## 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.

$b$ = ($L$2 $a$2).

So area of triangle, A = ½$ab$ = ½$a$($L$2 $a$2) = ½$a$($L$2 $a$2)½.

We may proceed from here via a calculus or non-calculus approach:

Calculus Method

 dA/da = ½($L$2 $a$2)½ + ½$a$½($L$2 $a$2)-½(-2$a$) = ½($L$2 $a$2)½ ½$a$2($L$2 $a$2)-½

At turning point, dA/da = 0.

Therefore, ½($L$2 $a$2)½ = ½$a$2($L$2 $a$2)

($L$2 $a$2) = $a$2/($L$2 $a$2)

Hence, $L$2 $a$2 = $a$2, leading to, $a$ = $L$/2.

As area of triangle, A = ½$a$($L$2 $a$2),

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

Non-calculus Method

As the area of the triangle, A = ½$a$($L$2 $a$2) is defined for positive values, the value of $a$ for which it maximises is the same for A2.

Therefore 4A2 = $a$2($L$2 $a$2) = $a$2$L$2 $a$4 = $L$2/4 (½$L$ $a$2)2.

Hence, 4A2 will maximise when (½$L$ $a$2)2 = 0 $a$ = $L$/2.

Proceeding as in previous method we deduce that Amax = $L$2/4.

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

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