Posted by **Valerie Pacetti** on Friday, January 21, 2011 at 4:02am.

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

- College Algebra -
**Dr Russ**, Friday, January 21, 2011 at 6:39am
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^2-4x^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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