We have, $$A B=2 r$$

and $\angle A C B=90^{\circ} \quad$ [since, angle in the semi-circle is always $90^{\circ}$ ]

Let $A C=x$ and $B C=y$

$$\begin{array}{ll} \therefore & (2 r)^2=x^2+y^2 \\ \Rightarrow & y^2=4 r^2-x^2 \\ \Rightarrow & y=\sqrt{4 r^2-x^2}\quad\text{.... (i)} \end{array}$$

$$\begin{aligned} \text { Now, } \quad \quad \text { area of } \triangle A B C, A & =\frac{1}{2} \times x \times y \\ & =\frac{1}{2} \times x \times\left(4 r^2-x^2\right)^{1 / 2}\quad\text{[using Eq. (i)]} \end{aligned}$$

$$\begin{aligned} &\text { Now, differentiating both sides w.r.t. } x \text {, we get }\\ &\begin{aligned} \frac{d A}{d x} & =\frac{1}{2}\left[x \cdot \frac{1}{2}\left(4 r^2-x^2\right)^{-1 / 2} \cdot(0-2 x)+\left(4 r^2-x^2\right)^{1 / 2} \cdot 1\right] \\ & =\frac{1}{2}\left[\frac{-2 x^2}{2 \sqrt{4 r^2-x^2}}+\left(4 r^2-x^2\right)^{1 / 2}\right] \end{aligned} \end{aligned}$$

$$\begin{aligned} & =\frac{1}{2}\left[\frac{-x^2}{\sqrt{4 r^2-x^2}}+\sqrt{4 r^2-x^2}\right] \\ & =\frac{1}{2}\left[\frac{-x^2+4 r^2-x^2}{\sqrt{4 r^2-x^2}}\right]=\frac{1}{2}\left[\frac{-2 x^2+4 r^2}{\sqrt{4 r^2-x^2}}\right] \end{aligned}$$

$$\begin{array}{lrl} \Rightarrow & \frac{d A}{d x} & =\left[\frac{\left(-x^2+2 r^2\right)}{\sqrt{4 r^2-x^2}}\right] \\ \text { Now, } & \frac{d A}{d x} & =0 \\ \Rightarrow & -x^2+2 r^2 & =0 \\ \Rightarrow & r^2 & =\frac{1}{2} x^2 \\ \Rightarrow & r & =\frac{1}{\sqrt{2}} x \\ \therefore & x & =r \sqrt{2} \end{array}$$

Again, differentiating both sides w.r.t. x, we get

$$\begin{aligned} \frac{d^2 A}{d x^2} & =\frac{\sqrt{4 r^2-x^2} \cdot(-2 x)+\left(2 r^2-x^2\right) \cdot \frac{1}{2}\left(4 r^2-x^2\right)^{-1 / 2}(-2 x)}{\left(\sqrt{4 r^2-x^2}\right)^2} \\ & =\frac{-2 x\left[\sqrt{4 r^2-x^2}+\left(2 r^2-x^2\right) \cdot \frac{1}{2 \sqrt{4 r^2-x^2}}\right]}{\left(\sqrt{4 r^2-x^2}\right)^2} \\ & =\frac{-4 x \cdot\left(\sqrt{4 r^2-x^2}\right)^2+\left(2 r^2-x^2\right)(-2 x)}{2 \cdot\left(4 r^2-x^2\right)^{3 / 2}} \\ & =\frac{-4 x\left(4 r^2-x^2\right)+\left(2 r^2-x^2\right) \cdot(-2 x)}{2 \cdot\left(4 r^2-x^2\right)^{3 / 2}} \\ & =\frac{-16 x r^2+4 x^3+\left(2 r^2-x^2\right)(-2 x)}{2 \cdot\left(4 r^2-x^2\right)^{3 / 2}} \end{aligned}$$

$$\begin{aligned} \left(\frac{d^2 A}{d x^2}\right)_{x=r \sqrt{2}} & =\frac{-16 \cdot r \sqrt{2} \cdot r^2+4 \cdot(r \sqrt{2})^3+\left[2 r^2-(r \sqrt{2})^2\right] \cdot(-2 \cdot r \sqrt{2})}{2 \cdot\left(4 r^2-2 r^2\right)^{3 / 2}} \quad[\because x=r \sqrt{2}] \\ & =\frac{-16 \sqrt{2} \cdot r^3+8 \sqrt{2} r^3}{2\left(2 r^2\right)^{3 / 2}}=\frac{8 \sqrt{2} r^2[r-2 r]}{4 r^3} \\ & =\frac{-8 \sqrt{2} r^3}{4 r^3}=-2 \sqrt{2}<0 \end{aligned}$$

For $x=r \sqrt{2}$, the area of triangle is maximum.

For $x=r \sqrt{2}\quad, y=\sqrt{4 r^2-(r \sqrt{2})^2}=\sqrt{2 r^2}=r \sqrt{2}$

Since, $\quad x=r \sqrt{2}=y$

Hence, the triangle is isosceles.

