How to solve 3d inequality or optimisation problems?

I want to find a value of $$x$$ for which these equations hold for the maximum number of values of $$y$$ and $$z$$.

$$0 $$0 $$0 $$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!

• Hi, welcome to Math.SE! Formatting formulas (MathJax) is easy, just put formulas between dollar symbols! Commented Dec 15, 2020 at 13:12
• It's a bit unclear what the optimization goal is. You say that you want the "equations hold for the maximum number of values of y and z" . But since they (I assume) are real numbers, the "number of solutions" is always infinite. So what do you want to optimize, really? Is there a function for which you want to find a maximum? Commented Dec 15, 2020 at 13:16
• @MattiP. I'm sorry I may be getting this wrong. But, I Commented Dec 15, 2020 at 13:19
• Here is something to try to visualize the problem: desmos.com/calculator/w4zht3nczs Commented Dec 15, 2020 at 13:23
• @MattiP. Thank you for this. I have been thinking about this problem again and perhaps i have been thinking with a wrong perspective. I will play around with this graph and get back to you if I find out where I'm going wrong. Thanks! Commented Dec 15, 2020 at 13:28

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 $$ax+by+cz one of the two half-spaces defined by this plane $$\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, $$\>0, $$\>0 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 $$y{x+y\over2},\qquad z>{x\over2}\ ,$$ 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 $$y and $$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$$.