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If x, y and z are positive with xy=24, xz=48 and yz=144, what is the value of x+y+z?

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\(xy=24\\ xz=48\\ yz=144\\ 24-y=48-z\\ \frac{144}{z}=z+24-48\\ z^2-24z-144=0\\ z=12(1+\sqrt{2})\\ \color{blue}z=12+12\sqrt{2}=28.9706\)

\(24-y=48-z\\ y=z-24\\ y=12(1+\sqrt{2})-24\\ \color{blue}y=-12+12 \sqrt{2}=4.9706\)

\(x = \frac{48}{z}=\frac{48}{12+12\sqrt{2}}\\ \color{blue}x=1.6569\)

\(\color{blue}x+y+z=35.5979\)

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