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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:enter image description here

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.

Ylyk Coitus
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1 Answers1

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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.

Cameron Buie
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  • My textbook states: If you find the value of the function in the vertexes and you find a point where the value is bigger (or smaller) you have the maximum (or minimum). But how is one supposed to tackle this? These methods seem like a bit of guessing? At first I thought only the vertexes could be a minimum/maximum, now everything is more complicated. To me, in this case it would seem logical that 0 at (0,0,0) is the minimum and 75 at (0,0,15) is the maximum.. – Ylyk Coitus Mar 31 '13 at 22:07
  • Use the constraint equation to rewrite $M$ in terms of 2 variables. Since none of the variables can be negative, it should quickly become clear where the maximum is achieved. – Cameron Buie Mar 31 '13 at 22:15
  • How can I rewrite M in terms of 2 variables? Am I missing something? – Ylyk Coitus Mar 31 '13 at 22:18
  • Since $x+y+z=15,$ then solve for $z$, then substitute into your expression for $M$. – Cameron Buie Mar 31 '13 at 22:21
  • So 75 is the maximum? – Ylyk Coitus Mar 31 '13 at 22:22
  • That's correct! – Cameron Buie Mar 31 '13 at 22:22
  • Thank you very much! I don't want to sound like a complete tool, but I think you must have overlooked me saying it in the first comment ;P. My final little question: Having multiple maximums is possible right? For example, if we have $M: x+y+z =15$ +1 btw – Ylyk Coitus Mar 31 '13 at 22:26
  • Yes, you're correct. That one would be maximized everywhere on your upper plane. – Cameron Buie Mar 31 '13 at 22:30