#### Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 38.

$h =\frac{l}{2}$

Hint: For maxima and minima value of S, we must have $\frac{dS}{dl}=0$

Given: Let l,h,v and S be the length, height, volume and surface area of the tank constructed

Since volume, v is constant  $l^2h=v\Rightarrow h=\frac{v}{l^2 }$                       →i

Solution:

Let S denote the total surface area, then we have $S=l^2+4lh$

$\Rightarrow S=l^2+4l\times \frac{v}{l^2 }$                             →fromi

$\Rightarrow S=l^2+\frac{4v}{l}$                              …(ii)

Differentiating S w.r.t 'l', we get

$\frac{dS}{dl}= \frac{d}{dl}\left (l^2+\frac{4v}{l} \right )$

\begin{aligned} &\Rightarrow \frac{d S}{d l}=\frac{d}{d l}\left(l^{2}\right)+\frac{d}{d l}\left(\frac{4 v}{l}\right) \quad\left[\because \frac{d}{d x}\left\{(f(x)+g(x)\}=\frac{d}{d x}(f(x))+\frac{d}{d x}(g(x))\right]\right. \\ &\begin{aligned} \Rightarrow \frac{d S}{d l}=\frac{d}{d l}\left(l^{2}\right)+4 v \frac{d}{d l}\left(\frac{1}{2}\right) \quad\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right] \\ \Rightarrow \frac{d S}{d l}=\frac{d}{d l}\left(l^{2}\right)+4 v \frac{d}{d l}\left(l^{-1}\right) \end{aligned} \\ &\Rightarrow \frac{d S}{d l}=2 l^{2-1}+4 v\left(-1 l^{-1-1}\right) \quad\left[\because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right] \end{aligned}

$\Rightarrow \frac{dS}{dl}=2l-\frac{4v}{l^2 }$                                    …(iii)

For maximum or minimum of S, we have

$\Rightarrow \frac{dS}{dl}=0$

$\Rightarrow 2l-\frac{4v}{l^2 }=0$                                 [using (iii)]⇒$\frac{2l^3-4v}{l^2}=0$]

$\Rightarrow 2l^3-4v=0$

$\Rightarrow 2l^3=4v\Rightarrow l^3=2v$                             …(iv)

Again, differentiating S w.r.t 'l', then $\frac{d^{2} S}{d l^{2}}=\frac{d}{d l}\left(\frac{d S}{d l}\right)=\frac{d}{d l}\left(2 l-\frac{4 v}{l^{2}}\right)$                     [using (iii)]

\begin{aligned} &\Rightarrow \frac{d^{2} S}{d l^{2}}=\frac{d}{d l}(2 l)+\frac{d}{d l}\left(\frac{-4 v}{l^{2}}\right) \quad\left[\because \frac{d}{d x}\left\{(f(x)+g(x)\}=\frac{d}{d x}(f(x))+\frac{d}{d x}(g(x))\right]\right. \\ &\begin{aligned} \Rightarrow \frac{d^{2} S}{d l^{2}}=2 \frac{d}{d l}(l)-4 v \frac{d}{d l}\left(\frac{1}{l^{2}}\right) &\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right] \\ & \Rightarrow \frac{d^{2} S}{d l^{2}}=2 \frac{d}{d l}\left(l^{1}\right)-4 v \frac{d}{d l}\left(l^{-2}\right) \end{aligned} \end{aligned}

\begin{aligned} &\begin{aligned} \Rightarrow \frac{d^{2} S}{d l^{2}}=2\left(1 l^{1-1}\right)-4 v\left(-2 l^{-2-1}\right) &\left[\because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right] \\ \Rightarrow \frac{d^{2} S}{d l^{2}}=2(1)+8 v\left(l^{-3}\right) \end{aligned} \\ &\begin{aligned} \Rightarrow \frac{d^{2} S}{d l^{2}} &=2+\frac{8 v}{l^{3}} \\ &=2+\frac{8}{2} \\ &=2+4 \end{aligned} \end{aligned}

$=6>0$

$\therefore \frac{d^2S}{dl^2}>0$ Hence the surface area is minimum, $h = \frac{v}{l^2}$ Substitute the value of $v=\frac{l^3}{2}$  from (iv) in eqni, we get $\frac{l^{3}}{2 l^{2}} \Rightarrow h=\frac{l}{2}$

Hence proved.