The surface is given:
$xyz = 2$
It is in the first octant so $x > 0, y > 0, z > 0$.
The tangent plane taken at any point of this surface binds with the coordinate axes to form a tetrahedron.
Task: prove that the volume of the solid is not dependent on the chosen point and calculate the volume.
I used the approximation formula to derive the tangent plane equation in the form $z-z_0 = \frac{d}{dx}(x-x_0)+\frac{d}{dy}(y-y_0)$. From this, I tried to calculate the volume with $V = \frac{1}{3}A_{base}\times height \,of\, tetrahedron$ (I tried to get the points of intersection with the axes and derive tetrahedron side lengths).
But in the end I come to the following: $0.5(zxy)^2+2zxy+\frac{1}{24}(zxy)^3+\frac{8}{3} = V$, where $x, y, z$ are just the coordinate of the point chosen for the tangent plane on the surface. And it does not seem to simplify to some universal constant. Moreover, it does not even seem that all tetrahedrons bound by tangent planes for this surface in the first octant are equal in volume.
How can I prove the task and find the volume? Any suggestions?
Best Answer
So you came up with a formula which is (with very little rearrangement) $$V = \frac{(xyz)^2}{2}+2 xyz+\frac{(xyz)^3}{24}+\frac83,$$ and you know from the very beginning that $$xyz = 2.$$ Having those two pieces of information, you are a very short step away from the solution. You should be able to use a simple substitution to find the numeric value of the first formula. The second formula tells you a substitution you can use.
I don't know exactly how you got the first formula; a comparison to a more general formula for $xyz = c,$ where $c$ is an arbitrary positive constant, makes me think you already used the specific fact that $xyz = 2.$ (Which is perfectly OK, because you were given that fact in the problem statement.) Since the resulting volume for $xyz = 2$ agrees with the general solution, however, I don't see any reason to suspect that you made any errors.
You can also look at the solutions and links mentioned in Show that product of x, y, and z intercents of tangent plane to surface xyz=1 is a constant. Your question is almost a duplicate of that one (up to a constant factor), except that you had gotten much closer to a solution at the point where you asked your question.