A square sheet of metal of side a has squares cut on all its corners and the sides bent to form an open box. What is the maximum volume of the box?
- print Print
- list Cite
Expert Answers
calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
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
=> 2a/27
The maximum volume of the box is 2a/27.
Related Questions
- Find the maximum volume of an open box made from 3ft by 8ft rectangular piece of sheet metal by...
- 1 Educator Answer
- What should be the size of the squares cut so that the volume of the open box is maximum?if we...
- 1 Educator Answer
- A 6×6 sq. sheet of metal is made into an open box by cutting out a sq. @ each corner and then...
- 1 Educator Answer
- Open-top boxes are constructed by cutting equal squares from the corners of cardboard sheets that...
- 1 Educator Answer
- This is a math project for class 10. Squares are cut from corners of a square sheet 10 cm and is...
- 1 Educator Answer
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.