Maximum volume of the pyramid 2

To find the volume of a tetrahedron, we may use the formula $V = \frac{bh}{3}$ , where $b$ is the area of a base and $h$ is the height from that base to the fourth vertex.

We may take any face to be the base. Let's take $\triangle ACD$ as the base. Then, given the constraint that $AC = 4$ and $AD = 5$ , the area of $\triangle ACD$ is maximized when it is a right triangle, with area $b = \frac{4*5}{2} = 10$ .

Now consider the height of the pyramid. We know that $AB = 3$ , thus the distance from the point $B$ to plane $\mathcal{P}$ containing $\{A,B,D\}$ is at most $3$ . But if we take $AB \perp \mathcal{P}$ , then $h = 3$ , and the volume of the pyramid is $V = \frac{3*10}{3} = 10$

I really liked your method of solution. So repeating that. Let the coordinates of $A$ be $(0,0,0)$ , $D$ be $(0,0,5)$ , of $B$ be $(4\sin α_b, 0, 4\cos α_b)$ , and of $C$ be $(3\sin α_c\cos β_c,3\sin α_c\sin β_c, 3\cos α_c)$ . Then the volume of the tetrahedron is $\dfrac{1}{6}\times {5\times 12\sin α_b\sin α_c\sin β_c}=10\sin α_b\sin α_c\sinβ_c$ . Obviously, the maximum value of the volume is $\boxed {10}$ when $α_b=α_c=β_c=90\degree$ . In general, if $\vec {AB}=\vec a, \vec {AC}=\vec b, \vec {AD}=\vec c$ , then the volume of the tetrahedron is $\dfrac{1}{6}\vec a\cdot (\vec b\times \vec c)$ , and it's maximum value is $\dfrac{abc}{6}$ when $\vec a, \vec b$ and $\vec c$ are mutually perpendicular.

This is a tetrahedron with all right angles at apex A. Volume is (3 x 4 x 5)/6=10