Chapter 6: Applications of Derivatives

Ans: Let r and h denote the cylinder’s radius and height respectively.

Then, the cylinder has the following volume (V),

$  V=\pi {{r}^{2}}h=100 $  

$  \therefore h=\frac{100}{\pi {{r}^{2}}} $  

The cylinder has the following surface area (S),

$  S=2\pi {{r}^{2}}+2\pi rh=2\pi {{r}^{2}}+\frac{200}{r} $  

$  \frac{dS}{dr}=0\Rightarrow 4\pi r=\frac{200}{{{r}^{2}}} $  

$  \Rightarrow {{r}^{3}}=\frac{200}{4\pi }=\frac{50}{\pi } $  

$  \Rightarrow r={{\left( \frac{50}{\pi } \right)}^{\frac{1}{3}}} $  

Now, it is observed that when $r={{\left( \frac{50}{\pi } \right)}^{\frac{1}{3}}},\frac{{{d}^{2}}S}{d{{r}^{2}}}>0.$ 

By second derivative test, when the radius of the cylinder is ${{\left( \frac{50}{\pi } \right)}^{\frac{1}{3}}}$ cm, the surface area is the minimum.

When $r={{\left( \frac{50}{\pi } \right)}^{\frac{1}{3}}},h=2{{\left( \frac{50}{\pi } \right)}^{\frac{1}{3}}}cm.$ 

Hence the required dimensions of the can which has the minimum surface area are given as $radius={{\left( \frac{50}{\pi } \right)}^{\frac{1}{3}}}cm$ and $height=2{{\left( \frac{50}{\pi } \right)}^{\frac{1}{3}}}cm.$ 

Ans: Let a piece of length l be cut from the given wire to make a square.

Then, the other piece of wire to be made into a circle is of length $\left( 28-l \right)m.$ 

Let r be the circle's radius. Then $2\pi r=28-l\Rightarrow r=\frac{1}{2\pi }\left( 28-l \right).$ 

The combined areas of the square and the circle (A) is given by,

$A={{\pi }^{2}}+{{r}^{2}}$ (side of the square)

$  =\frac{{{l}^{2}}}{16}+\pi {{\left[ \frac{1}{2\pi }\left( 28-l \right) \right]}^{2}} $  

$  =\frac{{{l}^{2}}}{16}+\frac{1}{4\pi }{{\left( 28-l \right)}^{2}} $  

$  \therefore \frac{dA}{dl}=\frac{2l}{16}+\frac{2}{4\pi }\left( 28-l \right)\left( -1 \right)=\frac{l}{8}-\frac{1}{2\pi }\left( 28-l \right) $  

$  \frac{{{d}^{2}}A}{d{{l}^{2}}}=\frac{l}{8}+\frac{1}{2\pi }>0 $  

$  Now,\frac{dA}{dl}=0\Rightarrow \frac{l}{8}-\frac{1}{2\pi }\left( 28-l \right)=0 $  

$  \Rightarrow \left( \pi +4 \right)l-112=0 $  

$  \Rightarrow l=\frac{112}{\pi +4} $  

Thus, when $l=\frac{112}{\pi +4},\frac{{{d}^{2}}A}{d{{l}^{2}}}>0$ 

By second derivative test, area (A) is the minimum when $l=\frac{112}{\pi +4}$ . Hence, when the length of the wire is$\frac{112}{\pi +4}$ cm in making the square the combined area is the minimum while the length of the wire in making the  circle is $28-\frac{112}{\pi +4}=\frac{28\pi }{\pi +4}cm.$

Ans: Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.

Let V be the volume of the cone.

