Question

What is the maximum volume of an open rectangular box (with no top face) if its surface area is 1 square foot?

Answer 1

Since there is a constraint im pretty sure you have to use Lagrange Multipliers

Answer 2

V(l,w,h) = lwh

Vl(l,w,h) = hw
Vh(l,w,h) = lw
Vw(l,w,h) = lh

Answer 3

l = length, w = width, h = height

Answer 4

so the constraint would be S(l,w,h) = 2(lh + wh) + wl since the base is only counted once

Answer 5

S being the surface area and V the volume

Answer 6

Sl(l,w,h) = 2h + w
Sh(l,w,h) = 2l + 2w
S(l,w,h) = 2h + l

Answer 7

so the equations would look like this
hw = (lamda)(2h + w)
lw = (landa)(2l + 2w)
lh = (lamda(2h + l)
2(lh + hw) + wl = 1

Answer 8

i dont know how to quite go from here to find l, w, and h

Answer 9

Any ideas?

Answer 10

so one panel is 1 m^2

Answer 11

panel?

Answer 12

it's a box that is missing the top face so it's open

Answer 13

ok

Answer 14

i cubic meter

Answer 15

one

Answer 16

i think multiply eq (1) with l and (2) with h to get a relation... multiply eq(1) with l and (3) with w to get a relation... and use those in equation (4)

See if that helps

Answer 17

when you say multiply do you mean like solve eq 1 for w and plug into eq 2?

Answer 18

lwh = (lambda)*l*(2h + w)
lwh = (lambda)*h*(2l + 2w)

solve these two :)

Answer 19

hmmm ok let me try that, i hadn't thought of doing it that way

Answer 20

sure...

Answer 21

ok got l = 2h from that

Answer 22

so basically i just do the same with 2 and 3 right?

Answer 23

yup

Answer 24

ok thank you and is there a general way to know when to do a trick like that?

Answer 25

or is it just a matter of doing a bunch of those types of problems and figuring it out?

Answer 26

i think if you practice a lot you would figure it out

Answer 27

ok thank you sir