In isosceles triangle AB=BC=n.

What value of AC warrants the largest area of the triangle ABC?
Solve by:
a) Plane geometry.
b) Trigonometry.
c) Calculus.
d) Any other way is welcome.

The magnitude of the cross product of two 3-dimensional vectors is equal to the are of the parallelogram having the vectors as adjacent sides.  The triangle sought is half of this parallelogram.

The first vector, representing BA can be
v = ni + 0j + 0k
The second vector BC, forming an angle θ with u can be written as
w = ncos(θ) i + nsin(θ) j + 0k

The cross product vXw = 0i - 0j + nēsin(θ)
and so the magnitude ||vXw|| = nēsin(θ)

Which is maximized when θ=90š, so w becomes
w = 0i + nj + 0k

AC can be called vector z, which satisfies the relationship v+z=w or
z = w - v = -ni + nj + 0k

which has magnitude
||z|| = √((-n)ē+nē+0ē) = √(2nē) = n√2

Posted by Jer on 2011-06-16 15:54:07