Question

A rectangular box without a lid is to be made from 12 m² of
cardboard. Find the maximum volume of such box using the method
of Lagrange's multiplier.​

Answer 1

Answer:

This problem involves three variables, length l, width w, and height h.

They can be reduced to two using the constraint that the surface area is 12 sq. m.

Out of the three, w and l are symmetrical, so assumption can be made that when w=l the volume is either a maximum or a minimum.

Assume therefore w=l and proceed with the volume calculation:

V=wlh

subject to 2h(w+l)+wl=12

When l=w, this reduces to

2h(2w)+w²=12

from which

h=(12-w²)/4w

Substitute in V:

V=wlh

=w²h

=w(12-w²)/4

=(12w-w³)/4

At maximum (or minimum) volume,

dV/dw=3-3w²/4=0

w²=4

w=2 (metres)

Check maximum or minimum:

d²V/dw²=-6w/4 <0 => maximum.

Therefore the dimensions of the box should be:

w=l=2m h=12/2²=3m.

Answer 2

Answer:

refer the attachment in the top