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yashar806

  • 2 years ago

I need help, please

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  1. yashar806
    • 2 years ago
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    I have attached the question

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  2. satellite73
    • 2 years ago
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    we can do this

  3. yashar806
    • 2 years ago
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    yes, could you explain it ?

  4. satellite73
    • 2 years ago
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    ok first off z is 4, so lets forget z and replace it by 4

  5. satellite73
    • 2 years ago
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    the volume of the first tube, the one with side \(x\) is \(4\times (\frac{1}{4}x)^2\)

  6. yashar806
    • 2 years ago
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    why is it 1/4 ?

  7. satellite73
    • 2 years ago
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    because your entire length is \(x\) and you are folding it in to a square so the side of the square is \(\frac{1}{4}x\)

  8. yashar806
    • 2 years ago
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    ok

  9. satellite73
    • 2 years ago
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    and therefore the area of the square is \((\frac{1}{4}x)^2=\frac{x^2}{16}\) and since \(z=4\) the volume is \(4\times \frac{x^2}{16}=\frac{x^2}{4}\)

  10. yashar806
    • 2 years ago
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    ok?

  11. satellite73
    • 2 years ago
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    the volume of the second one is easy, since \(z=4\) it is just \(y\)

  12. yashar806
    • 2 years ago
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    ok

  13. yashar806
    • 2 years ago
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    then we add them togther?

  14. satellite73
    • 2 years ago
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    now we use the fact that \(x+y=8\) and so \(y=8-x\)

  15. satellite73
    • 2 years ago
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    now we add them together, because now we have an expression for the volume in one variable it will be \[V(x)=\frac{x^2}{4}+8-x\]

  16. satellite73
    • 2 years ago
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    now if this is a calculus question you can take the derivative, set it equal to zero and solve if not, you can say the vertex is at \(-\frac{b}{2a}\) with \(a=\frac{1}{4}\) and \(b=-1\)

  17. yashar806
    • 2 years ago
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    got it

  18. satellite73
    • 2 years ago
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    ok good this has a min at the vertex, and i guess a max at the end point of the interval, so be careful about what that is

  19. satellite73
    • 2 years ago
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    i guess it is \([0,8]\) since \(x\) must be in there somewhere

  20. yashar806
    • 2 years ago
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    ok, thanks alot `

  21. yashar806
    • 2 years ago
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    Hi just quick question, why we need interval there? how did you know that there should be an inteval?

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