Posted by **Carlee** on Sunday, December 16, 2012 at 4:33pm.

you have a rectangular piece of cardboard 40X48 inches and need to create an open box with the maximum volume(ignore flaps). what would the dimensions of the box need to be to maximize the volume?

- math -
**Steve**, Sunday, December 16, 2012 at 5:40pm
v = x(40-2x)(48-2x) = 4x(20-x)(24-x)

v=0 at x=0,20,24, so you know there will be a max for 0 < x < 20

If you haven't had some calculus, this will be hard to find; you may have to do some probing around x=10 and try some values.

If calculus is in your bag of tools, then you know that

dv/dx = 4(x^2 - 88x + 480)

dv/dx=0 when x = 4/3 (11-√31) = 7.24

So, the box is 32.76 x 40.76 x 7.24

## Answer This Question

## Related Questions

- calculus optimization problem - by cutting away identical squares from each ...
- Math - a rectangular piece of cardboard is twice as long as it is wide . from ...
- calc - by cutting away identical squares from each corner of a rectangular piece...
- calculus - By cutting away identical squares from each corner of a rectangular ...
- Math - An open-topped box is constructed from a piece of cardboard with a length...
- math - By cutting away identical squares from each corner of a rectangular ...
- Calculus - By cutting away identical squares from each corner of a rectangular ...
- math - a piece of cardboard is twice as it is wide. It is to be made into a box ...
- Algebra - A rectangular piece of cardboard is 15 inches longer than it is wide. ...
- Math - Arlan needs to create a box from a piece of cardboard. The dimensions of...

More Related Questions