Question

The frame for a kite is to be made from six pieces of wood. The four exterior pieces have been cut with the lengths indicated in the figure. To maximize the area of the kite, how long should the diagonal pieces be?

Short answer

The diagonal pieces should be \(\frac{{2ab}}{{\sqrt {{a^2} + {b^2}} }}\) and \(\sqrt {{a^2} + {b^2}} \) long to maximize the area.

Step-by-step solution

Given data

The side lengths of the kite are \(a\) and \(b\).

Maximization condition

The conditions for maxima of a function is

\({\frac{{df\left( x \right)}}{{dx}}_{x = {x_{\max }}}} = 0\;\;\;\;\;.....\left( 1 \right)\)

Determination of diagonal lengths

Label the figure as follows

The lengths of the diagonal parts are related as

\(\begin{aligned}{c}{a^2} &= {d^2} + {f^2}\\d &= \sqrt {{a^2} - {f^2}} \\{b^2} &= {e^2} + {f^2}\\e &= \sqrt {{b^2} - {f^2}} \end{aligned}\)

The area of the kite is the sum of area of the four triangles given by

\(\begin{array}{c}A &= df + ef\\ &= f\sqrt {{a^2} - {f^2}} + f\sqrt {{b^2} - {f^2}} \end{array}\)

From equation (I), the area can be maximized by

\(\begin{aligned}{c}{\frac{{dA}}{{df}}_{f = {f_{\max }}}} &= 0\\\frac{d}{{df}}{\left( {f\sqrt {{a^2} - {f^2}} + f\sqrt {{b^2} - {f^2}} } \right)_{f = {f_{\max }}}} &= 0\\\sqrt {{a^2} - f_{\max }^2} - \frac{{f_{\max }^2}}{{\sqrt {{a^2} - f_{\max }^2} }} + \sqrt {{b^2} - f_{\max }^2} - \frac{{f_{\max }^2}}{{\sqrt {{b^2} - f_{\max }^2} }} &= 0\\\left( {{a^2} - 2f_{\max }^2} \right)\sqrt {{b^2} - f_{\max }^2} = - \left( {{b^2} - 2f_{\max }^2} \right)\sqrt {{a^2} - f_{\max }^2} \end{aligned}\)

Square both sides to get

\(\begin{aligned}{c}\left( {{a^4} - 4{a^2}f_{\max }^2 + 4f_{\max }^4} \right)\left( {{b^2} - f_{\max }^2} \right) &= \left( {{b^4} - 4{b^2}f_{\max }^2 + 4f_{\max }^4} \right)\left( {{a^2} - f_{\max }^2} \right)\\{a^4}{b^2} - {a^4}f_{\max }^2 - 4{a^2}{b^2}f_{\max }^2 + 4{a^2}f_{\max }^4 + 4{b^2}f_{\max }^4 - 4f_{\max }^6 &= {b^4}{a^2} - {b^4}f_{\max }^2 - 4{b^2}{a^2}f_{\max }^2 + 4{b^2}f_{\max }^4 + 4{a^2}f_{\max }^4 - 4f_{\max }^6\\f_{\max }^2\left( {{a^4} - {b^4}} \right) &= {a^4}{b^2} - {b^4}{a^2}\\f_{\max }^2\left( {{a^2} - {b^2}} \right)\left( {{a^2} + {b^2}} \right) &= {a^2}{b^2}\left( {{a^2} - {b^2}} \right)\end{aligned}\)

If \({a^2} \ne {b^2}\)

\(\begin{aligned}{c}f_{\max }^2 &= \frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}\\{f_{\max }} &= \frac{{ab}}{{\sqrt {{a^2} + {b^2}} }}\end{aligned}\)

The other lengths are

\(\begin{aligned}{c}d &= \sqrt {{a^2} - f_{\max }^2} \\ &= \sqrt {{a^2} - \frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}} \\ &= \frac{{{a^2}}}{{\sqrt {{a^2} + {b^2}} }}\end{aligned}\)

and

\(\begin{aligned}{c}e &= \sqrt {{b^2} - f_{\max }^2} \\ &= \sqrt {{b^2} - \frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}}} \\ &= \frac{{{b^2}}}{{\sqrt {{a^2} + {b^2}} }}\end{aligned}\)

Thus, the diagonal lengths are

\(\begin{array}{c}2f = \frac{{2ab}}{{\sqrt {{a^2} + {b^2}} }}\\d + e = \sqrt {{a^2} + {b^2}} \end{array}\)