Chapter 6: Applications of Derivatives

Ans: Let $f\left( x \right)={{x}^{4}}-62{{x}^{2}}+ax+9$ 

$\therefore {f}'\left( x \right)=4{{x}^{2}}-124x+a$ 

It is given that function $f$ attains its maximum value on the interval 

$\left[ 0,2 \right]$ at $x=1$.

$  \therefore {f}'\left( 1 \right)=0 $  

$  \Rightarrow 4-124+a=0 $  

$  \Rightarrow a=120 $  

Hence, the value of $a$ is $120.$

Ans: Let $f\left( x \right)=x+\sin 2x$ 

$  \therefore {f}'\left( x \right)=1+2\cos 2x $  

$  Now,{f}'\left( x \right)=0\Rightarrow \cos 2x=-\frac{1}{2}=\cos \frac{2\pi }{3} $  

$  2x=2\pi \pm \frac{2\pi }{3},n\in Z $  

$  \Rightarrow x=\pi \pm \frac{\pi }{3},n\in Z $  

$  \Rightarrow x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}\in \left[ 0,2\pi  \right] $  

Then, we evaluate the value of at $f$critical points $\Rightarrow x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}$ and the end points of the interval $\left[ 0,2\pi  \right]$.

$  f\left( \frac{\pi }{3} \right)=\frac{\pi }{3}+\sin \frac{2\pi }{3}=\frac{\pi }{3}+\frac{\sqrt{3}}{2} $  

$  f\left( \frac{2\pi }{3} \right)=\frac{2\pi }{3}+\sin \frac{4\pi }{3}=\frac{2\pi }{3}-\frac{\sqrt{3}}{2} $  

$  f\left( \frac{4\pi }{3} \right)=\frac{4\pi }{3}+\sin \frac{8\pi }{3}=\frac{4\pi }{3}+\frac{\sqrt{3}}{2} $  

$  f\left( \frac{5\pi }{3} \right)=\frac{5\pi }{3}+\sin \frac{10\pi }{3}=\frac{5\pi }{3}-\frac{\sqrt{3}}{2} $  

$  f\left( 0 \right)=0+\sin 0=0 $  

$  f\left( 2\pi  \right)=2\pi +\sin 4\pi =2\pi  $  

Hence, we can conclude that the absolute maximum value of $f\left( x \right)$ in 

the interval $\left[ 0,2\pi  \right]$ is $2\pi $ occurring at $x=2\pi $ and the absolute maximum value of $f\left( x \right)$ in the interval $\left[ 0,2\pi  \right]$ is $0$ occurring at $x=0.$

Ans: Let one number be $x$. Then, the other number is $\left( 24-x \right)$.

Let $p\left( x \right)$ denote the product of the two numbers. Thus, we have:

$  P\left( x \right)=x\left( 24-x \right)=24x-{{x}^{2}} $  

$  \therefore {P}'\left( x \right)=24-2x $  

$  {P}''\left( x \right)=-2 $  

$  Now, $  

$  {P}'\left( x \right)=0\Rightarrow x=12 $  

$  Also, $  

$  {P}''\left( 12 \right)=-2<0 $  

By second derivative test, $x=12$ is the point of local maxima of P.

Hence, the product of the numbers is the maximum when the numbers are  $12$ and $24-12=12.$ 

Ans: It is assumed that the two numbers are $x$ and $y$ such that $x+y=60$.

$y=60-x$ 

Let $f\left( x \right)=x{{y}^{3}}$ 

$  \Rightarrow f\left( x \right)=x{{\left( 60-x \right)}^{3}} $  

$  \therefore {f}'\left( x \right)={{\left( 60+x \right)}^{3}}-3x{{\left( 60-x \right)}^{2}} $  

$  ={{\left( 60+x \right)}^{3}}\left[ 60-4x \right] $  

$  And, $  

$  {f}''\left( x \right)=-2\left( 60-x \right)\left( 60-4x \right)-4{{\left( 60-x \right)}^{2}} $  

$  =-2\left( 60-x \right)\left( 180-6x \right) $  

$  =-12\left( 60-x \right)\left( 30-x \right) $  

$  Now,{f}'\left( x \right)=0 $  

$  \Rightarrow x=60,15 $  

When $x=60,{f}''\left( x \right)=0$ 

When $x=15,{f}''\left( x \right)=-12\left( 60-15 \right)\left( 30-15 \right)<0.$ 

By second derivative test, $x=15$ is a point of local maxima of f.

Thus, function $x{{y}^{3}}$ maximum when $x=15$ and $y=60-15=45$.

Hence, the required numbers are 15 and 45.

