Suppose you have two polytopes $P_1, P_2 \in \Bbb{R}^n$ given by

$$ P_1 = \lbrace x: A_1 x \le b_1\rbrace$$ $$ P_2 = \lbrace x: A_2 x \le b_2\rbrace $$

I wish to find their convex hull, that is a matrix $A_3$ and vector $b_3$ such that

$$ A_3 x \le b_3$$

Is the convex hull of $P_1$ and $P_2$. Now I have two questions,

Is there a bound on the number of n-1 facets of the convex hull?, that is suppose $P_1$ has $\alpha$ facets (so $A_1$ has $\alpha$ inequalities) and $P_2$ has $\beta$ facets (so $A_2$ has $\beta$ inequalities) , then can we say for example that $A_3$ has no more than $\gamma(\alpha + \beta)$ inequalities itself? Where $\gamma$ is some constant.

What is the best way to actually compute $A_3,b_3$ given $A_1,b_1$ and $A_2,b_2$?

Now I have a sketch of a procedure to do (2) but it seems far from elegant. I am also noticing there seems to be a lack of literature (at least after some surface level googling and discussing with two professors) regarding this specific problem. I'm pretty certain many people have considered it so I would also be requesting references about it.

## Some Ideas for (1)

So $\gamma = 2$ seems plausible. If you take both polytopes, lift them to a higher dimensional space and orient them to be parallel. You can create a high dimensional antiprism with one facet for every face on each of the original polytopes (that is $\alpha + \beta$ new facets), and if you add it's to the current facet count you end up with twice of what you started with. I feel if you go back to the original dimension $n$ from this higher dimensional setting its only natural that some number of faces becomes "unnecessary". But this is a loose hand wavey kind of argument.

A definite upper bound is $\begin{pmatrix} \alpha \\ 2 \end{pmatrix}+ \begin{pmatrix} \beta \\ 2 \end{pmatrix} + \alpha + \beta $ and this is trivial from considering every n-1 facet that can be generated from an $n-2$ face of one polytope connected to a point on the other. But I have very sincere doubts about it being quadratic.