College Algebra
posted by Valerie Pacetti .
On a rectangular piece of cardboard with perimeter 11 inches, three parallel and equally spaced creases are made. The cardboard is then folded along the creases to make a rectangular box with open ends. Letting x represent the distance (in inches) between the creases, use a graphing calculator to find the value of that maximizes the volume enclosed by this box. Then give the maximum volume. Round your responses to two decimal places.
values of x that maximizes volume = in
Maximum volume= in^3

I would start with a drawing as I do for most problems. Draw a reactangle 4x one side and b the other.
The perimeter is then 4x+b+4x+b=11
8x+2b=11
if this is folded to a tube then the volume of the tube is bx^2, i.e. a tube with cross sectional area x^2 and length b.
so V=bx^2
rearrangen and substitute for b into the equation above gives
8x+2V/(x^2) = 11
or
8x^3+2V=11x^2
or
V=5.5x^24x^3
which you can plot to find max V
I got 1.54 in^3 as the max volume
but check the maths!
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