Question

SOSS ANSWER ASAP!! 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 16 in. long and 6 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

The dimensions yield maximum volume are 13.66 in x 3.66 in x 1.17 in.

How the dimensions of the cardboard is calculated.

Dimensions of box

16 in long, 6 in wide, H in high.

Let x represents size of square x

4 squares were cut out from the vertices of the cardboard. Two from each sides.

New length = (16 -x -x) = 16 - 2x.

New width = (6 - x - x) = 6 - 2x

New height = x

So V=(16-2x)(6-2x)x

= (96 - 12x -32x + 4x²)x

= 4x³ - 48x² + 96x

Maximum (critical points) gotten by differentiating V

So V'= 12x² - 96x + 96

At critical point V' = 0

x ²- 8x + 8 =0

x = [-(-8)±√(-8² - 4*1*8)]/(2*1)

= 8 +-√(64 - 32)/2

= 8 +- √32/2

= (8+-5.66)/2

= (8 + 5.66)/2 or (8 - 5.66)/2

= 6.83 or 1.17

Since the width is 6 in we cannot use 6.83,

So using 1.17

The dimensions of rectangle cardboard

Length =(16-2x)=(16-2(1.17))= 13.66 in

Width = (6-2x)=(6-2(0.81))= 3.66 in

Height = 1.17 in

Therefore the dimensions yield maximum volume are 13.66 in x 3.66 in x 1.17 in.