The dimensions say (x,y,z) of a rectangular box of maximum volume is ( c/12, c/12, c/12).
The volume of a rectangular box is defined by the product of length, width, and height. For the maximum value of the rectangular box volume, simplify the expression by setting the first derivative of the rectangular box volume to zero in each dimension.
Let x, y, and z be the dimensions of the rectangular box. Since, the sum of the lengths of its 12 edges is a constant c. This implies
4x + 4y + 4z = c
=> x + y + z = c/4 --(1)
The volume of rectangular box = xyz
Consider xyz + lamda(x + y + z - c/4) where λ is Lagrange multiplier.
let f( x,y,z) = xyz + λ(x + y + z - c/4) --(*)
compute the partial derivatives of f( x,y,z) as
fx = yz + λ(1) = yz + λ
fy = xz + λ(1) = xz + λ
fz = xy + λ(1) = xy + λ
For critical points, set the partial derivatives equals to zero.
fx = yz + λ = 0
fy = xz + λ = 0
fz = xy + λ =0
=> xz = yz = xy = - λ
So, x = y = z
then, from equation (1) we get,
x + x + x = c/4
=> 3x = c/4 => x = c/12
thus, x = c/12, y = c/12, z = c/12
Thus, the dimensions of the rectangular box are x = y = z = c/12 or ( c/12, c/12, c/12) and
Maximum volume of rectangular box = xyz
= c³/1728
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