Then, $V=\frac{1}{3}\pi {{r}^{2}}h$ 

Height of the cone is given by,

$h=R+AB=R+\sqrt{{{R}^{2}}-{{r}^{2}}}$                                                  (ABC is a right triangle)

$  V=\frac{1}{3}\pi {{r}^{2}}\left( R+\sqrt{{{R}^{2}}-{{r}^{2}}} \right) $  

$  =\frac{1}{3}\pi {{r}^{2}}R+\frac{1}{3}\pi {{r}^{2}}\sqrt{{{R}^{2}}-{{r}^{2}}} $  

$  \frac{dV}{dr}=\frac{2}{3}\pi rR+\frac{2}{3}\pi r\sqrt{{{R}^{2}}-{{r}^{2}}}+\frac{1}{3}\pi {{r}^{2}}.\frac{\left( -2r \right)}{2\sqrt{{{R}^{2}}-{{r}^{2}}}} $  

$  =\frac{2}{3}\pi rR+\frac{2}{3}\pi r\sqrt{{{R}^{2}}-{{r}^{2}}}-\frac{1}{3}\pi .\frac{{{r}^{3}}}{\sqrt{{{R}^{2}}-{{r}^{2}}}} $  

$  =\frac{2}{3}\pi rR+\frac{2\pi {{R}^{2}}-3\pi {{r}^{3}}}{3\sqrt{{{R}^{2}}-{{r}^{2}}}} $  

$  \frac{{{d}^{2}}V}{d{{r}^{2}}}=\frac{2}{3}\pi rR+\frac{9\left( {{R}^{2}}-{{r}^{2}} \right)\left( 2\pi {{R}^{2}}-9\pi {{r}^{2}} \right)+2\pi r{{R}^{2}}-3\pi {{r}^{4}}}{27{{\left( {{R}^{2}}-{{r}^{2}} \right)}^{\frac{3}{2}}}} $  

$  Now,\frac{dV}{dr}=0\Rightarrow \pi \frac{2}{3}rR=\frac{3\pi {{r}^{3}}-2\pi {{R}^{2}}}{3\sqrt{{{R}^{2}}-{{r}^{2}}}} $  

$  \Rightarrow 2R=2R\sqrt{{{R}^{2}}-{{r}^{2}}}=3{{r}^{2}}-2{{R}^{2}} $  

$  \Rightarrow 4{{R}^{4}}-4{{R}^{2}}{{r}^{2}}=9{{r}^{4}}+4R $  

$  \Rightarrow 9{{r}^{4}}=8{{R}^{2}}{{r}^{2}} $  

$  \Rightarrow {{r}^{2}}=\frac{8}{9}{{R}^{2}} $  

When ${{r}^{2}}=\frac{8}{9}{{R}^{2}}$ then $\frac{{{d}^{2}}V}{d{{r}^{2}}}<0$ 

By second derivative test, the volume of the cone is the maximum when   

${{r}^{2}}=\frac{8}{9}{{R}^{2}}$.

When ${{r}^{2}}=\frac{8}{9}{{R}^{2}},h=R+\sqrt{{{R}^{2}}-\frac{8}{9}{{R}^{2}}}=R+\frac{R}{3}=\frac{4}{3}R$ 

Therefore, 

$  =\frac{1}{3}\pi \left( \frac{8}{9}{{R}^{2}} \right)\left( \frac{4}{3}R \right) $  

$  =\frac{8}{27}\left( \frac{4}{3}\pi {{R}^{3}} \right) $  

$  =\frac{8}{27}\times \left( Volume \right) $  

Hence, the volume of the largest cone that can be inscribed in the sphere is $\frac{8}{27}$ the volume of the sphere.

Ans: Let r and h be the radius and the altitude of the cone respectively.

Then, the cone has the following volume (V):

$V=\frac{1}{3\pi }\pi {{r}^{2}}h\Rightarrow h=\frac{3V}{{{r}^{2}}}$ 

The cone has the following surface area (S),

$S=\pi rl$  (l = slant height)

$  =\pi r\sqrt{{{r}^{2}}+{{h}^{2}}} $  

$  =\pi r\sqrt{{{r}^{2}}+\sqrt{\frac{9{{V}^{2}}}{{{\pi }^{2}}{{r}^{4}}}}}=\pi \frac{r\sqrt{{{9}^{2}}{{r}^{2}}+{{V}^{2}}}}{\pi {{r}^{2}}} $  

