Thread: Maximum volume of a box

I was asked this question in an exam, Could you please check my working and see if I was on the right track.

A square sheet of cardboard is 16cm on each side. Squares of side 'x' cm are cut from each corner and the sodes bent up to form an open box. What is the maximum volume of the box?

My working

Volume = L x B x H

= (16-2x)(16-2x)(x)
= x(16-2x)²
= x(256 - 64x + 4x²)
= 4x³ - 64x² + 256x *

to find maximum find v'=0
12x² -128x + 256 = 0
4(3x² -32x + 64) = 0

Need help from here

Originally Posted by Joel

I was asked this question in an exam, Could you please check my working and see if I was on the right track.

A square sheet of cardboard is 16cm on each side. Squares of side 'x' cm are cut from each corner and the sodes bent up to form an open box. What is the maximum volume of the box?

My working

Volume = L x B x H

= (16-2x)(16-2x)(x)
= x(16-2x)²
= x(256 - 64x + 4x²)
= 4x³ - 64x² + 256x *

to find maximum find v'=0
12x² -128x + 256 = 0
4(3x² -32x + 64) = 0

Need help from here

You are correct so far . Now solve for x . Then you will get 2 values of x so you will need to verify which of them would give a max volume .

Use second order : d^2y/dx^2=6x-32

6(8/3)-32 =

6(8)-32 =

Whichever which gives a negative value is the one .

Then substitute this value of x back into the original equation

V=4x³ - 64x² + 256x

You are OK so far I think. The quadratic is probably going to factorise and x=8 will be a root, since the volume of the box is clearly 0 then and will then start increasing again.
In fact (3x^-2-32x+64) = (x-8)(3x-8).
So now you can easily calculate the volume.

Roots are 8 and 8/3

back into the equation 8 gives a volume of 0, so max volum is where x=8/3

thanks heaps