## 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})^{-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}),

A_{max} | = | 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 A^{2}.

Therefore 4A^{2} = $a$^{2}($L$^{2} $a$^{2}) = $a$^{2}$L$^{2} $a$^{4} = $L$^{2}/4 (½$L$ $a$^{2})^{2}.

Hence, 4A^{2} 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.