Packaging By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made. If the cardboard is 21 in. long and 5 in. wide, find the dimensions (in inches) of the box that will yield the maximum volume. (Round your answers to two decimal places if necessary.)

The volume of a box is the amount of space in the box.

The dimensions that yield the maximum volume are: 18.60 by 2.60 by 1.20 inches

The dimension of the cardboard is:

Assume the cut-out is x.

So, the dimension of the box becomes

The volume of the box is:

Open brackets

Expand

Differentiate

Set to 0

Rewrite as:

Using a calculator, the value of x is:

When x = 7.50;

The width cannot be negative.

When x = 1.20;

So, the dimensions that yield the maximum volume are: 18.60 by 2.60 by 1.20 inches

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The dimension of the box is 18.66 in by 2.66 in by 1.17 in.

Step-by-step explanation:

Given that, The cardboard is 21 in. long and 5 in. wide.

Assume,x be the length of the each sides of the square .

Then the length of the box is = (21-2x) in.

The breadth of the box is =(5-2x) in.

The height of the box is  = length of side of the square

                                         = x in.

The volume of the box = length × wide × height

                                      =(21-2x)(5-2x)x

                                      =(105-52x+4x²)x

                                      =()

Let

V=

Differentiating with respect to x

Again differentiating with respect to x

To find the dimension of the box, we set V'=0

Applying the quadratic formula , here a=12, b= -104 and c =105

      =7.5, 1.17

For x= 7.5 , the wide of the box will be negative.

∴x=1.17 in.

Since at x= 1.17, V''<0 , therefore at x=1.17 in. the volume of the given box will be maximum.

The length of the box is = [21-(1.17×2)]=18.66 in.

The wide of the box is = [5-(1.17×2)]=2.66 in.

The height of the box is = 1.17 in.

The dimension of the box is 18.66 in by 2.66 in by 1.17 in.