supporting hyperplanes We have the closed convex set $\{x \in \mathbb{R}^{2}_{+}:x_{1}x_{2}\ge 1\}$ and we are interested to know the supporting hyperplane in the point $x=(t,\frac{1}{t})$. The result is that the supporting hyperplane at the point is given by $\frac {x_{1}} {t^{2}}+ x_{2}=\frac {2}{t}.$ Why? Thank you for your help.
 A: I am supposing $n=2$. In that case, you have the hyperbola $x_1x_2=1$ as boundary, so the supporting hyperplane will simply be the tangent line at $(t,t^{-1})$. Working out the derivative ($\frac{dx_2}{dx_1}$) at $x_1=t$  gives you $-t^{-2}$ and the equation of the line through it becomes $ x_2=-t^{-2}x_1+c$, where $c$ is obtained by the substitution $(x_1,x_2)=(t,t^{-1})$, which gives $c= 2t^{-1}$; hence, we arrive at
$$x_2= -\frac{1}{t^2}x_1+\frac{2}{t}\,,$$
as desired.
A: The general equation for a (hyper) plane is
$$
n \cdot x = C
$$
where $n$ is some "normal vector" to the plane and $C$ is a constant; alternatively, we can write
$$
n \cdot (x - x_0) = 0
$$
where $x_0$ is some point of the plane.
In this case, we are working in $\Bbb R^2$, and we know that the point $(t, 1/t)$ lies on the plane. (For some fixed constant $t$). So  our equation has the form
$$
n \cdot \pmatrix{x_1 - t\\x_2 - \frac1t} = 0.
$$
The only question that remains is "what is $n$"?
Well, the boundary of the set we're interested in is given by the function $F(x_1, x_2) = x_1 x_2 - 1$ being $0$. And in general, the normal vector to that boundary is given by the gradient of $F$, i.e.,
$$
\nabla F(x_1, x_2) = (x_2, x_1).
$$
We're looking at the point $(t, \frac1t)$, so our gradient is
$$
n = \nabla F(t, \frac1t) = (\frac1t, t)
$$
and the overall equation is then
$$
\pmatrix{\frac1t\\t} \cdot \pmatrix{x_1 - t\\x_2 - \frac1t} = 0.
$$
Writing out the dot-product, we get
$$
\frac1t (x_1 - t) + t (x_2 - \frac1t) = 0
$$
which simplifies to
$$
\frac1t x_1 - 1 + t x_2 - 1 = 0 \\
\frac1t x_1 + t x_2 = 2 \\
$$
Dividing through by $t$ gives your formula.
