## 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} + ½a½(L^{2} a^{2})^{-½}(-2a) |

= | ½(L^{2} a^{2})^{½} ½a^{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.