Maximum volume for tetrahedron

Let $P$ , $Q$ , $R$ , and $S$ be the vertices of the tetrahedron, where $PQ = SR = a$ and $PS = PR = QS = QR = 2$ . Let $O$ be the midpoint of $PQ$ so that $OP = OQ = \frac{1}{2}a$ , and let $T$ be on $OR$ so that $ST \perp OT$ .

By Pythagorean's Theorem on $\triangle OQR$ , $OR = \frac{1}{2}\sqrt{16 - a^2}$ . The area of $\triangle RPQ$ , which is also the base of the tetrahedron, is then $B = \frac{1}{4}a\sqrt{16 - a^2}$ .

On $\triangle OSR$ , $OS = OR = \frac{1}{2}\sqrt{16 - a^2}$ and $SR = a$ . Since the altitude of a triangle from a side $a$ is $h = \frac{2\sqrt{s(s-a)(s-b)(s-c)}}{a}$ , the altitude $ST$ from $OR$ , which is also the height of the tetrahedron, is $h = \frac{a\sqrt{16-2a^2}}{\sqrt{16-a^2}}$ .

Since the volume of a tetrahedron is $V = \frac{1}{3}Bh$ and $B = \frac{1}{4}a\sqrt{16 - a^2}$ and $h = \frac{a\sqrt{16-2a^2}}{\sqrt{16-a^2}}$ , the volume simplifies to $V = \frac{1}{12}a^2\sqrt{16-2a^2}$ .

This equation can be rearranged to $144V^2 = a^2 \cdot a^2 \cdot (16 - 2a^2)$ . By the AM-GM inequality, $a^2 \cdot a^2 \cdot (16 - 2a^2) \leq (\frac{a^2 + a^2 + (16 - 2a^2)}{3})^3 = \frac{2^{12}}{3^3}$ . Therefore, $144V^2 \leq \frac{2^{12}}{3^3}$ , which solves to $V \leq \boxed{\frac{16\sqrt{3}}{27}}$ .

I use Wolfram

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Setting up a suitable coordinate system, the tetrahedron vertices have coordinates $P\;(-\tfrac12a,0,0)\hspace{0.5cm} Q\;(\tfrac12a,0,0) \hspace{0.5cm} R\;(0,b,-\tfrac12a)\hspace{0.5cm} S\;(0,b,\tfrac12a)$ where $\tfrac12a^2 + b^2 = 4$ . If we define the vectors $\mathbf{u} \; =\; \overrightarrow{PQ} = \left(\begin{array}{c} a \\ 0 \\ 0 \end{array}\right) \hspace{1cm} \mathbf{v} \; =\; \overrightarrow{PR} = \left(\begin{array}{c}\tfrac12a \\ b \\ -\tfrac12a\end{array}\right) \hspace{1cm} \mathbf{w} = \overrightarrow{PS} = \left(\begin{array}{c} \tfrac12a \\ b \\ \tfrac12a\end{array}\right)$ Then $(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} \; = \; \left(\begin{array}{c} 0 \\ \tfrac12a^2 \\ ab\end{array}\right) \cdot \mathbf{w} \; = \; a^2b \; = \; 2b(4-b^2)$ and hence the volume of the tetrahedron is $V \; =\; \tfrac16\big|(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}\big| \; = \; \tfrac13b(4-b^2)$ which is maximized when $b = \tfrac{2}{\sqrt{3}}$ , so the maximum volume is $\boxed{\frac{16\sqrt{3}}{27}}$ .