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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 :)

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  1. zaphod
    • 3 years ago
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    @satellite73 @Callisto @Omniscience @amistre64

  2. amistre64
    • 3 years ago
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    is this roll call?

  3. anonymous
    • 3 years ago
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    here!

  4. zaphod
    • 3 years ago
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    yes?

  5. amistre64
    • 3 years ago
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    |dw:1349789270142:dw| \[b*b*h=4000\]

  6. amistre64
    • 3 years ago
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    and theres some calculus involved to find minimum stuff eh

  7. amistre64
    • 3 years ago
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    \[b^2h=4000\] \[D[b^2h=4000]\] \[D[b^2h]=D[4000]\] \[D[b^2]h+b^2D[h]=D[4000]\] \[2bb'h+b^2h'=0\] its been awhile since i tried figuring this one out, but hheres my first idea :)

  8. zaphod
    • 3 years ago
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    what does D mean?

  9. amistre64
    • 3 years ago
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    D is just notation for "take the derivative of" with respect to some arbitrary independant variable; at the moment

  10. amistre64
    • 3 years ago
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    another equation we need to consider it "amount of material"

  11. Zekarias
    • 3 years ago
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    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
    • 3 years ago
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    i dont understand ><

  13. Zekarias
    • 3 years ago
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    |dw:1349789802165:dw|

  14. zaphod
    • 3 years ago
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    now i get it

  15. Zekarias
    • 3 years ago
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    what did u get?

  16. amistre64
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    yes dA/db = 0 \[1600b^{-1} + b^{2}\] \[0 = -1600b^{-2} + 2b\] solve for b

  19. Zekarias
    • 3 years ago
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    That is what I am saying. Thanks

  20. zaphod
    • 3 years ago
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    b = 20 thanks everyone:)

  21. amistre64
    • 3 years ago
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    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
    • 3 years ago
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    im sure that was not the simplest route to take lol

  23. zaphod
    • 3 years ago
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    hehe anyways thanks:)

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