$  =\frac{1}{r}\sqrt{{{\pi }^{2}}{{r}^{6}}+9{{V}^{2}}} $  

$  \frac{dS}{dr}=\frac{r.\frac{6{{\pi }^{2}}{{r}^{5}}}{2\sqrt{{{\pi }^{2}}{{r}^{6}}+9{{V}^{2}}}}-\sqrt{{{\pi }^{2}}{{r}^{6}}+9{{V}^{2}}}}{{{r}^{2}}} $  

$  =\frac{2{{\pi }^{2}}{{r}^{6}}-9{{V}^{2}}}{{{r}^{2}}\sqrt{{{\pi }^{2}}{{r}^{6}}+9{{V}^{2}}}} $  

$Now,\frac{dS}{dr}=0\Rightarrow 2{{\pi }^{2}}{{r}^{6}}=9{{V}^{2}}\Rightarrow {{r}^{6}}=\frac{9{{V}^{2}}}{2{{\pi }^{2}}}$ 

Thus, it can be easily verified that when ${{r}^{6}}=\frac{9{{V}^{2}}}{2{{\pi }^{2}}},\frac{{{d}^{2}}S}{d{{r}^{2}}}<0$ 

By second derivative test, the surface area of the cone is the least when ${{r}^{6}}=\frac{9{{V}^{2}}}{2{{\pi }^{2}}}$

When, ${{r}^{6}}=\frac{9{{V}^{2}}}{2{{\pi }^{2}}},h=\frac{3V}{\pi {{r}^{2}}}{{\left( \frac{2{{\pi }^{2}}{{r}^{6}}}{9} \right)}^{\frac{1}{2}}}=\frac{3}{\pi {{r}^{2}}}\frac{\sqrt{2\pi {{r}^{3}}}}{3}=\sqrt{2}r$

Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to $\sqrt{2}$ times the radius of the base.

Ans: Let $\theta $ be the semi-vertical angle of the cone.

It is clear that $\theta \in \left[ 0,\frac{\pi }{2} \right]$ 

Let r, h, and l be the radius, height, and the slant height of the cone respectively.

The slant height of the cone is given as constant.

Now r = l $\sin \theta $ and h = l $\cos \theta $ 

The volume (V) of the cone is given by,

$  V=\frac{1}{3}\pi {{r}^{2}}h $  

$  =\frac{1}{3}\pi \left( {{l}^{2}}{{\sin }^{2}}\theta  \right)\left( l\cos \theta  \right) $  

$  =\frac{1}{3}\pi {{l}^{3}}{{\sin }^{2}}\theta \cos \theta  $  

$  \therefore \frac{dV}{d\theta }=\frac{{{l}^{3}}\pi }{3}\left[ {{\sin }^{2}}\theta \left( -\sin \theta  \right)+\cos \theta \left( 2\sin \theta \cos \theta  \right) \right] $  

$  =\frac{{{l}^{3}}\pi }{3}\left[ -{{\sin }^{3}}\theta +2\sin \theta {{\cos }^{2}}\theta  \right] $  

$  \frac{{{d}^{2}}V}{d{{\theta }^{2}}}=\frac{{{l}^{3}}\pi }{3}\left[ -3{{\sin }^{2}}\theta \cos \theta +2{{\cos }^{3}}\theta -4{{\sin }^{2}}\theta \cos \theta  \right] $  

$  =\frac{{{l}^{3}}\pi }{3}\left[ 2{{\cos }^{2}}\theta -7{{\sin }^{2}}\theta \cos \theta  \right] $  

$  Now,\frac{dV}{d\theta }=0 $  

$  \Rightarrow {{\sin }^{3}}\theta =2\sin \theta {{\cos }^{2}}\theta  $  

$  \Rightarrow \tan \theta =2 $  

$  \Rightarrow \theta ={{\tan }^{-1}}\sqrt{2} $  

Now, when $\theta ={{\tan }^{-1}}\sqrt{2},$then ${{\tan }^{2}}\theta =2$ or ${{\sin }^{2}}\theta =2{{\cos }^{2}}\theta $ 

