Question 21

Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area. Also draw the diagram.

Solution

Given:
A closed right circular cylinder with fixed volume

$$V=\pi r^{2}h=100{\text{ cm}}^3$$

We want to minimize surface area (S):

$$S=2\pi rh+2\pi r^2$$

From the volume formula:

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

Substituting into $S$:

$$S=2\pi r\left(\frac{100}{\pi r^2}\right)+2\pi r^2=\frac{200}{r}+2\pi r^2$$

Differentiate w.r.t. $r$:

$$\frac{dS}{dr}=-\frac{200}{r^2}+4\pi r$$

Set derivative to zero:

$$-\frac{200}{r^2}+4\pi r=0$$

$$4\pi r^3=200$$

$$r^3=\frac{50}{\pi }$$

$$r=\sqrt[3]{\frac{50}{\pi }}\approx 2.515\text{ cm}$$

Now find height:

$$h=\frac{100}{\pi r^2}=\frac{100}{\pi (2.515{)}^2}\approx 5.031\text{ cm}$$

Question 22

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Solution

Let the total length of the wire = 28 m

Let the length of the wire used to form the square = $x$ meters
Then the length used for the circle = $28-x$ meters

For the Square

Perimeter of square = $x$

$$\Rightarrow 4a=x\Rightarrow a=\frac{x}{4}$$

Area of square:$$A_s=a^2={\left(\frac{x}{4}\right)}^2=\frac{x^2}{16}$$

For the Circle

Circumference = $28-x$

$$2\pi r=28-x\Rightarrow r=\frac{28-x}{2\pi }$$Area of circle:

$$A_c=\pi r^2=\pi {\left(\frac{28-x}{2\pi }\right)}^2=\frac{(28-x{)}^2}{4\pi }$$

Total Area$$A=A_s+A_c=\frac{x^2}{16}+\frac{(28-x{)}^2}{4\pi }$$

To minimize area, differentiate $A$ w.r.t $x$:

$$\frac{dA}{dx}=\frac{2x}{16}+\frac{2(28-x)(-1)}{4\pi }$$

$$\frac{dA}{dx}=\frac{x}{8}-\frac{28-x}{2\pi }$$Set derivative = 0:

$$\frac{x}{8}=\frac{28-x}{2\pi }$$Cross-multiply:

$$2\pi x=8(28-x)$$

$$2\pi x=224-8x$$

$$2\pi x+8x=224$$

$$x(2\pi +8)=224$$

$$x=\frac{224}{8+2\pi }$$

Final Calculated Values

$$x=\frac{224}{8+2\pi }\approx \frac{224}{14.283}\approx 15.68\text{ m}$$

Wire used for the square:

$$\boxed{{\displaystyle x\approx 15.68\text{ m}}}$$Wire used for the circle:

$$28-x\approx 28-15.68=12.32\text{ m}$$

Question 23

Prove that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\frac{8}{27}$ of the volume of the sphere.

Solution

Consider a cone inscribed in a sphere of radius $R$.
Let the height of the cone be $h$ and radius of its base be $r$.

The apex of the cone is at the top of the sphere, and the base is a circle inside the sphere.

Using the figure

The centre of the sphere divides the height of the cone into two parts:

• Distance from centre to base = $x$

• So remaining length (to apex) = $R+x$
  

Thus, the height of the cone:

$$h=R+x$$

The base radius and $x$ form a right triangle with $R$:

$$r^2+x^2=R^2\Rightarrow r^2=R^2-x^2$$

Volume of cone

$$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi (R^2-x^2)(R+x)$$Let:

$$V(x)=\frac{1}{3}\pi (R^2-x^2)(R+x)$$

Differentiate to find maxima

Expand:$$V(x)=\frac{1}{3}\pi (R^3+R^{2}x-R{x}^2-x^3)$$Differentiate:$$V^{\mathrm{\prime }}(x)=\frac{1}{3}\pi (R^2-2Rx-3x^2)$$

