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.
Best Answer
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 $n$.