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

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

Answers (1)

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.

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