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October 26, 2014

October 26, 2014

Posted by **daniel** on Sunday, April 17, 2011 at 3:55am.

- calculuc -
**MathMate**, Sunday, April 17, 2011 at 9:40amThis 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.

If the assumption of symmetry cannot be made, Lagrange multipliers can be used to determine the three dimensions.

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