- 23 Nov '09 07:21Hello all, this one's a bit of a deusy.

Applications of Differentiation/Optimization

A painting in an art gallery has height 4 meters and is hung so that its lower edge is one-half meter above the eye of the observer. Show that the observer should stand three-half meters from the wall to get the best view.

I'm at a loss. I'm not even sure where to begin. Any assistance would be greatly appreciated.

Regards,

Drew - 25 Nov '09 05:13

I'm guessing the "best view" is the one in which the painting takes up the most space in your field of vision. Is that right? You'd need to take into account how looking at it from close up angles it too much to see properly, while the farther you go the smaller it gets in your vision. Hmmm.*Originally posted by Drew***Hello all, this one's a bit of a deusy.**

Applications of Differentiation/Optimization

A painting in an art gallery has height 4 meters and is hung so that its lower edge is one-half meter above the eye of the observer. Show that the observer should stand three-half meters from the wall to get the best view.

I'm at a loss. I'm not even sure where to begin. Any assistance would be greatly appreciated.

Regards,

Drew

Well, you worked it out, so I won't bother. Interesting though. - 25 Nov '09 16:04

I think what you are saying is correct.*Originally posted by AThousandYoung***I'm guessing the "best view" is the one in which the painting takes up the most space in your field of vision. Is that right? You'd need to take into account how looking at it from close up angles it too much to see properly, while the farther you go the smaller it gets in your vision. Hmmm.**

Well, you worked it out, so I won't bother. Interesting though.

so

you let the horizontal distance the observer is from the painting be "x" (the base of the right triangle)

then draw a line from the observer's eye to the top of the painting

followed by one to the bottom of the painting

then maximize the angle bounded by the two hypotenuse( or the largest angle-smallest in the subdivided angle)by using arccot functions to solve. - 26 Nov '09 20:19Well here's what I came up with:

Angle alpha is the angle from eye level to the bottom edge of the painting,

angle theta is the angle from the bottom edge of the paining to the top.

Symbology:

=> implies

<=> if and only if

tan(alpha) = .5/x => alpha = tan^-1(.5/x)

tan(alpha + theta) = (4+.5)/x => alpha + theta = tan^-1(4.5/x)

=> theta = tan^-1(4.5/x) - alpha

theta = tan^-1(4.5/x) - tan^-1(.5/x)

derivative of theta = theta prime = theta'

theta' = (-4.5 / (x^2 + 4.5^2)) + (.5 / (x^2 +.5^2))

theta' = 0 implies max angle

theta' = 0 <=> (-4.5 / (x^2 + 4.5^2)) + (.5 / (x^2 +.5^2))

=> (4.5 / (x^2 + 4.5^2)) = (.5 / (x^2 +.5^2))

x = 3/2