Question

Andre wants to make an open-top box by cutting out corners of a 22 inch by 28 inch piece of poster board and then folding up the sides. The volume  in cubic inches of the open-top box is a function of the side length  in inches of the square cutouts.
Write an expression for .

What is the volume of the box when ?

What is a reasonable domain for  in this context?​

Answer 1

Answer:

Step-by-step explanation:

THE box volume is V = (length)(width)(height).  Representing the cutout side length by x, we get the formula V = (28 - 2x)(22 - 2x)(x).  

Explanation:  We are to cut out x by x squares, one at each corner of the 28 by 22 inch board; thus, the bottom length is 28 - 2x, the bottom width 22 - 2x and the height just x.

You may either leave this expression as is, or perform the multiplication and simplify the result.

If x = 2 (which I have chosen arbitrarily because you have not specified x), then the box volume is

V = (28 - 2*2)(22 - 2*2)(2), or

V = (24)(18)(2) cubic inches.

Since x is a measurement of length, it cannot be less than zero.  Likewise, since the width of the bottom of the box cannot be less than zero, we have the following inequality for x:  22 - 2x > 0, or

11 - x > 0, or x < 11.

Suppose that x = 10, as a check.  Then V = (28 - 20)(22 - 20)(10).  Is this greater than zero?  YES

Therefore x < 11 is a readonable domain in this context.