## zaphod Group Title 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 :) one year ago one year ago

1. zaphod Group Title

@satellite73 @Callisto @Omniscience @amistre64

2. amistre64 Group Title

is this roll call?

3. satellite73 Group Title

here!

4. zaphod Group Title

yes?

5. amistre64 Group Title

|dw:1349789270142:dw| $b*b*h=4000$

6. amistre64 Group Title

and theres some calculus involved to find minimum stuff eh

7. amistre64 Group Title

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

8. zaphod Group Title

what does D mean?

9. amistre64 Group Title

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

10. amistre64 Group Title

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

11. Zekarias Group Title

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.

12. zaphod Group Title

i dont understand ><

13. Zekarias Group Title

|dw:1349789802165:dw|

14. zaphod Group Title

now i get it

15. Zekarias Group Title

what did u get?

16. amistre64 Group Title

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

17. Zekarias Group Title

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?

18. zaphod Group Title

yes dA/db = 0 $1600b^{-1} + b^{2}$ $0 = -1600b^{-2} + 2b$ solve for b

19. Zekarias Group Title

That is what I am saying. Thanks

20. zaphod Group Title

b = 20 thanks everyone:)

21. amistre64 Group Title

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 $A'=4\frac{4000}{b^2}-2\frac{b^2\frac{-8000^2}{b^6}}{\frac{4000}{b^2}}-\frac{b^2\frac{-8000}{b^3}}{\frac{4000}{b^2}}$ $A'=\frac{16000}{b^2}-\frac{32000}{b^2}+2b$ $A'=2b-\frac{16000}{b^2}$ $A'=\frac{2b^3-16000}{b^2}=0$ $b=2(1000)^{1/3}=20$

22. amistre64 Group Title

im sure that was not the simplest route to take lol

23. zaphod Group Title

hehe anyways thanks:)