Ans: Let one number be x. Then, the other number is $y=\left( 35-x \right)$ 

Let $p\left( x \right)={{x}^{2}}{{y}^{5}}$. Then, we have:

$  P\left( x \right)={{x}^{2}}{{\left( 35-x \right)}^{5}} $  

$  \therefore {P}'\left( x \right)=2x{{\left( 35-x \right)}^{5}}-5{{x}^{2}}{{\left( 35-x \right)}^{4}} $  

$  =x{{\left( 35-x \right)}^{4}}\left[ 2\left( 35-x \right)-5x \right] $  

$  =7x{{\left( 35-x \right)}^{4}}\left( 10-x \right) $  

$  And, $  

$  {P}''\left( x \right)=7{{\left( 35-x \right)}^{4}}\left( 10-x \right)+7x\left[ -{{\left( 35-5 \right)}^{4}}-4{{\left( 35-x \right)}^{3}}\left( 10-x \right) \right] $  

$  =7{{\left( 35-x \right)}^{4}}\left( 10-x \right)-7x{{\left( 35-x \right)}^{4}}-28x{{\left( 35-x \right)}^{3}}\left( 10-x \right) $  

$  =7{{\left( 35-x \right)}^{3}}\left[ 350-45x+{{x}^{2}}-35x+{{x}^{2}}-40x+4{{x}^{2}} \right] $  

$  =7{{\left( 35-x \right)}^{3}}\left( 6{{x}^{2}}-120x+350 \right) $  

$  Now,{P}'\left( x \right)=0\Rightarrow x=0,x=35,x=10 $  

$  When,x=35,{f}'\left( x \right)=f\left( x \right)=0 $  

$  y=35-35=0. $  

This makes the product equal to 0.

When $x=0,y=35-0=35$ and the product ${{x}^{2}}{{y}^{5}}$ will be 0.

$x=0$ and $x=35$ cannot be the possible values of x.

When $x=10$ we have:

${P}''\left( x \right)=7{{\left( 35-10 \right)}^{3}}\left( -250 \right)<0$ 

By second derivative test, P(x) will be the maximum when $x=10$ and 

$y=35-10=25$.

Hence, the required numbers are 10 and 25.

Ans: Let one number be x. Then, the other number is $\left( 16-x \right)$.

Let the sum of the cubes of these numbers be denoted by S(x). Then, 

$  S\left( x \right)={{x}^{3}}+{{\left( 16-x \right)}^{3}} $  

$  \therefore {S}'\left( x \right)=3{{x}^{2}}-3{{\left( 16-x \right)}^{2}}, $  

$  {S}''\left( x \right)=6x+6\left( 16-x \right) $  

$  Now,{S}'\left( x \right)=0\Rightarrow 3{{x}^{2}}-3{{\left( 16-x \right)}^{2}}=0 $  

$  \Rightarrow {{x}^{2}}-{{\left( 16-x \right)}^{2}}=0 $  

$  \Rightarrow x=\frac{256}{32}=8 $  

$  Now, $  

$  {S}''\left( 8 \right)=6\left( 8 \right)+6\left( 16-8 \right)=96>0 $  

By second derivative test, x = 8 is the point of local minima of S.

Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 –8= 8.

Ans: Let the side of the square be cut off by x cm. 

Then, the length and the breath of the box will be (18 – 2x) cm each.

The box has a height of x cm. As a result, the box's volume V(x) is given by,

$  V\left( x \right)=x{{\left( 18-2x \right)}^{2}} $  

$  \therefore {V}'\left( x \right)={{\left( 18-2x \right)}^{2}}-4x\left( 18-2x \right) $  

$  =\left( 18-2x \right)\left[ 18-2x-4x \right] $  

$  =12\left( 9-x \right)\left( 3-x \right) $  

$  And, $  

$  {V}'\left( x \right)=12\left[ -\left( 9-x \right)-\left( 3-x \right) \right] $  

$  =-24\left( 6-x \right) $  

$  Now,{v}'\left( x \right)=0\Rightarrow x=9,3 $  

If x = 9, then the length and the breadth will become 0.

$  \therefore x=9 $  

$  \Rightarrow x=3 $  

$  Now,{V}''\left( 3 \right)=-24\left( 6-3 \right)=-72<0 $  

Therefore, by second derivative test, x = 3 is the point of maxima of V.

As a result, the volume of the box obtained is the biggest achievable by removing a square of side 3 cm from each corner of the square tin and building a box from the remaining sheet.

Ans: Let the side of the square be cut by x cm.

Then, the height of the box is x, the length is $45-2x$ , and the breadth is $24-2x$.

$  V\left( x \right)=x\left( 45-2x \right)\left( 24-2x \right) $  

$  =4{{x}^{3}}-138{{x}^{2}}+1080x $  

$  {V}'\left( x \right)=12{{x}^{2}}-276+1080 $  

$  =12\left( {{x}^{2}}-23x+90 \right) $  

$  =12\left( x-18 \right)\left( x-5 \right) $  

$  {V}''\left( x \right)=24x-276=12\left( 2x-23 \right) $  

$  Now,{V}'\left( x \right)=0\Rightarrow x=18,5 $  

It is not possible to cut off a square of side 18 cm from each corner of the rectangular sheet,

Thus, x cannot be equal to 18.

