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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    I think its 4000 but I'm not sure

  2. misty1212
    • one year ago
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    HI!!

  3. misty1212
    • one year ago
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    calc question?

  4. anonymous
    • one year ago
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    Hello and Yes

  5. misty1212
    • one year ago
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    |dw:1443580022426:dw|

  6. misty1212
    • one year ago
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    area is \(x^2+4xh\) with no top

  7. misty1212
    • one year ago
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    set \[x^2+4xh=1200\] and solve for\(h\)

  8. misty1212
    • one year ago
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    i get \[h=\frac{1200-x^2}{x}\]

  9. anonymous
    • one year ago
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    Okay

  10. misty1212
    • one year ago
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    oops i made a mistake

  11. misty1212
    • one year ago
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    \[x^2+4xh=1200\\ h=\frac{1200-x^2}{4x}\]

  12. misty1212
    • one year ago
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    volume is \[V=x^2h\] or now \[V(x)=x^2\left(\frac{1200-x^2}{4x}\right)\]or \[V(x)=\frac{1}{4}x(1200-x^2)\]

  13. misty1212
    • one year ago
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    find the critical points by taking the derivative, setting it equal to zero, etc

  14. anonymous
    • one year ago
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    Okay 1200x^2-x^4/4x^3 =0

  15. misty1212
    • one year ago
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    hmm lets multiply out first

  16. misty1212
    • one year ago
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    \[V(x)=300x-\frac{1}{4}x^3\]

  17. misty1212
    • one year ago
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    the derivative is a lot simpler than that, it is just \[V'(x)=300-\frac{3}{4}x^2\]

  18. anonymous
    • one year ago
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    Okay and then plug my numbers in?

  19. misty1212
    • one year ago
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    set that equal to zero and solve

  20. misty1212
    • one year ago
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    \[300-\frac{3}{4}x^2=0\]

  21. misty1212
    • one year ago
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    don't fret the solution, solve \[\frac{3}{4}x^2=300\] forget the negative solution

  22. misty1212
    • one year ago
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    you get \(x=20\) yet?

  23. anonymous
    • one year ago
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    Yes

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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