Finding the Distance
Where the Maximum Angle Occurs
On the previous page we found the equation for
our viewing angle which was y = arctan (6/x) - arctan (2/x). This gives
us the following graph:
If we find the derivative of this equation and solve it for y' = 0, this
will give us the maximum point.
y = arctan (6/x) - arctan (2/x)
y' = (-6/x^2)/(1+(6/x)^2) - (-2/x^2)/(1+(2/x)^2)
We can simplify this quite a bit:
y' = -6/(x^2 + 36) + 2/(x^2 + 4)
= [-6(x^2+4)+2(x^2+36)]/[(x^2+36)(x^2+4)]
= [-6x^2-24+2x^2+72]/[(x^2+36)(x^2+4)]
= [48-4x^2]/[(x^2+36)(x^2+4)]
= [4(12-x^2)]/[(x^2+36)(x^2+4)]
We now want to solve for y' = 0. To do this we only need to set the
numerator equal to zero:
4(12 - x^2) = 0
This gives us x = ± √(12)
We only want positive values. Therefore, the viewer must stand x =
2√3 feet from the picture to get the best angle.
The angle at this point is y = arctan (6/2√3) - arctan (2/2√3) = 30
degrees.
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