Q:

Drawing in 3 dimensions: Feasible region

We have the following constraints:
$$ 0 \leq x \leq 10$$
$$ 0 \leq y \leq 20$$
$$ 0 \leq z \leq 25$$
$$ 0 \leq x + y + z \leq 15$$
We have to draw this in an 0xyz - coordinate system, I did the following, I don't know if it's correct:
Mainly the part where the piramid is cut off by the cube is confusing me.
Then we have to find the vertices.
So these ones I know for sure:
$$(0,0,0),(0,0,15), (0,15,0),(10,0,0)$$
But the other 3 (well, 3 if my drawing is correct) I'm not sure how to find. Intuitively I'd say you have to get the intersects of for example $x=10$ with $y=-x+15$ which would give us another vertex $(10,5,0)$. Is this correct?
And finally, we have to find the maximum value $M = 2x+3y+5z$ gets within this region. I have absolutely no idea at all how to do that.
I would appreciate help/feedback.  

A:

Actually, a the bit that is cut off the end of your pyramid will be geometrically similar to your pyramid, so instead of $3$ other vertices, there are merely two: one on the $xy$-plane, one on the $xz$-plane. $(10,5,0)$ is one of them. I'll let you find the other.
Note that within the region, as you hold $x,y$ constant, then $M$ increases with $z$. Similarly, $M$ increases with $x$ when we hold $y,z$ constant, and increases with $y$ when we hold $x,z$ constant. Thus, $M$ will not be maximized inside the region. By similar reasoning, $M$ will not be maximized in the middle of your surfaces lying on the $xy$-plane, the $xz$-plane, the $yz$-plane, or the plane $x=10$. Thus, $M$ will be maximized on one of your edges, or in the middle of your surface on the plane $x+y+z=15$. Once again, though, we can rule out maximization on all edges but the edges of the "slanted" surface.
Your task, then, is to maximize $M$ subject to the constraints $x,y,z\ge 0,x\le10,$ and $x+y+z=15$, which I leave to you.

