Consider a cuboid of sides 2x,4x and 5x and a closed hemisphere of radius r. If the sum of their surface areas is constant k, then the ratio x:r, for…

Let sum of surface area

$$k=3\pi {r}^{2}+76{x}^{2}$$k=3πr2+76x2

$\because k$∵k is constant so $\frac{dk}{dx}=0$dxdk​=0

$$\Rightarrow 6\pi r\frac{dr}{dx}+2\times 76x=0$$⇒6πrdxdr​+2×76x=0

$$\Rightarrow \frac{dr}{dx}=-\frac{76x}{3\pi r}\cdots (1)$$⇒dxdr​=−3πr76x​⋯(1)

Now total volume $V=\frac{2}{3}\pi {r}^{3}+40{x}^{3}$V=32​πr3+40x3

For maximum volume $\frac{dV}{dx}=0$dxdV​=0

$$\Rightarrow 2\pi {r}^{2}\frac{dr}{dx}+120{x}^{2}=0\cdots (2)$$⇒2πr2dxdr​+120x2=0⋯(2)

$$\Rightarrow 2\pi {r}^{2}(-\frac{76x}{3\pi r})+120{x}^{2}=0$$⇒2πr2(−3πr76x​)+120x2=0

$$\Rightarrow 8x[-\frac{19r}{3}+15x]=0$$⇒8x[−319r​+15x]=0

$$\Rightarrow \frac{x}{r}=\frac{19}{45}$$⇒rx​=4519​