Question

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.)

Answer 1

Answer:

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

                                      =($105 x-52x^2+4x^3$)

Let

V=$105 x-52x^2+4x^3$

Differentiating with respect to x

$V'=105-104x+12x^2$

Again differentiating with respect to x

$V''=-104+24x$

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

$105-104x+12x^2=0$

Applying the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , here a=12, b= -104 and c =105

$\Rightarrow x=\frac{-(-104)\pm\sqrt{(-104)^2-4.12.105}}{2.12}$

      $=\frac{104\pm 76}{24}$

      =7.5, 1.17

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

∴x=1.17 in.

$V''_{x=1.17}=-104+(24\times 1.17)$

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.