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.

$b$ = ($L$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)^-½

($L$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 A_max = $L$²/4.