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anonymous

  • 5 years ago

Find the area of the largest rectangle with a base on the positive x axis, its right side on the line x=9 and which is inscribed under the curve f(x) = root(x)

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  1. anonymous
    • 5 years ago
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    I don't get it... can anybody help please?

  2. mattfeury
    • 5 years ago
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    you have to integrate on root(x) from x=0 to x=9.

  3. mattfeury
    • 5 years ago
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    actually that doesn't give you a rectangle. hmmm....

  4. anonymous
    • 5 years ago
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    elaborate please

  5. mattfeury
    • 5 years ago
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    this is the graph. it is strange though because at x=0, y = 0

  6. anonymous
    • 5 years ago
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    screen shot is nothing

  7. anonymous
    • 5 years ago
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    hmm, i can't see anything

  8. mattfeury
    • 5 years ago
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    *hopefully*

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  9. mattfeury
    • 5 years ago
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    you have to find the largest rectangle possible. looking at the graph i would guess (4,2) but that's just a though....

  10. anonymous
    • 5 years ago
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    apparently we have to do it through optimization - derivatives and all that fun stuff

  11. anonymous
    • 5 years ago
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    This is one of those questions you ask an instructor whose got over 20 yrs cal experience.

  12. mattfeury
    • 5 years ago
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    find a function for the area of the square and optimize. the area of the square at a given x up until 9 = x*y = sqrt(x) * (9-x)

  13. mattfeury
    • 5 years ago
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    try deriving that and see if find an extrema

  14. anonymous
    • 5 years ago
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    not really :) as long as you have the skill then it's no prob

  15. anonymous
    • 5 years ago
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    it's one of those questions that are a challenge ^_^

  16. anonymous
    • 5 years ago
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    loving the enthusiasm sstarica, but i'm getting mercilessly owned in cal a

  17. anonymous
    • 5 years ago
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    LOL, nah, let it be the opposite. It's quite simple, believe me :)

  18. anonymous
    • 5 years ago
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    logically, I'd say the following: since the lenght of the rectangle is = 9, and is under the sqrt(x) now, let's first draw sqrt of (x) and notice that when you take x = 9, you'llhave y = 3. Logically, the rectangle won't exceed this limit, so the width is equal = 3

  19. anonymous
    • 5 years ago
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    length*

  20. myininaya
    • 5 years ago
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    matt has ir right! good job

  21. anonymous
    • 5 years ago
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    since it's inscribed under it

  22. mattfeury
    • 5 years ago
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    yes the maxima of that graph is 3!

  23. anonymous
    • 5 years ago
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    :) no need for all this.

  24. myininaya
    • 5 years ago
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    now we need to find A'

  25. anonymous
    • 5 years ago
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    all you had to do is draw sqrt(x) and see the limit of the rectangle. ^_^

  26. anonymous
    • 5 years ago
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    why A'? doesn't he want A?

  27. anonymous
    • 5 years ago
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    A = 3 x 9 = 27 that if he asked for A right? ^_^

  28. myininaya
    • 5 years ago
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    to maximize you find A'

  29. mattfeury
    • 5 years ago
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    so it's 3 * sqrt(3)

  30. mattfeury
    • 5 years ago
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    or no. 6 * sqrt(3)

  31. anonymous
    • 5 years ago
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    oh right lol my bad

  32. myininaya
    • 5 years ago
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    yes x=3 is right matt thats are only critcal number thats in the domain of A

  33. anonymous
    • 5 years ago
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    wait wait, why is 9=xy? doesn't A=xy?

  34. anonymous
    • 5 years ago
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    since he wants the largest area, then yes maximize, proceed matt and hope you understood what was going on Ire :)

  35. myininaya
    • 5 years ago
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    i didnt see A=9 up there no

  36. anonymous
    • 5 years ago
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    you want the largest area, so you have to maximize in this case, they didn't ask for the "area" alone, but largest one

  37. mattfeury
    • 5 years ago
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    right. to find where the area is a maximum, take a function for the area and find the maxima point. take that point to make your rectangle. since you go up to 9 the x width is (9-3) = 6. the height of the rectangle is sqrt(3)

  38. anonymous
    • 5 years ago
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    could you remind me how to find the maxima point please?

  39. anonymous
    • 5 years ago
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    which means "optimize" lol, alight matt will take it from here ^_^ good luck

  40. mattfeury
    • 5 years ago
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    take the derivative of the area function 'A'. find where the derivative = 0.

  41. mattfeury
    • 5 years ago
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    normally, you'd have to do the second derivative test to find out if it is a max/min too. but i know secrets.

  42. anonymous
    • 5 years ago
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    hmm

  43. anonymous
    • 5 years ago
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    1. A = (9-x) * sqrt(x)

  44. myininaya
    • 5 years ago
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  45. anonymous
    • 5 years ago
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    wow, thanks guys

  46. mattfeury
    • 5 years ago
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    very nice.

  47. myininaya
    • 5 years ago
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    also i never use the second derivative test it is pointless you can use A' to see if it is increasing to decreasing at x=3 to find that it is a max

  48. myininaya
    • 5 years ago
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    The first derivative rocks!

  49. anonymous
    • 5 years ago
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    thanks for the help guys, really appreciated

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