I want to find a value of $x$ for which these equations hold for the maximum number of values of $y$ and $z$.
$$ 0<x<1 $$ $$ 0<y<1 $$ $$ 0<z<1 $$ $$ 0< x < 2z $$ $$ y<(x+y)/2 < z $$
I am a high schooler who has never come across such '3d inequality/optimisation'. I tried graphing these but could not figure out how. Can someone please guide me on how to solve these kind of problems, it would provide me great satifaction. Further is there a book or a course I could take which would help me be able to solve these kind of problems? Thanks!
A linear equation $ax+by+cz=d$ determines a plane $$\pi:=\bigl\{(x,y,z\in{\mathbb R}^3\bigm|ax+by+cz=d\bigr\}$$ in our $(x,y,z)$-space, and the linear inequality $ax+by+cz<d$ one of the two half-spaces defined by this plane $\pi$. You are given $10$ such inequalities, and are told to draw the set $S$ of points that fulfill all of them simultaneously. This set $S$ then is the intersection of ten half-spaces. This intersection is a certain convex polytope, it could also be empty.
The inequalities $0<x<1$, $\>0<y<1$, $\>0<z<1$ obviously define the open unit cube $C:=\>]0,1[\>\times\>]0,1[\>\times]0,1[\>$. Hence we know that $S\subset C$, and a nice drawing should be possible. We then are given the additional inequalities $$y<x,\qquad z>{x+y\over2},\qquad z>{x\over2}\ ,$$ the first of which is a rewriting of $y<{x+y\over2}$. We already know that $y>0$ in $S$. It follows that the inequality $z>{x\over2}$ is automatically fulfilled when $z>{x+y\over2}$.
In all, we have to draw the part of $C$ where $y<x$ and $z>{x+y\over2}$. The equality $y=x$ splits $C$ diagonally in two halves, and we are left with a triangular prism $P$ standing on the $(x,y)$-plane. This prism has three vertical edges of length $1$. The last plane $\pi:\ z={x+y\over2}$ intersects these edges in the points $(0,0,0)$, $\bigl(1,0,{1\over2}\bigr)$, and $(1,1,1)$. The set $S$ then consists of all points in $P\subset C$ lying above $\pi$.