The area of the sheet metal is 800 m^2. The length is twice as long as the the width. Let the length be L. If the width is W, L = 2W.
800 = 2*W*W
=> W = sqrt(400)
=> W = 20
L = 40
Four squares are cut from the corners of the sheet of 40 x 20 cm and a box is constructed without a lid. Let the squares that are cut have a side of length s.
The volume of the box is : V = L*W*H, where L, W and H are the length, width and height of the the box.
The Length is (40 - 2s)
Width = 20 - 2s
Height = s
Volume = (40 - 2s)(20 - 2s)(s)
=> (40s - 2s^2)(20 - 2s)
=> 800s - 40s^2 - 80s^2 + 4s^3
=> 4s^3 - 120s^2 + 800s
To maximize volume find the derivative and solve for s.
12s^2 - 240s + 800 = 0
=> 3s^2 - 60s + 200 = 0
s1 = 15.77
s2 = 4.22
There are two roots of the equation, one giving a minimum point and the other the maximum point. The second derivative of the volume which is 24s - 240 is used to determine which of the values of s give the maximum. 24s - 240 has a negative value for the value of s that gives a maximum value.
For s = 15.7, 24s - 240 = 138.48
and for s = 4.2 , 24s - 240 = -138.72
This indicates that volume is minimum if s = 15.77 and maximum if s = 4.2.
For s = 4.2, the dimensions of the box are: 31.6, 11.6 and 4.2
The box has a maximum volume when the dimensions are 31.6, 11.6 and 4.2 cm