$  x=5 $  

$  Now,{V}''\left( 5 \right)=12\left( 10-23 \right)=-156<0 $  

Therefore, by second derivative test x = 5 is the point of maxima.

Hence, the side of the square to be cut off to make the volume of the box maximum possible is 5cm.

Ans: Let a rectangle of length l and breadth b be inscribed in the given circle of radius a.

Then, the diagonal of length 2a cm passes through the center.

By Pythagoras theorem, we have:

$  {{\left( 2a \right)}^{2}}={{l}^{2}}+{{b}^{2}} $  

$  \Rightarrow {{b}^{2}}=4{{a}^{2}}-{{l}^{2}} $  

$  \Rightarrow b=\sqrt{4{{a}^{2}}-{{l}^{2}}} $  

Area of the rectangle,$A=l\sqrt{4{{a}^{2}}-{{l}^{2}}}$ 

$  \therefore \frac{dA}{dl}=\sqrt{4{{a}^{2}}-{{l}^{2}}}+l\frac{1}{2\sqrt{4{{a}^{2}}-{{l}^{2}}}}\left( -2l \right)=\sqrt{4{{a}^{2}}-{{l}^{2}}}+\frac{{{l}^{2}}}{\sqrt{4{{a}^{2}}-{{l}^{2}}}}=\frac{4{{a}^{2}}-2{{l}^{2}}}{\sqrt{4{{a}^{2}}-{{l}^{2}}}} $  

$  \frac{{{d}^{2}}A}{d{{l}^{2}}}=\frac{\sqrt{4{{a}^{2}}-{{l}^{2}}}\left( -4l \right)-\left( 4{{a}^{2}}-2{{l}^{2}} \right)\frac{\left( -2l \right)}{2\sqrt{4{{a}^{2}}-{{l}^{2}}}}}{\left( 4{{a}^{2}}-{{l}^{2}} \right)} $  

$  =\frac{\left( 4{{a}^{2}}-{{l}^{2}} \right)\left( -4l \right)+1\left( 4{{a}^{2}}-2{{l}^{2}} \right)}{{{\left( 4{{a}^{2}}-{{l}^{2}} \right)}^{\frac{3}{2}}}} $  

$  =\frac{-2l\left( 6{{a}^{2}}-{{l}^{2}} \right)}{{{\left( 4{{a}^{2}}-{{l}^{2}} \right)}^{\frac{3}{2}}}} $  

$Now,\frac{dA}{dl}=0$ gives $4{{a}^{2}}=2{{l}^{2}}\Rightarrow l=\sqrt{2a}$ 

$\Rightarrow b=\sqrt{4{{a}^{2}}-2{{a}^{2}}}=\sqrt{2{{a}^{2}}}=\sqrt{2a}$ 

Now, when $l=\sqrt{2a}$ 

$\frac{{{d}^{2}}A}{d{{l}^{2}}}=\frac{-2\left( \sqrt{2a} \right)\left( -6{{a}^{2}}-2{{a}^{2}} \right)}{2\sqrt{2{{a}^{3}}}}=\frac{-8\sqrt{2{{a}^{3}}}}{2\sqrt{2{{a}^{3}}}}=-4<0$ 

By second derivative test, when $l=\sqrt{2a}$ then the area of the rectangle is the maximum.

Since $l=b=\sqrt{2a}$the rectangle is a square.

Hence, it has been proved that the square has the maximum area of all the rectangles inscribed in the given fixed circle. 

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

Then, the cylinder has the following Surface Area (S):

$  S=2\pi {{r}^{2}}+2\pi rh $  

$  \Rightarrow h=\frac{S-2\pi {{r}^{2}}}{2\pi r} $  

$  =\frac{S}{2\pi }\left( \frac{1}{r} \right)-r $  

Let V be the volume of the cylinder. Then,

$  V=\pi {{r}^{2}}h=\pi {{r}^{2}}\left[ \frac{S}{2\pi }\left( \frac{1}{r} \right)-r \right]=\frac{Sr}{2} $  

$  =-\pi {{r}^{3}}=\frac{S}{2\pi }\left( \frac{1}{r} \right)-r $  

$  Then,\frac{dV}{dr}=\frac{S}{2}-3\pi {{r}^{2}},\frac{{{d}^{2}}V}{d{{r}^{2}}}=-6\pi r $  

$  Now,\frac{dV}{dr}=0\Rightarrow \frac{S}{2}=3\pi {{r}^{2}}\Rightarrow {{r}^{2}}=\frac{S}{6\pi } $  

When ${{r}^{2}}=\frac{S}{6\pi }$ then $h=\frac{6\pi {{r}^{2}}}{2\pi }\left( \frac{1}{r} \right)-r=3r-r=2r$ 

Hence, the volume is the maximum when the height is twice the radius i.e. when the height is equal to the diameter.