## zaphod 3 years ago 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 :)

1. zaphod

@satellite73 @Callisto @Omniscience @amistre64

2. amistre64

is this roll call?

3. satellite73

here!

4. zaphod

yes?

5. amistre64

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

6. amistre64

and theres some calculus involved to find minimum stuff eh

7. amistre64

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

8. zaphod

what does D mean?

9. amistre64

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

10. amistre64

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

11. Zekarias

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

i dont understand ><

13. Zekarias

|dw:1349789802165:dw|

14. zaphod

now i get it

15. Zekarias

what did u get?

16. amistre64

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

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

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

19. Zekarias

That is what I am saying. Thanks

20. zaphod

b = 20 thanks everyone:)

21. amistre64

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

im sure that was not the simplest route to take lol

23. zaphod

hehe anyways thanks:)