## Triangle In Square

#### 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)

(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 L 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.

Problem ID: 147 (Jan 2004)     Difficulty: 3 Star

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