Answer:
- for fixed AC, x = 45° maximizes the area
- for fixed AB, x → 90° maximizes the area
Step-by-step explanation:
Call the point of intersection of AA' and BC point X. Then ...
 CX = AC·cos(90°-x) = AC·sin(x)
and the area of AA'C is ...
 area = AC·CX·sin(90°-x) = AC²·sin(x)cos(x) = (1/2)AC²·sin(2x)
Obviously, area is maximized for 2x = π/2, or x = π/4 when AC is fixed.
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On the other hand, ...
 AC = AB·tan(x)
so the area of the triangle is ...
 area = (1/2)AC²·sin(2x) = (1/2)(AB·tan(x))²·sin(2x) = AB²·sin(x)³/cos(x)
For fixed AB, area approaches infinity as x approaches 90°.
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Comment on the attachments
The attached diagrams show AC=1 and B free to move. Values of x around 45° are shown. The number in the middle of the figure is the approximate area of ΔAA'C.