Then,

$\frac{{{d}^{2}}V}{d{{\theta }^{2}}}=\frac{{{l}^{3}}\pi }{3}\left[ 2{{\cos }^{3}}\theta -14{{\cos }^{3}}\theta  \right]=-4\pi {{l}^{3}}{{\cos }^{3}}\theta <0$     (For $\theta \in \left[ 0,\frac{\pi }{2} \right]$)

By second derivative test, the volume (V) is the maximum when $\theta ={{\tan }^{-1}}\sqrt{2}$. Hence, the semi-vertical angle of the cone of the maximum volume is  ${{\tan }^{-1}}\sqrt{2}$ for a given slant height.

Ans: Let’s assume x be the radius and y be the height of the cone and semi-vertical angle be $\Theta$. 

And, Total Surface area of cone (S) = $\pi x \sqrt{x^{2}+y^{2}}+\pi x^{2}$

$x\sqrt{x^{2}+y^{2}} +x^{2}=\frac{S}{\pi}$

Let’s say $\frac{S}{\pi}=k$

$x\sqrt{x^{2}+y^{2}} = k-x^{2}$

Squaring on both sides

$x^{2}\left ( x^{2}+y^{2} \right )=\left ( k-x^{2} \right )^{2}$

$x^{2}y^{2} = k^{2}-2kx^{2}$

$x^{2}=\frac{k^{2}}{y^{2}+2k}$   ……(1)

Volume of cone (V) = $\frac{1}{3}\pi x^{2}y$

= $\frac{1}{3}\pi \left ( \frac{k^{2}}{y^{2}+2k} \right )y$

= $\frac{1}{3}\pi k^{2}\left ( \frac{y}{y^{2}+2k} \right )$

Derive on both sides

$\frac{dV}{dy}= \frac{1}{3}\pi k^{2}\frac{d}{dy}\frac{y}{y^{2}+2k}$

= $\frac{1}{3}\pi k^{2}\left [ \frac{\left ( y^{2}+2k \right ).1-y.2y}{\left ( y^{2}+2k \right )^{2}} \right ]$    [Using quotient rule]

$\frac{dV}{dy}=\frac{1}{3}\pi k^{2}\frac{\left ( 2k-y^{2} \right )}{\left ( y^{2}+2k \right )^{2}}$   …….(2)

Now $\frac{dV}{dy}=0$

$\frac{1}{3}\pi k^{2}\frac{\left ( 2k-y^{2} \right )}{\left ( y^{2}+2k \right )^{2}}=0$

$2k-y^{2}=0$

$y^{2}=2k$

$y=\pm \sqrt{2k}$

Height can’t be negative so we neglect $y=- \sqrt{2k}$

So, $y=\sqrt{2k}$

As, $\frac{dV}{dy}> 0$, therefore, Volume is maximum at $y = \sqrt{2k}$

From equation (1),  We get,

$x^{2}=\frac{k^{2}}{2k+2k}=\frac{k^{2}}{4k}=\frac{k}{4}$

$x=\frac{\sqrt{k}}{2}$

Now, Semi-vertical angle of the cone  $sin\theta = \frac{x}{\sqrt{x^{2}+y^{2}}}$

= $\frac{\frac{\sqrt{k}}{2}}{\sqrt{\frac{k}{4}+2k}}=\frac{\sqrt{k}}{2}\times \sqrt{\frac{4}{9k}} = \frac{1}{3}$

Which implies

$\theta  = sin^{-1}\frac{1}{3}$

Ans: The given curve is ${{x}^{2}}=2y$ 

For each value of x, y will be $\frac{{{x}^{2}}}{2}$ the position of the point will be $\left( x,\frac{{{x}^{2}}}{2} \right)$

The distance d(x) between the points $\left( x,\frac{{{x}^{2}}}{2} \right)$ and $\left( 0,5 \right)$ is given by

