Given that, $$f(x)=x^5-5 x^4+5 x^3-1$$

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

$f^{\prime}(x)=5 x^4-20 x^3+15 x^2$

For maxima or minima, $f^{\prime}(x)=0$

$$\begin{aligned} \Rightarrow \quad& 5 x^4-20 x^3+15 x^2 =0 \\ \Rightarrow \quad& 5 x^2\left(x^2-4 x+3\right) =0 \\ \Rightarrow \quad& 5 x^2\left(x^2-3 x-x+3\right) =0 \\ \Rightarrow \quad& 5 x^2[x(x-3)-1(x-3)] =0 \\ \Rightarrow \quad& 5 x^2[(x-1)(x-3)] =0 \\ \therefore \quad& x =0,1,3 \end{aligned}$$

Sign scheme for $\frac{d y}{d x}=5 x^2(x-1)(x-3)$

So, $y$ has maximum value at $x=1$ and minimum value at $x=3$.

At $x=0, y$ has neither maximum nor minimum value.

$$\begin{aligned} & \therefore \quad \text { Maximum value of } y=1-5+5-1=0 \\ & \text { and } \\ & \text { minimum value }=(3)^5-5(3)^4+5(3)^3-1 \\ & =243-81 \times 5-27 \times 5-1=-298 \end{aligned}$$

Consider that company increases the annual subscription by ₹ $x$.

So, $x$ subscribes will discontinue the service.

$\therefore$ Total revenue of company after the increment is given by

$$\begin{aligned} R(x) & =(500-x)(300+x) \\ & =15 \times 10^4+500 x-300 x-x^2 \\ & =-x^2+200 x+150000 \end{aligned}$$

$$\begin{aligned} &\text { On differentiating both sides w.r.t. } x \text {, we get }\\ &\begin{array}{rlrl} & R^{\prime}(x) =-2 x+200 \\ \text { Now, } & R^{\prime}(x) =0 \\ \Rightarrow & 2 x =200 \Rightarrow x=100 \\ \therefore & R^{\prime \prime}(x) =-2<0 \\ \text { So, } R(x) \text { is maximum when } \quad x & =100 \end{array} \end{aligned}$$

Hence, the company should increase the subscription fee by ₹ $100$, so that it has maximum profit.

$$\begin{array}{lr} \text { Given, } & \text { line is } x \cos \alpha+y \sin \alpha=p \quad\text{... (i)}\\ \text { and } & \text { curve is } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \Rightarrow & b^2 x^2+a^2 y^2=a^2 b^2\quad\text{.... (ii)} \end{array}$$

$$\begin{aligned} &\text { Now, differentiating Eq. (ii) w.r.t. } x \text {, we get }\\ &b^2 \cdot 2 x+a^2 \cdot 2 y \cdot \frac{d y}{d x}=0 \end{aligned}$$

$$\begin{array}{lr} \Rightarrow & \frac{d y}{d x}=\frac{-2 b^2 x}{2 a^2 y}=\frac{-x b^2}{y a^2} \quad\text{.... (iii)}\\ \text { From Eq. (i), } & y \sin \alpha=p-x \cos \alpha \\ \Rightarrow & y=-x \cot \alpha+\frac{p}{\sin \alpha} \end{array}$$

Thus, slope of the line is $(-\cot \alpha)$.

So, the given equation of line will be tangent to the Eq. (ii), if $\left(-\frac{x}{y} \cdot \frac{b^2}{a^2}\right)=(-\cot \alpha)$

$$\begin{aligned} \Rightarrow \quad \frac{x}{a^2 \cos \alpha} & =\frac{y}{b^2 \sin \alpha}=k \quad\text{[say]}\\ \Rightarrow \quad x & =k a^2 \cos \alpha \\ \text { and } \quad y & =b^2 k \sin \alpha \end{aligned}$$

So, the line $x \cos \alpha+y \sin \alpha=p$ will touch the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ at point $\left(k a^2 \cos \alpha, k b^2 \sin \alpha\right)$

$\begin{aligned} & \text { From Eq. (i), } \quad k a^2 \cos ^2 \alpha+k b^2 \sin ^2 \alpha=p \\ & \Rightarrow \quad a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha=\frac{p}{k}\end{aligned}$

$\Rightarrow \quad\left(a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha\right)^2=\frac{p^2}{k^2}\quad\text{.... (iv)}$

$$\begin{array}{lrl} \text { From Eq. (ii), } & b^2 k^2 a^4 \cos ^2 \alpha+a^2 k^2 b^4 \sin ^2 \alpha & =a^2 b^2 \\ \Rightarrow & k^2\left(a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha\right) & =1 \\ \Rightarrow & \left(a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha\right) & =\frac{1}{k^2}\quad\text{.... (v)} \end{array}$$

$$\begin{aligned} &\text { On dividing Eq. (iv) by Eq. (v), we get }\\ &a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha=p^2 \end{aligned}$$

Hence proved.

