Triangle In Square
A line segment is placed on top of a unit square so as to form a triangle region.
Given the length of the line segment, L, find the maximum area of the triangle.
Clearly for L 2, the maximum area will be ½, when the segment is placed along the diagonal of the square.
For L 2, consider the following diagram.
b = (L2 a2).
So area of triangle, A = ½ab = ½a(L2 a2) = ½a(L2 a2)½.
We may proceed from here via a calculus or non-calculus approach:
|dA/da||=||½(L2 a2)½ + ½a½(L2 a2)-½(-2a)|
|=||½(L2 a2)½ ½a2(L2 a2)-½|
At turning point, dA/da = 0.
Therefore, ½(L2 a2)½ = ½a2(L2 a2)-½
(L2 a2) = a2/(L2 a2)
Hence, L2 a2 = a2, leading to, a = L/2.
As area of triangle, A = ½a(L2 a2),
As the area of the triangle, A = ½a(L2 a2) is defined for positive values, the value of a for which it maximises is the same for A2.
Therefore 4A2 = a2(L2 a2) = a2L2 a4 = L2/4 (½L a2)2.
Hence, 4A2 will maximise when (½L a2)2 = 0 a = L/2.
Proceeding as in previous method we deduce that Amax = L2/4.