0
$\begingroup$

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 $xyz =48$.
  • 0
    @David Benjamim Lim Well, I got that exercise like I wrote before.In the first exercise the numbers are real, as I said, and in the second one they are real as well.2011-04-15

2 Answers 2

3

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

  • 0
    @tom: sorry, brain glitch. See my answer below.2011-04-15
3

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 $0$...

  • 0
    Very interesting, thank you so much. It really taught me a great idea.2011-04-15