So I basically need some help people. Math nerds unite!?

A painting in an art gallery has height (h) and is hung so that its lower edge is a distance (d) above the eye of an observer. How far from the wall should the observer stand to get the best view? (In other words, where should the observer stand so as to maximize the angle 胃 (theta) subtended at his eye by the painting?)

I need help understanding all the steps in solving this problem.
I have the answer...I'm just not 100% sure how to get there.

Okay, so here goes.

Let:

x=horizontal distance from observer to wall

Since the lower edge of the painting is distance (d) above the height of the observer's eye, you can draw a right triangle where:

side1 = x
side2 = d
hypotenuse = distance from eye to bottom of painting.
Call the angle at his eye for this right triangle angle A.

tan(A)=opposite/adjacent= d/x
Solve for angle A:
A = arctan(d/x)

We know the height of the painting is h, so the vertical distance from the height of the observer's eye to the top of the painting is (d+h).

You can draw another right triangle, where:

side1 = x
side2 = (d+h)
hypotenuse = distance from eye to top of painting
Call the angle at the eye for this right triangle angle B.

tan(B) = opposite/adjacent = (d+h)/x
Solve for angle B:
B=arctan((d+h)/x)

So,

angle 螛 = angle B - angle A

Substituting the values we got for angle A and B above,

螛 = arctan((d+h)/x) - arctan(d/x)

To find the max 螛, we need to take the derivative of 螛 as a function of x, and set equal to zero.

derivative of arctan(x) = 1/(1+x虏)
derivative of (d+h)/x = -(d+h)/x虏
derivative of d/x = -d/x虏

By the chain rule:
derivative of f(g(x)) = f'(g(x))g'(x)
So:

螛'(x) = [-(d+h)/x虏][1/[1+[(d+h)/x]虏]] - [-d/x虏][1/[1+[d/x]虏]] = 0

Simplify this:
-(d+h)/(x虏+(d+h)虏) + d/(x虏+d虏) = 0
Get rid of the fractions:
-(d+h)(x虏+d虏) + d(x虏+(d+h)虏) = 0
Combine like terms and solve for x:
x虏(-d-h+d) + (-d虏(d+h)+d(d+h)虏) = 0
(-h)x虏 + (d虏h + dh虏) = 0

x虏 = (d虏h + dh虏)/h = d(d+h)

x = 鈭?d(d+h)

_
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| h
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| d
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|______x_______

Finish this diagram by drawing line segments from the right endpoint of the horizontal line segment of length x to the left side of the two "tick marks" (on the vertical line) that indicate the top and bottom of the painting. The angle between these two lines is 胃. You should have no trouble showing that

胃 = arctan[(d+h)/x] - arctan(d/x)

Now maximize 胃.

You should find that x = 鈭歔d(d+h)] maximizes 胃