1 Answer | Add Yours
The sheet of metal we have has sides equal to a. Let the sides of the squares that are cut from all the corners be x.
Now we have a length equal to a – 2x, to fold on all the sides and create the box.
The volume of the box is (a – 2x)^2*x = (a^2 + 4x^2 – 4ax)*x
V = a^2x + 4x^3 – 4ax^2
To maximize V, we find its derivative.
V’ = a^2 + 12x^2 – 8ax
This is equated to 0
=> 12x^2 – 8ax + a^2= 0
=> 12x^2 – 6ax -2ax + a^2 = 0
=> 6x (2x – a) – a (2x – a) = 0
=> (6x – a) (2x – a) = 0
=> x = a/6 and x = a/2
Now, we have two values of x and we have to determine which value provides the maximum volume.
V’’ = 24x – 8a,
at x = a/2, V’’ = 12a – 8a = 4a.
As the second derivative is positive at x = a/2, we have a minimum value here.
The volume of the box is maximum at x = a/6 and the volume is equal to a^2*a/6 + 4(a/6)^3 – 4a*(a/6)^2
=> a^3 / 6 + a^3(4/6^3) – (4/36)*a^3
The maximum volume of the box is 2a/27.
We’ve answered 319,817 questions. We can answer yours, too.Ask a question