mathschallenge.net

Problem

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.

Solution

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:

Calculus Method

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),

Amax	=	½(L/2)(L2 L2/2)
	=	½(L/2)(L2/2)
	=	½(L/2)(L/2)
	=	L2/4

Non-calculus Method

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 A_max = L²/4.