Since, volume of the box $=1024 \mathrm{~cm}^3$

Let length of the side of square base be $x \mathrm{~cm}$ and height of the box be $y \mathrm{~cm}$.

$\therefore \quad$ Volume of the box $(V)=x^2 \cdot y=1024$

Since, $\quad x^2 y=1024 \Rightarrow y=\frac{1024}{x^2}$

$$\begin{aligned} &\text { Let } C \text { denotes the cost of the box. }\\ \therefore\quad &C=2 x^2 \times 5+4 x y \times 2.50 \end{aligned}$$

$$\begin{aligned} & =10 x^2+10 x y=10 x(x+y) \\ & =10 x\left(x+\frac{1024}{x^2}\right) \\ & =\frac{10 x}{x^2}\left(x^3+1024\right) \\ \Rightarrow\quad C & =10 x^2+\frac{10240}{x}\quad\text{.... (i)} \end{aligned}$$

$$\begin{aligned} &\text { On differentiating both sides w.r.t. } x \text {, we get }\\ &\begin{aligned} \frac{d C}{d x} & =20 x+10240(-x)^{-2} \\ & =20 x-\frac{10240}{x^2}\quad\text{.... (ii)} \end{aligned} \end{aligned}$$

$$\begin{array}{lrl} \text { Now, } & \frac{d C}{d x} & =0 \\ \Rightarrow & 20 x & =\frac{10240}{x^2} \\ \Rightarrow & 20 x^3 & =10240 \\ \Rightarrow & x^3 & =512=8^3 \Rightarrow x=8 \end{array}$$

$$\begin{aligned} &\text { Again, differentiating Eq. (ii) w.r.t. } x \text {, we get }\\ &\begin{aligned} \frac{d^2 C}{d x^2} & =20-10240(-2) \cdot \frac{1}{x^3} \\ & =20+\frac{20480}{x^3}>0 \\ \therefore \quad\left(\frac{d^2 C}{d x^2}\right)_{x=8} & =20+\frac{20480}{512}=60>0 \end{aligned} \end{aligned}$$

For $x=8$ cost is minimum and the corresponding least cost of the box,

$$\begin{aligned} C(8) & =10 \cdot 8^2+\frac{10240}{8} \\ & =640+1280=1920 \end{aligned}$$

$$\therefore$$ Least cost ₹ $=1920$

We have given that, the sum of the surface areas of a rectangular parallelopiped with sides $x, 2 x$ and $\frac{x}{3}$ and a sphere is constant.

Let $S$ be the sum of both the surface area.

$\therefore \quad S=2\left(x \cdot 2 x+2 x \cdot \frac{x}{3}+\frac{x}{3} \cdot x\right)+4 \pi r^2=k$

$$\begin{aligned} k & =2\left[2 x^2+\frac{2 x^2}{3}+\frac{x^2}{3}\right]+4 \pi r^2 \\ & =2\left[3 x^2\right]+4 \pi r^2=6 x^2+4 \pi r^2 \end{aligned}$$

$$\begin{array}{lr} \Rightarrow & 4 \pi r^2=k-6 x^2 \\ \Rightarrow & r^2=\frac{k-6 x^2}{4 \pi} \\ \Rightarrow & r=\sqrt{\frac{k-6 x^2}{4 \pi}}\quad\text{.... (i)} \end{array}$$

Let V denotes the volume of both the parallelopiped and the sphere.

Then,

$$\begin{aligned} V & =2 x \cdot x \cdot \frac{x}{3}+\frac{4}{3} \pi r^3=\frac{2}{3} x^3+\frac{4}{3} \pi r^3 \\ & =\frac{2}{3} x^3+\frac{4}{3} \pi\left(\frac{k-6 x^2}{4 \pi}\right)^{3 / 2} \\ & =\frac{2}{3} x^3+\frac{4}{3} \pi \cdot \frac{1}{8 \pi^{3 / 2}}\left(k-6 x^2\right)^{3 / 2} \\ & =\frac{2}{3} x^3+\frac{1}{6 \sqrt{\pi}}\left(k-6 x^2\right)^{3 / 2}\quad\text{.... (ii)} \end{aligned}$$

On differentiating both sides w.r.t. $x$, we get

$$\begin{aligned} \frac{d V}{d x} & =\frac{2}{3} \cdot 3 x^2+\frac{1}{6 \sqrt{\pi}} \cdot \frac{3}{2}\left(k-6 x^2\right)^{1 / 2} \cdot(-12 x) \\ & =2 x^2-\frac{12 x}{4 \sqrt{\pi}} \sqrt{k-6 x^2} \\ & =2 x^2-\frac{3 x}{\sqrt{\pi}}\left(k-6 x^2\right)^{1 / 2}\quad\text{.... (iii)} \end{aligned}$$

$$\begin{array}{ll} \because & \frac{d V}{d x}=0 \\ \Rightarrow & 2 x^2=\frac{3 x}{\sqrt{\pi}}\left(k-6 x^2\right)^{1 / 2} \end{array}$$

