thickness = x.
Volume of the folded shape = (8-2x)(3-2x)(x)
After expansion, V = 4[tex]x^{3}[/tex]-[tex]22x^{2}[/tex] +24x
for turning point of the function, dv/dx = 0
dv/dx = 12[tex]x^{2}[/tex] -44x + 24
lowest term = 3[tex]x^{2}[/tex] - 11x + 6
3[tex]x^{2}[/tex] - 11x + 6 = 0
3[tex]x^{2}[/tex] - 9x -2x +6 = 0
3x(x-3) -2(x-3) = 0
(3x-2)(x-3) = 0
x = 2/3 or x = 3
To test for maximum point, we differentiate dv/dx again
we have 6x - 11
for x = 3, 6(3) - 11 = 18 - 11 = 7 which is positive x= 3 is a minimum
for x = 2/3 6(2/3) - 11 = 4 - 11 = -7, x = 2/3 is a maximum.
Therefore for maximum volume, the length to be cut out is 2/3 which is 0.66 inches.
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