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dessyj1

  • one year ago

I need help visualizing this question. (Calc 1)

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  1. anonymous
    • one year ago
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    Type le question

  2. dessyj1
    • one year ago
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    there was supposed to be an attachment

  3. dessyj1
    • one year ago
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    A metal through with equal semicircluar ends and open top is to have a capacity of 128pi cubic feet. Determine its radius r and length h is the trough is to require the least material for its construction.

  4. dessyj1
    • one year ago
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    trough*

  5. triciaal
    • one year ago
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    |dw:1433651448727:dw|

  6. dessyj1
    • one year ago
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    so the ends of the are semi-circle are closed?

  7. triciaal
    • one year ago
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    |dw:1433651580013:dw|

  8. anonymous
    • one year ago
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    Now just write an expression for the volume, Then one for the Surface area. Use the volume formula to turn the surface area formula into a single variable. Then Differentiate the last formula, set it equal to zero and solve. This will give you either h or r (depending on how you get on) Then substitute back into first formula to find the other variable.

  9. triciaal
    • one year ago
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    I see a cylinder split in half then turned on the side

  10. dessyj1
    • one year ago
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    I see a cylinder split in half too.

  11. dessyj1
    • one year ago
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    volume of a cylinder/2= 128pi?

  12. dessyj1
    • one year ago
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    then 2h+4r+2pi(r)=perimeter of the trough?

  13. zepdrix
    • one year ago
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    perimeter? hmm we're looking to minimize `surface area`, ya? :) remember how to find surface area of a cylinder?

  14. dessyj1
    • one year ago
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    the materials would be sum the dimensions of all the sides would it not?

  15. zepdrix
    • one year ago
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    no, that would take care of the .. edges i suppose, not the surfaces though.

  16. zepdrix
    • one year ago
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    |dw:1433648747657:dw|For a cylinder, we have two of these panels, one on top and one on bottom, ya? area of a circle is pi r^2. So to get the surface area of the top and bottom combined, we have 2pi r^2

  17. dessyj1
    • one year ago
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    SA of a cylinder= 2pi(r)h+ 2pi(r)^2

  18. zepdrix
    • one year ago
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    Or you can jump right to your formula :) yah that's fine hehe.

  19. triciaal
    • one year ago
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    |dw:1433652364767:dw|

  20. zepdrix
    • one year ago
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    Our trough will have exactly half of that surface area, so the formula we'll use is:\[\Large\rm A=\pi r^2+\pi r h\]I divided everything by 2 to cut it in half.

  21. dessyj1
    • one year ago
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    alright

  22. dessyj1
    • one year ago
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    the cylinder volume formula would have to be cut in half as well?

  23. zepdrix
    • one year ago
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    You're trying to minimize the area function. So eventually you want to get \(\Large\rm A'\) and set it equal to zero to look for `critical points`. But first, you want A in terms of ONE VARIABLE so it's easier to work with.

  24. zepdrix
    • one year ago
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    Yes, tric put the 1/2 in front of the pi r^2 h to show that :)

  25. zepdrix
    • one year ago
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    the Area formula is what we're trying to minimize, the volume formula is our constraint. It relates r to h. It allows us to make a substitution in our Area formula. Do you see how that will work? :)

  26. dessyj1
    • one year ago
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    yup

  27. dessyj1
    • one year ago
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    would isolate for h in the volume formula then sub it into the SA formula

  28. zepdrix
    • one year ago
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    good good good

  29. dessyj1
    • one year ago
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    then derive it.

  30. dessyj1
    • one year ago
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    \[SA \prime =4Pir-(512\Pi/r ^{2})?\]

  31. dessyj1
    • one year ago
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    r=\[\sqrt[3]{128}\]

  32. zepdrix
    • one year ago
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    Mmm that's what I'm coming up with also! yay

  33. dessyj1
    • one year ago
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    okay, so when i originally tried to figure out the perimeter i was trying to get just the sides?

  34. dessyj1
    • one year ago
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    |dw:1433649816855:dw|

  35. zepdrix
    • one year ago
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    um um um :) ya perimeter gives us the distance across all of those edges i guess.

  36. dessyj1
    • one year ago
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    Okay, thank you so much.

  37. zepdrix
    • one year ago
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    Were you able to find the corresponding height using that radial value? :)

  38. zepdrix
    • one year ago
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    Remember that they want the radius `and` height that minimize the area :D

  39. dessyj1
    • one year ago
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    yes h=10.07 inches

  40. dessyj1
    • one year ago
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    or

  41. dessyj1
    • one year ago
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    \[h= 256\div \sqrt[3]{128}^2\]

  42. zepdrix
    • one year ago
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    cool :) that will simplify a lil bit if you want both your height and radius in exact value,\[\Large\rm h=\frac{256}{(128)^{2/3}}=\frac{2\cdot128}{(128)^{2/3}}=2(128)^{1/3}\]

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