Question

An open rectangular box is to be made wit a square base, and its capacity is to be 4000cm^3. find the length of the side of the base when the amount of material used to make the box is as small as possible...

PLEASE HELP :)

Answer 1

@satellite73 @Callisto @Omniscience @amistre64

Answer 2

is this roll call?

Answer 3

here!

Answer 4

yes?

Answer 5

\[b*b*h=4000\]

Answer 6

and theres some calculus involved to find minimum stuff eh

Answer 7

\[b^2h=4000\]
\[D[b^2h=4000]\]
\[D[b^2h]=D[4000]\]
\[D[b^2]h+b^2D[h]=D[4000]\]
\[2bb'h+b^2h'=0\]
its been awhile since i tried figuring this one out, but hheres my first idea :)

Answer 8

what does D mean?

Answer 9

D is just notation for "take the derivative of" with respect to some arbitrary independant variable; at the moment

Answer 10

another equation we need to consider it "amount of material"

Answer 11

The amount of the material can be expressed through total surface area.
Thus \[A_{t}=4bh+b^{2}\]But \[h=\frac{ 4000 }{ b^{2} }\] Now insert the expression of h in At, the take derivative.

Answer 12

i dont understand ><

Answer 13

now i get it

Answer 14

what did u get?

Answer 15

using the surface area formula pforvided by Zek
\[A=4bh+b^2\]
\[D[A=4bh+b^2]\]
\[D[A]=D[4bh]+D[b^2]\]
\[A'=4(D[bh])+2bb'\]
\[A'=4(D[b]h+bD[h])+2bb'\]
\[A'=4(b'h+bh')+2bb'\]
\[A'=4b'h+4bh'+2bb'\]
this gives us 2 equations that work together

\[A'=4b'h+4bh'+2bb'\]\[0=2bb'h+b^2h'\]

just trying to recall if i needed to go thru all that ... if not just for the practice

Answer 16

It is not what I am saying. What I am trying too explain is
we have the volume 4000 = hbb and At = 4hb+bb.
Now combining this two equations we get\[A_{t}=\frac{ 1600 }{ b }+b^{2}\]Therfore take the derivative at this stage.
Will u do that?

Answer 17

yes dA/db = 0
\[1600b^{-1} + b^{2}\]
\[0 = -1600b^{-2} + 2b\]

solve for b

Answer 18

That is what I am saying.
Thanks

Answer 19

b = 20 thanks everyone:)

Answer 20

we can prolly assume we can work this by adjusting the base with respect to itself so b'=1
\[A'=4h+4bh'+2b\]\[0=2bh+b^2h'\]
\[2b=\frac{-b^2h'}{h}\]
\[A'=4h-2\frac{b^2h'}{h}h'-\frac{b^2h'}{h}\]using h=4000b^-2, h'=-8000b^-3
\[A'=4\frac{4000}{b^2}-2\frac{b^2\frac{-8000^2}{b^6}}{\frac{4000}{b^2}}-\frac{b^2\frac{-8000}{b^3}}{\frac{4000}{b^2}}\]
\[A'=\frac{16000}{b^2}-\frac{32000}{b^2}+2b\]
\[A'=2b-\frac{16000}{b^2}\]
\[A'=\frac{2b^3-16000}{b^2}=0\]
\[b=2(1000)^{1/3}=20\]

Answer 21

im sure that was not the simplest route to take lol

Answer 22

hehe anyways thanks:)