How high should the box be ?

The maximum and minimum values of $ux + vy + wz$ , where $u^2 + v^2 + w^2 = 1$ , subject to the constraint $\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} + \tfrac{z^2}{c^2} = 1$ , are $\pm\sqrt{a^2u^2 + b^2v^2 + c^2w^2}$ . Thus the width of the ellipse $\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} + \tfrac{z^2}{c^2} = 1$ in the direction of the unit vector $(u\;v\; w\;)^T$ is $2\sqrt{a^2u^2 + b^2v^2 + c^2w^2}$ . We need to find three mutually perpendicular unit vectors such that the widths of the given ellipse in the directions given by two of these vectors are $33$ and $25$ ; then the desired height of the box is the width of the ellipse in the direction of the third vector.

Thus we need to find mutually perpendicular unit vectors $\mathbf{u}$ and $\mathbf{v}$ such that $2\sqrt{400u_1^2 + 225u_2^2 + 100u_3^2} \; = \; 33 \hspace{2cm} 2\sqrt{400v_1^2 + 225v_2^2 + 100v_3^2} \; = \; 25 \hspace{3cm} (\star)$ If we consider the equations $(\star)$ , together with the equations $u_1^2 + u_2^2 + u_3^2 \; = \; v_1^2 + v_2^2 + v_3^2 \; = \; 1 \hspace{2cm} u_1v_1 + u_2v_2 + u_3v_3 \; = \;0$ then there are many possible different solutions; for example we can obtain $\mathbf{u} \; = \; \left(\begin{array}{c} \alpha \\ \beta\sqrt{11} \\ -3\beta \end{array}\right) \hspace{2cm} \mathbf{v} \; = \; \frac{1}{\sqrt{20}}\left(\begin{array}{c} 0 \\ 3 \\ \sqrt{11} \end{array}\right)$ where $\alpha^2 \; = \; \frac{414}{925} \hspace{2cm} \beta^2 \; = \; \frac{511}{18500}$ Then the height of the box is $2\sqrt{400w_1^2 + 225w_2^2 + 100w_3^2}$ , where $\mathbf{w} = \mathbf{u} \times \mathbf{v}$ . We calculate $\mathbf{w} \; = \; \frac{1}{5\sqrt{370}}\left(\begin{array}{c} \sqrt{5110} \\ -3\sqrt{253} \\ 9\sqrt{23}\end{array}\right)$ and the height of the box is therefore $\sqrt{1186} = \boxed{34.4384}$ .

The simplest solution of this problem depends on the fact that if an ellipsoid with semi-axes $a, b,$ and $c$ , is placed such that it is tangent to the three perpendicular planes, (and let the origin of the Cartesian Coordinates System be at their intersection point), then, the coordinates of the ellipsoid center $C = (C_x, C_y, C_z)$ satisfy,

$C_x^2 + C_y^2 + C_z^2 = a^2 + b^2 + c^2$

Now, from symmetry, we know that after placing the ellipsoid in the box, with it tangent to all four vertical walls of the box, then the center of the ellipsoid $x$ and $y$ coordinates satisfy

$C_x = \dfrac{L}{2}$ and $C_y = \dfrac{W}{2}$

And, hence, we can very easily now, compute the $z$ -coordinate of the ellipsoid center $C_z$ , and it follows that the height of the box will be $H = 2 C_z$ , because the center of the ellipsoid is at the center of the box.