Alternate Method

We know that, if a line $y=m x+c$ touches ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then the required condition is $c^2=a^2 m^2+b^2$

Here, given equation of the line is

$x \cos \alpha+y \sin \alpha=p$

$$ \begin{aligned} \Rightarrow \quad &y =\frac{p-x \cos \alpha}{\sin \alpha} \\ & =-x \cot \alpha+\frac{p}{\sin \alpha} \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & c=\frac{p}{\sin \alpha} \\ \text { and } & m=-\cot \alpha \end{array}$$

$\therefore\quad\left(\frac{p}{\sin \alpha}\right)^2=a^2(-\cot \alpha)^2+b^2$

$$\begin{array}{ll} \Rightarrow & \frac{p^2}{\sin ^2 \alpha}=a^2 \frac{\cos ^2 \alpha}{\sin ^2 \alpha}+b^2 \\ \Rightarrow & p^2=a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha\quad\text{Hence proved.} \end{array}$$

Let the length of side of the square base of open box be $x$ units and its height be $y$ units.

$$\begin{array}{lr} \therefore & \text { Area of the metal used }=x^2+4 x y \\ \Rightarrow & x^2+4 x y=c^2 \quad\text{[given]}\\ \Rightarrow & y=\frac{c^2-x^2}{4 x}\quad\text{.... (i)} \end{array}$$

$$\begin{aligned} &\quad \text { Now, } \quad \text { volume of the box }(V)=x^2 y\\ &\begin{aligned} \Rightarrow \quad V & =x^2 \cdot\left(\frac{c^2-x^2}{4 x}\right) \\ & =\frac{1}{4} x\left(c^2-x^2\right) \\ & =\frac{1}{4}\left(c^2 x-x^3\right) \end{aligned} \end{aligned}$$

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

$$\begin{aligned} & \frac{d V}{d x}=\frac{1}{4}\left(c^2-3 x^2\right) \\ \text{Now,}\quad & \frac{d V}{d x}=0 \Rightarrow c^2=3 x^2 \end{aligned}$$

$$ \begin{aligned} &\begin{array}{ll} \Rightarrow & x^2=\frac{c^2}{3} \\ \Rightarrow & x=\frac{c}{\sqrt{3}}\quad\text{[using positive sign]} \end{array}\\ &\text { Again, differentiating Eq. (ii) w.r.t. } x \text {, we get } \end{aligned}$$

$$\begin{aligned} \frac{d^2 V}{d x^2} & =\frac{1}{4}(-6 x)=\frac{-3}{2} x<0 \\ \therefore \quad\left(\frac{d^2 v}{d x^2}\right)_{\text {at } x=\frac{c}{\sqrt{3}}} & =-\frac{3}{2} \cdot\left(\frac{c}{\sqrt{3}}\right)<0 \end{aligned}$$

Thus, we see that volume $(V)$ is maximum at $x=\frac{c}{\sqrt{3}}$.

$\therefore$ Maximum volume of the box, $(V)_{x=\frac{c}{\sqrt{3}}}=\frac{1}{4}\left(c^2 \cdot \frac{c}{\sqrt{3}}-\frac{c^3}{3 \sqrt{3}}\right)$

$$\begin{aligned} & =\frac{1}{4} \cdot \frac{\left(3 c^3-c^3\right)}{3 \sqrt{3}}=\frac{1}{4} \cdot \frac{2 c^3}{3 \sqrt{3}} \\ & =\frac{c^3}{6 \sqrt{3}} \text { cu units } \end{aligned}$$

Let breadth and length of the rectangle be $x$ and $y$, respectively.

$$\begin{array}{lr} \because & \text { Perimeter of the rectangle }=36 \mathrm{~cm} \\ \Rightarrow & 2 x+2 y=36 \\ \Rightarrow & x+y=18 \\ \Rightarrow & y=18-x \quad\text{.... (i)} \end{array}$$

Let the rectangle is being revolved about its length $y$.

Then, volume $(V)$ of resultant cylinder $=\pi x^2 \cdot y$

$\Rightarrow \quad V=\pi x^2 \cdot(18-x) \quad\left[\because V=\pi r^2 h\right][$ [using Eq. (i) $]$

$=18 \pi x^2-\pi x^3=\pi\left[18 x^2-x^3\right]$

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

$$\begin{array}{llrl} & \frac{d V}{d x} =\pi\left(36 x-3 x^2\right) \\ \text { Now, } & \frac{d V}{d x} =0 \\ \Rightarrow & 36 x =3 x^2 \end{array}$$

$$\begin{aligned} \Rightarrow \quad& 3 x^2-36 x =0 \\ \Rightarrow \quad& 3\left(x^2-12 x\right) =0 \\ \Rightarrow \quad& 3 x(x-12) =0 \\ \Rightarrow \quad& x =0, x=12 \\ \therefore \quad& x =12\quad [\because, x \neq 0] \end{aligned}$$

$$\begin{aligned} &\text { Again, differentiating w.r.t. } x \text {, we get }\\ &\begin{aligned} \frac{d^2 V}{d x^2} & =\pi(36-6 x) \\ \Rightarrow \quad\left(\frac{d^2 V}{d x^2}\right)_{x=12} & =\pi(36-6 \times 12)=-36 \pi<0 \end{aligned} \end{aligned}$$

At $x=12$, volume of the resultant cylinder is the maximum.

So, the dimensions of rectangle are 12 cm and 6 cm, respectively. $\quad$ [using Eq. (i)]

$$\begin{aligned} &\therefore \text { Maximum volume of resultant cylinder, }\\ &\begin{aligned} (V)_{x=12} & =\pi\left[18 \cdot(12)^2-(12)^3\right] \\ & =\pi\left[12^2(18-12)\right] \\ & =\pi \times 144 \times 6 \\ & =864 \pi \mathrm{~cm}^3 \end{aligned} \end{aligned}$$