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Determing the best distance to view a picture

  • Thread starter Emethyst
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  • #1
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Homework Statement


A large painting is 5m in height. The painting is hung on a wall so that its lower edge is 1m above the eye level of an observer. How far from the wall should the observer stand to get the best view? (Assume that the best view means that the angle subtended by the painting in the observer's eye is maximum.)



Homework Equations


Implicit differentation, trig differentiations



The Attempt at a Solution


Sort of lost on this question. I thought it would simply be a right triangle where the opposite side is 6 and the adjacent side is the one im looking for, where θ is the angle subtended by the observer's eye (making the adjacent side 6cotθ). I now know this way is wrong, so if someone could point me along here it would be greatly appreciated, thank you. (for reference note the answer should be sqrt6.)
 

Answers and Replies

  • #2
Redbelly98
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Your θ is for a painting 6m high, and with bottom edge level with the observer.

You'll need to subtract the angle due to the 1m that is not part of the painting.
 
  • #3
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So the 6cotθ i obtained from my diagram would have to be -6cotθ, or 6cotθ-θ?
 
  • #4
Redbelly98
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Sorry for not replying sooner.

So the 6cotθ i obtained from my diagram would have to be -6cotθ, or 6cotθ-θ?
No, the θ=arccot(adjacent side / 6) you obtained needs to become

θ=arccot(adjacent side / 6) - arccot(adjacent side / 1)​

Now minimize θ.
 

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