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BC Calculus Help!

  1. Jan 15, 2010 #1
    A picture 1.4 meters high stands on a wall so that its lower edge is 1.8 meters above the eye of an observer. What is the most favorable distance from the wall for this observer to stand (that is, to maximize his or her angle of vision)?

    I've tried using the tan (pheta+beta) = (1.4+1.8)/x equation
    but it still doesn't work out...
     
  2. jcsd
  3. Jan 16, 2010 #2

    ehild

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    Draw a picture first! What is "angle of vision"? Show it in the picture!

    ehild
     
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