A open box to b made by cutting squares from the corners of a piece of cardboard 40 by 40 inches n folding up the sides. What dimensions maximize...

An open box has to be made by cutting squares from the
sides of a piece of cardboard with dimensions 40 inches by 40 inches and folding up the
sides. Let the length of the side of the squares cut be
s.

After the squares are cut and the sides folded up, the
base of the box has dimensions 40 - 2s by 40 - 2s. The height of the box is
s.

This gives the volume of the box as (40 -
2s)^2*s

=> (1600 + 4s^2 -
160s)*s

=> 4s^3 - 160s^2 +
1600s

This has to be
maximised.

Solve for s by equating the first derivative of
4s^3 - 160s^2 + 1600s with respect to s to 0

12s^2 - 320s +
1600 = 0

=> 12s^2 - 240s - 80s + 1600 =
0

=> 12s(s - 20) - 80(s - 20) =
0

=> (s - 20)(12s - 80) =
0

=> s = 20 and s =
20/3

The second derivative of 4s^3 - 160s^2 + 1600s is 24s
- 320. This is equal to 160 for s = 20 and -160 for s = 20/3. The maximum value lies at
s = 20/3

The sides of the squares that are cut should be
20/3 inches.

The dimensions of the box are
(80/3) by (80/3) by (20/3)