$$\begin{aligned} \Rightarrow \quad & 4 x^4 =\frac{9 x^2}{\pi}\left(k-6 x^2\right) \\ \Rightarrow \quad & 4 \pi x^4 =9 k x^2-54 x^4 \\ \Rightarrow \quad & 4 \pi x^4+54 x^4 =9 k x^2 \\ \Rightarrow \quad & x^4[4 \pi+54] =9 \cdot k \cdot x^2 \\ \Rightarrow \quad& x^2 =\frac{9 k}{4 \pi+54} \\ \Rightarrow \quad & x =3 \cdot \sqrt{\frac{k}{4 \pi+54}}\quad\text{.... (iv)} \end{aligned}$$

$$\begin{aligned} &\text { Again, differentiating Eq. (iii) w.r.t. } x \text {, we get }\\ &\begin{aligned} \frac{d^2 V}{d x^2} & =4 x-\frac{3}{\sqrt{\pi}}\left[x \cdot \frac{1}{2}\left(k-6 x^2\right)^{-1 / 2} \cdot(-12 x)+\left(k-6 x^2\right)^{1 / 2} \cdot 1\right] \\ & =4 x-\frac{3}{\sqrt{\pi}}\left[-6 x^2 \cdot\left(k-6 x^2\right)^{-1 / 2}+\left(k-6 x^2\right)^{1 / 2}\right] \\ & =4 x-\frac{3}{\sqrt{\pi}}\left[\frac{-6 x^2+k-6 x^2}{\sqrt{k-6 x^2}}\right] \\ & =4 x-\frac{3}{\sqrt{\pi}}\left[\frac{k-12 x^2}{\sqrt{k-6 x^2}}\right] \end{aligned} \end{aligned}$$

Now, $\left(\frac{d^2 V}{d x^2}\right)_{x=3 \cdot \sqrt{\frac{k}{4 \pi+54}}}=4 \cdot 3 \sqrt{\frac{k}{4 \pi+54}}-\frac{3}{\sqrt{\pi}}\left[\frac{k-12 \cdot 9 \cdot \frac{k}{4 \pi+54}}{\sqrt{k-\frac{6 \cdot 9 \cdot k}{4 \pi+54}}}\right]$

$$\begin{aligned} & =12 \sqrt{\frac{k}{4 \pi+54}}-\frac{3}{\sqrt{\pi}}\left[\frac{k-\frac{108 k}{4 \pi+54}}{\sqrt{k-\frac{54 k}{4 \pi+54}}}\right] \\ & =12 \sqrt{\frac{k}{4 \pi+54}}-\frac{3}{\sqrt{\pi}}\left[\frac{4 k \pi+54 k-108 k / 4 \pi+54}{\sqrt{4 k \pi+54 k-54 k / 4 \pi+54}}\right] \\ & =12 \sqrt{\frac{k}{4 \pi+54}}-\frac{3}{\sqrt{\pi}}\left[\frac{4 k \pi-54 k}{\sqrt{4 k \pi} \sqrt{4 \pi+54}}\right] \\ & =12 \sqrt{\frac{k}{4 \pi+54}}-\frac{6}{\sqrt{\pi}}\left[\frac{k(2 \pi-27)}{\sqrt{k} \sqrt{16 \pi^2+216 \pi}}\right] \end{aligned}$$

For $x=3 \sqrt{\frac{k}{4 \pi+54}}$, the sum of volumes is minimum.

For $x=3 \sqrt{\frac{k}{4 \pi+54}}$, then $\quad r=\sqrt{\frac{k-6 x^2}{4 \pi}}\quad\text{[using Eq. (i)]}$

$$\begin{aligned} & =\frac{1}{2 \sqrt{\pi}} \sqrt{k-6 \cdot \frac{9 k}{4 \pi+54}} \\ & =\frac{1}{2 \sqrt{\pi}} \cdot \sqrt{\frac{4 k \pi+54 k-54 k}{4 \pi+54}} \\ & =\frac{1}{2 \sqrt{\pi}} \cdot \sqrt{\frac{4 k \pi}{4 \pi+54}}=\frac{\sqrt{k}}{\sqrt{4 \pi+54}}=\frac{1}{3} x \end{aligned}$$

$$ \begin{aligned} & \Rightarrow \quad x=3 r \quad\text{Hence proved.}\\ & \therefore \text { Minimum sum of volume, } \end{aligned}$$

$\therefore$ Minimum sum of volume,

$$\begin{aligned} \left.V_{\left(x=3 \cdot \sqrt{\frac{k}{4 \pi+54}}\right.}\right) & =\frac{2}{3} x^3+\frac{4}{3} \pi r^3=\frac{2}{3} x^3+\frac{4}{3} \pi \cdot\left(\frac{1}{3} x\right)^3 \\ & =\frac{2}{3} x^3+\frac{4}{3} \pi \cdot \frac{x^3}{27}=\frac{2}{3} x^3\left(1+\frac{2 \pi}{27}\right) \end{aligned}$$