The first step to solving this problem is to create the equation. The volume of any open-top box is the length times width times height. Let the dimensions of the square being cut out of each corner of the rectangular cardboard be x by x. The length of the formed...

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The first step to solving this problem is to create the equation. The volume of any open-top box is the length times width times height. Let the dimensions of the square being cut out of each corner of the rectangular cardboard be x by x. The length of the formed box will be (32-2x). The width will be (28-2x). The height will be x. The equation to solve becomes:

V=(32-2x)(28-2x)x

1920=(32-2x)(28-2x)x, or

(32-2x)(28-2x)x-1920=0

The equation can be solved by finding the zeros, x-intercepts, on a graph.

The graph shows that x=4, 6, or 20.

If x=4, then the length is 32-2(4)=24 cm, the width is 28-2(4)=20 cm and the height is 4 cm.

If x=6, then the length is 32-2(6)=20 cm, the width is 28-2(6)=16, cm and the height is 6 cm.

The value for x=20 is not possible since it would make the box edges negative.