Minimizing over an ellipsoid

$f = 5 x + 10 y + 15 z = 5 (x - 2) + 10 + 10 (y - 3) + 30 + 15 (z - 8) + 120 = 10 \dfrac{(x - 2)}{2} + 30 \dfrac{(y-3)}{3} + 75 \dfrac{(z - 8)}{5 }+ 160$

From Cauchy-Schwarz inequality,

$f - 160 \ge - \sqrt{ 10^2 + 30^2 + 75^2} (1)$

Thus $f_{\text{min} } = 160 - \sqrt{6625} = 78.606$

Same as before. Let

$x=2+2\cos α\cos β,y=3+3\cos α\sin β,z=8+5\sin α$ .

Then $5x+10y+15z\geq 5(32-\sqrt {40}\cos α+15\sin α)\geq 5(32-\sqrt {265})\approx 78.606$ .

Therefore the minimum value of $5x+10y+15z$ is $\boxed {78.606}$ .

The minimum is attained when

$\sin \left (α-\tan^{-1} \frac{\sqrt {40}}{15}\right ) =-1$ ,

$\implies \sin α=-\dfrac {15}{\sqrt {265}}$ ,

$\cos α=\sqrt {\frac{40}{265}}$

$\sin \left (β+\tan{-1} \frac 13\right )=-1$

$\implies \sin β=-\dfrac {3}{\sqrt {10}}$ ,

$\cos β=-\dfrac {1}{\sqrt {10}}$

Substituting values we get

$x=\dfrac {2(\sqrt {265}-2)}{\sqrt {265}}\approx 1.754$

$y=\dfrac {3(\sqrt {265}-6)}{\sqrt {265}}\approx 1.894$

$z=\dfrac {8\sqrt {265}-75}{\sqrt {265}}\approx 3.393$ .