Show that product of $x$, $y$, and $z$ intercepts of tangent plane to surface $xyz=1$ is a constant I am studying for my math final and I just wrote the practice final. Unfortunately there are no solutions and I am completely lost on how to do this problem. If anyone could help I would really appreciate it.
Question: Show that the product of the $x$, $y$, and $z$ intercepts of any tangent plane to the surface $xyz = 1$ in the first octant is a constant.
I tried rearranging the equation to $z=\frac1{xy}$ then I tried to find the tangent plane using the formula $$z=f(a,b)+f_1(a,b)(x-a) + f_2(a,b)(y-b)$$ but I got confused and it ended up being a big mess. Anyways if anyone could lend a hand here I would really appreciate it.
 A: The gradient of $f(x,y,z)=xyz$ at $(a,b,c)$ is $\langle bc,ac,ab\rangle$ which is the same as $\langle 1/a,1/b,1/c \rangle$. Therefore, the tangent plane has equation
$$\frac{x-a}{a}+\frac{y-b}{b}+\frac{z-c}{c}=0$$
Rearrange it into 
$$\frac{x }{a}+\frac{y }{b}+\frac{z }{c}=3$$
and then divide by $3$ to obtain the intercept form of the plane equation:
$$\frac{x }{3a}+\frac{y }{3b}+\frac{z }{3c}=1$$
(The intercepts are what you see in the denominators.)
This generalizes to $\mathbb R^n$: the product of $n$ intercepts of tangent hyperplanes to $x_1\cdots x_n=1$ is $n$.
A: Look at question two in this set of examples: http://www.math.ubc.ca/~malabika/teaching/ubc/fall08/math263/problem-set1-solution.pdf
A: Non-calculus solution just for fun:
The product in question is 6 times the volume of the tetrahedron cut out from the positive octant by the tangent plane. So we want to show that this volume is constant.
Well, consider the case when $x_0=y_0=z_0=1$ (then the tangent plane is $x+y+z=3$, each intercept is $3$, the product is $27$). Now fix $(x_1, y_1, z_1)$ on the surface (meaning $x_1y_1z_1=1$). Apply the linear map sending $x$ to $x_1x$, $y$ to $y_1 y$ and $z$ to $z_1z$. This map takes $(1,1,1)$ to $(x_1, y_1, z_1)$; it also takes the surface $xyz=1$ to itself, and being linear, takes planes to planes. Hence it takes the tangent plane to $xyz=1$ at $(1,1,1)$ to a plane that intersects this surface exactly once at $(x_1, y_1, z_1)$. That is of course the tangent plane to this surface at $(x_1, y_1, z_1)$. This map also takes coordinate planes to themselves. We conclude that it takes the tetrahedron cut out by the tangent plane at $(1,1,1)$ from the positive orthant to the tetrahedron cut out by the tangent plane at $(x_1,y_1,z_1)$ from the positive orthant. But the map is volume preserving (since  $x_1y_1z_1=1$), so we have what we want.
(Well, of course we also know that $(3,0,0)$ is taken to $(3x_1, 0,0)$ so that's the new intercept, and same for the $y$ and $z$ axis intercepts, so the product is simply $27x_1y_1z_1=27$, but that's just extra info).
This also generalizes to the hypersurface $\prod x_i=1$ in $\mathbb{R}^n$.