$  d\left( x \right)=\sqrt{{{\left( x-0 \right)}^{2}}+{{\left( \frac{{{x}^{2}}}{2}-5 \right)}^{2}}}=\sqrt{\frac{{{x}^{4}}}{4}-4{{x}^{2}}+25} $  

$  \therefore {d}'\left( x \right)=\frac{\left( {{x}^{3}}-8x \right)}{\sqrt{{{x}^{4}}-16{{x}^{2}}+100}} $  

$  Now,{d}'\left( x \right)=0\Rightarrow {{x}^{3}}-8x=0 $  

$  \Rightarrow x=0,\pm 2\sqrt{2} $  

And ${d}''\left( x \right)=\frac{\left( {{x}^{4}}-16{{x}^{2}}+100 \right)\left( 3{{x}^{2}}-8 \right)-2{{\left( {{x}^{3}}-8x \right)}^{2}}}{{{\left( {{x}^{4}}-16{{x}^{2}}+100 \right)}^{\frac{3}{2}}}}$ 

When $x=0$ then ${d}''\left( x \right)=\frac{36\left( -8 \right)}{{{6}^{3}}}<0$ 

When $x=\pm 2\sqrt{2}$ then $d\left( x \right)>0$ 

By second derivative test, d(x) is the minimum at $x=\pm 2\sqrt{2}$.

When $x=\pm 2\sqrt{2},y=4.$

Hence, the point on the curve ${{x}^{2}}=2y$ which is nearest to the point $\left( 0,5 \right)$ 

is $\left( \pm 2\sqrt{2},4 \right).$ 

Therefore, the correct option is A.

Ans: Let $f\left( x \right)=\frac{1-x+{{x}^{2}}}{1+x+{{x}^{2}}}$ 

$  {f}'\left( x \right)=\frac{\left( 1+x+{{x}^{2}} \right)\left( -1-2x \right)-\left( 1-x+{{x}^{2}} \right)\left( 1+2x \right)}{{{\left( 1+x+{{x}^{2}} \right)}^{2}}} $  

$  =\frac{2\left( {{x}^{2}}-1 \right)}{{{\left( 1+x+{{x}^{2}} \right)}^{2}}} $  

$  \therefore {f}'\left( x \right)=0\Rightarrow {{x}^{2}}=1\Rightarrow x=\pm 1 $  

$  Now,{f}''\left( x \right)=4\frac{\left( 1+3x-{{x}^{3}} \right)}{{{\left( 1+x+{{x}^{2}} \right)}^{3}}} $  

$  And,{f}''\left( 1 \right)=\frac{4\left( 1+3-1 \right)}{{{\left( 1+1+1 \right)}^{3}}}=\frac{4}{9}>0 $  

$  And,{f}''\left( -1 \right)=\frac{4\left( 1-3+1 \right)}{{{\left( 1+1+1 \right)}^{3}}}=\frac{-4}{27}<0 $  

By second derivative test, f is the minimum at $x=1$ and the minimum 

value is given by $f\left( 1 \right)=\frac{1}{3}$ 

The correct option is D.

Ans: Let $f\left( x \right)={{\left[ x\left( x+1 \right)+1 \right]}^{\frac{1}{3}}}.$ 

$  \therefore {f}'\left( x \right)=\frac{2x+1}{3{{\left[ x\left( x+1 \right)+1 \right]}^{\frac{2}{3}}}} $  

$  Now,f\left( x \right)=0\Rightarrow x=-\frac{1}{2} $  

But, x = -1/2 is not part of the interval [0,1]

Now, evaluate the value of f at the endpoints of the interval [0, 1] {i.e., 

at x = 0 and x =1}.

$  f\left( 0 \right)={{\left[ 0\left( 0-1 \right)+1 \right]}^{\frac{1}{3}}}=1 $  

$  f\left( 0 \right)={{\left[ 1\left( 1-1 \right)+1 \right]}^{\frac{1}{3}}}=1 $  

Hence, we can conclude that the maximum value of  f is 1 in the 

interval [0, 1].

The correct option is 13.