Consider a cuboid of sides 2x, 4x and 5x and a closed hemisphere of...

Let sum of surface area

$k = 3\pi{r}^{2} + 76{x}^{2}$

$∵ k$ is constant so $\dfrac{dk}{dx} = 0$

$⇒ 6\pir\dfrac{dr}{dx} + 2 \times 76x = 0$

$⇒ \dfrac{dr}{dx} = - \dfrac{76x}{3\pir} ⋯ \left(1\right)$

Now total volume $V = \dfrac{2}{3}\pi{r}^{3} + 40{x}^{3}$

For maximum volume $\dfrac{dV}{dx} = 0$

$⇒ 2\pi{r}^{2}\dfrac{dr}{dx} + 120{x}^{2} = 0 ⋯ \left(2\right)$

$⇒ 2\pi{r}^{2}\left(- \dfrac{76x}{3\pir}\right) + 120{x}^{2} = 0$

$⇒ 8x\left[- \dfrac{19r}{3} + 15x\right] = 0$

$⇒ \dfrac{x}{r} = \dfrac{19}{45}$