Maximization problem - version 3

Using Cauchy-Schwarz inequality as suggested. Rewrite $5x+3y+4z$ as follows and square it:

$\begin{aligned} \left(10 \times \frac x2 + 9 \times \frac y3 + 20 \times \frac z5 \right)^2 & \le \left(10^2 + 9^2 + 20^2\right) \left(\frac {x^2}4 + \frac {y^2}9 + \frac {z^2}{25} \right) = 581(1) \\ \implies 5x + 3y + 4z & \le \sqrt{581} \approx \boxed{24.1} \end{aligned}$

Equality occurs when $\dfrac x{20} = \dfrac y{27} = \dfrac z{100} = \dfrac 1{\sqrt{581}}$ .

Let $x=2\cos α\cos β,y=3\cos α\sin β,z=5\sin α$

Then $5x+3y+4z=10\cos α\cos β+9\cos α\sin β+20\sin α\leq \sqrt {181}\cos α+20\sin α\leq \sqrt {581}$

Therefore the maximum of $5x+3y+4z$ is $\sqrt {581}\approx \boxed {24.1039}$ .

The maximum is attained when

$\sin \left (α+\tan^{-1} \frac{\sqrt {181}}{20}\right ) =1$

$\implies \sin α=\dfrac {20}{\sqrt {581}}$

$\cos α=\sqrt {\dfrac {181}{581}}$

$\sin \left (β+\tan^{-1} \frac {10}{9}\right ) =1$

$\implies \sin β=\dfrac {9}{\sqrt {181}}$

$\cos β=\dfrac {10}{\sqrt {181}}$

Substituting values we get

$x=\dfrac {20}{\sqrt {581}}\approx 0.8297$

$y=\dfrac {27}{\sqrt {581}}\approx 1.1201$

$z=\dfrac {100}{\sqrt {581}}\approx 4.1487$ .