Maximisation and min question (1 Viewer)

[enigma_1]

A cardboard box is to have square ends and a volume of 768cm^3. It is to be sealed
using two pieces of tape, one passing entirely around the length and width of the box and
the other passing entirely around the height and width of the box. Find the dimensions
of the box so that the least amount of tape is used.

Answer: 12 x 8 x 8

Worked solution please thanks

 

[bottleofyarn]

It's a pretty standard question, just tricky wording and nothing too tricky really.
Here is a diagram oriented nicely (yes MS paint) to clear things up and each piece of tape goes over all four sides. Simultaneous, sub in one and differentiate to find the minimum and that should be it.

 

[NerdSlayer]

Begin with drawing a box with dimensions of x*x*h

Therefore the Volume of the box is V=768cm^2

Making 'h' the subject, h=768/x^2

Now to find the amount of tape used, let it be 'T'

T=8x+4x+4h (8x is the sum of the width and length and 4x+4h is the sum of the height and width)

Therefore, T=12x+4h
subbing the earlier equation in:

T=12x+(4*768)/x^2

Now differentiate the equation and it should be,

dT/dx=12-6144x^-3

when dT/dx=0

12-6144x^-3=0

12x^3=1664

x^3=512

Therefore x=8

Substitute that back in to find the rest of the dimensions of the box.