## Task 3A - 271233A

Determine the maximum value that the area of the image of an arbitrarily oriented cube $Q$ with given edge lengths $a, b, c$ can take when projected perpendicularly onto a plane.
The cuboid lies with its three edges on the coordinate axes, where $a$ lies along the $x$-axis, $b$ along the $y$-axis, and $c$ along the $z$-axis. The projection plane is defined by the normal vector $\vec{n}=\left(\begin{array}{l}n_{1} \\ n_{2} \\ n_{3}\end{array}\right)$ with $|\vec{n}|=1$. The area $A_{i}$, when projected onto the inclined plane, is proportional to the cosine of the angle between $\vec{n}$ and the normal vector $\vec{n}_{i}$ of the original area $A_{i}$. It is $\overrightarrow{n_{a b}}=\vec{e}_{z}, \overrightarrow{n_{b c}}=\vec{e}_{x}$ and $\overrightarrow{n_{a c}}=\vec{e}_{y}$. This gives:

$$
\begin{gathered}
A_{\text {proj }}=a b\left(\vec{n} \vec{n}_{a b}\right)+b c\left(\vec{n} \vec{n}_{b c}\right)+a c\left(\vec{n} \vec{n}_{a c}\right) \\
A_{\text {proj }}=a b\left(\vec{n} \vec{e}_{z}\right)+b c\left(\vec{n} \vec{e}_{x}\right)+a c\left(\vec{n} \vec{e}_{y}\right) \\
A_{\text {proj }}=a b n_{3}+b c n_{1}+a c n_{2}=\vec{n}\left(\begin{array}{c}
b c \\
a c \\
a b
\end{array}\right) \\
A_{\text {proj }}=a b c \cdot \vec{n}\left(\begin{array}{c}
\frac{1}{a} \\
\frac{1}{b} \\
\frac{1}{c}
\end{array}\right)
\end{gathered}
$$

The scalar product of two vectors of given length is maximized exactly when the two vectors are parallel. The normal vector must therefore be a multiple of $\left(\begin{array}{c}\frac{1}{a} \\ \frac{1}{b} \\ \frac{1}{c}\end{array}\right)$. Since its length must also be 1, we can conclude:

$$
\vec{n}=\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}}\left(\begin{array}{c}
\frac{1}{a} \\
\frac{1}{b} \\
\frac{1}{c}
\end{array}\right)
$$

For the maximum projected area, we get:

$$
\begin{aligned}
& A_{p r o j}=\frac{a b c\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}} \\
& A_{p r o j}=a b c \sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}
\end{aligned}
$$