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I've tried unsuccessfully to solve these two problems. I'd grateful for any help here.

  1. What is the biggest real number $z$ that obeys these two conditions: $$x + y + z=5 \quad\text{and}\quad xy +xz + yz=3\quad\text{?}$$
  2. Find the lower positive number for $$xy + 2xz + 3yz$$ if $xyz =48$.
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What level are we talking about? Calculus, algebra? Is there a specific method you are expected to use? (E.g., the first problem can be solved in several ways, Lagrange multipliers among them). – Arturo Magidin Apr 14 '11 at 13:44
@tom: It seems you have received useful answers to at least some of your questions. Please accept the ones you like by clicking the gray checkmark to the left. – Ross Millikan Apr 14 '11 at 13:46
If they are homework, they seem standard problems of Lagrange multipliers. – N. S. Apr 14 '11 at 13:52
I'm just a high school student, so I think the simplest method. I suppose that it could be solved by arithmetic but I'm not completely sure of that. – tom Apr 14 '11 at 15:07
@tom Please add and state what $x$, $y$ and $z$ are. Complex numbers? Real numbers? Positive real numbers? Please state!! – user38268 Apr 14 '11 at 23:58

For 1: Observe that $(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)$. From here you can get $x^2+y^2+z^2=19$.

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So do I have to solve that system to find out the lower number? – tom Apr 14 '11 at 15:10
@tom: Now you just need the largest allowable $z$ that satisfies the last. What should $x$ and $y$ be for this? – Ross Millikan Apr 14 '11 at 15:28
@Ross Millikan hmmm, I don't know because the numbers are real. I know how to solve that kind of system though. But it's very boring. – tom Apr 15 '11 at 1:18
@tom: The greatest $z$ is $\sqrt{19}$ when $x=y=0$ – Ross Millikan Apr 15 '11 at 3:29
@Ross Millikan: I don't think so because it is true for just one of the equations. We should remember that the sum of values must be five, then what you think it is not the answer. IMHO. – tom Apr 15 '11 at 12:35

For 1, it is "obvious" (and can be proven using calculus) that to make $z$ big, you want $x, y$ small and from symmetry they should be equal. Then your equations become $2x+z=5, x^2+2xz=3$ which you can combine, eliminating $z$ to $x=\frac{10\pm \sqrt{100-36}}{6}=\frac{1}{3},3$, choose the smaller root, and $x=y=\frac{1}{3}, z=\frac{13}{3}$

The more rigorous approach would be to write $x=\frac{3-yz}{y+z}$ and substitute in to get $(y+z)^2+3-yz=5(y+z), z=\frac{5-y\pm \sqrt{13+10y-3y^2}}{2}$, take the derivative with respect to $y$, set to $0$...

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Very interesting, thank you so much. It really taught me a great idea. – tom Apr 15 '11 at 20:28

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