# word prob

• March 9th 2009, 09:42 AM
william
word prob
a rectangular piece of cardboard that has dimensions 50 cm by 30cm will be formed into the base of a cardboard box. the corners of the piece of cardboard in the shaape of a square will be removed. if the desired volume of the cardboard box is to be 1056cm^3 what will be the dimensions of the cardboard box?
• March 9th 2009, 10:53 AM
u2_wa
Quote:

Originally Posted by william
a rectangular piece of cardboard that has dimensions 50 cm by 30cm will be formed into the base of a cardboard box. the corners of the piece of cardboard in the shaape of a square will be removed. if the desired volume of the cardboard box is to be 1056cm^3 what will be the dimensions of the cardboard box?

Length $= 50-2x$
Width $= 30-2x$
Height $= x$

Volume of box $= x(30-2x)(50-2x)$
$1056 = x(30-2x)(50-2x)$

Find the value of x.
Hope this helps!
• March 9th 2009, 10:55 AM
HallsofIvy
Quote:

Originally Posted by william
a rectangular piece of cardboard that has dimensions 50 cm by 30cm will be formed into the base of a cardboard box. the corners of the piece of cardboard in the shaape of a square will be removed. if the desired volume of the cardboard box is to be 1056cm^3 what will be the dimensions of the cardboard box?

The problem as you stated it makes no sense. If you formed the cardboard into the base that says nothing about the height. What I think you mean is that squares are cut out of the corners and then sides are folded up so that the cardboard forms the bottom and sides of a box.

Assuming that, let x be the length, in cm., of a side of the square cut out. The height of the box will be x. Since one side of the cardboard was 50 cm and x cm was cut from each end, the length of the base will be 50- 2x while the width will be 30- 2x. So the volume of the box will be the product of those: x(50-2x)(30-2x)= 1056. Can you solve that equation.