Angle maximization problem

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)?

Any help is appreciated!

Let the required distance = x m.
Let A = angle of elevation of the top of the picture from the eyes of the observer
and B = angle of elevation of the bottom of the picture from the eyes of the observer
=> tanA = 3.2/x and tanB = 1.8/x
=> tan(A - B) = (3.2/x - 1.8/x) / [1 + (3.2)*(1.8)/x^2] = 1.4x/(x^2 + 5.76)
To maximise A - B,
d/dx tan(A - B) = 0
=> (x^2 + 5.76) * 1.4 - (1.4x) * (2x) = 0
=> 1.4x^2 = (1.4) * (5.76)
=> x = 2.4 m.

Use A squared + B squared = C squared.

1.4 + 1.8= 3.2 Meters away.....
^
(Stupid Yahoo saying i have too much punctuation.)