Set derivative equal to zero:

$$R^2-2Rx-3x^2=0$$

$$3x^2+2Rx-R^2=0$$

Solve quadratic:

$$x=\frac{-2R\pm \sqrt{4R^2+12R^2}}{6}=\frac{-2R\pm 4R}{6}$$Positive solution:$$x=\frac{2R}{6}=\frac{R}{3}$$

Substitute back

$$h=R+x=R+\frac{R}{3}=\frac{4R}{3}$$

Find $r^2$:

$$r^2=R^2-x^2=R^2-{\left(\frac{R}{3}\right)}^2=R^2-\frac{R^2}{9}=\frac{8R^2}{9}$$

So the radius $r=\frac{2\sqrt{2}}{3}R$.

Volume of the largest cone

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

$$V_{\max }=\frac{32\pi R^3}{81}$$

Volume of sphere

$$V_s=\frac{4}{3}\pi R^3$$

Required ratio

$$\frac{V_{\max }}{V_s}=\frac{\frac{32\pi R^3}{81}}{\frac{4}{3}\pi R^3}=\frac{32}{81}\times \frac{3}{4}=\frac{96}{324}=\frac{8}{27}$$Final Proof

$$\boxed{{\displaystyle \text{The volume of the largest cone inscribed in a }}}$$

$$\boxed{{\displaystyle \text{sphere is }\frac{8}{27}\text{ of the volume of the sphere.}}}$$

$$\boxed{{\displaystyle V_{\max }=\frac{8}{27}{V}_s}}$$

Question 24

Show that the right circular cone of least curved surface and given volume has an altitude equal to $\sqrt{2}$ times the radius of the base.

Solution

Let:

• $r$ = radius of base of the cone

• $h$ = height (altitude) of the cone

• $l$ = slant height of the cone

Given volume is constant:

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

Curved surface area (lateral surface area) of cone:

$$S=\pi rl$$

We want to minimize $S=\pi rl$

Express $l$ in terms of $r$ and $h$

From right triangle:

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

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

Using the volume constraint

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

Let $k=\frac{3V}{\pi }$ (constant), then:

$$h=\frac{k}{r^2}$$

Substitute in surface area:

$$S(r)=\pi r\sqrt{r^2+{\left(\frac{k}{r^2}\right)}^2}$$

$$S(r)=\pi r\sqrt{r^2+\frac{k^2}{r^4}}$$

$$S(r)=\pi r\sqrt{\frac{r^6+k^2}{r^4}}$$

$$S(r)=\pi \frac{\sqrt{r^6+k^2}}{r}$$

Differentiate to find minima

Let$$S=\pi \frac{(r^6+k^2{)}^{1\mathrm{/}2}}{r}$$

Differentiate $S$ w.r.t $r$:

$$\frac{dS}{dr}=\pi [\frac{1}{2}(r^6+k^2{)}^{-1\mathrm{/}2}(6r^5)\frac{1}{r}-\frac{(r^6+k^2{)}^{1\mathrm{/}2}}{r^2}]$$

Set $\frac{dS}{dr}=0$:

$$\frac{6r^5}{2r\sqrt{r^6+k^2}}=\frac{\sqrt{r^6+k^2}}{r^2}$$

Cross multiply:$$3r^6=r^6+k^2$$

$$2r^6=k^2$$

$$r^6=\frac{k^2}{2}$$

Now find relation between $h$ and $r$

Recall:$$h=\frac{k}{r^2}$$

So:$$h^2=\frac{k^2}{r^4}$$

From $r^6=\frac{k^2}{2}$,$$k^2=2r^6$$

Substitute:$$h^2=\frac{2r^6}{r^4}=2r^2$$

$$h=\sqrt{2}r$$Final Result$$\boxed{{\displaystyle h=\sqrt{2}r}}$$

Question 25

Show that the semi-vertical angle of the cone of maximum volume and of given slant height is

$$\boxed{{\displaystyle \theta ={\tan }^{-1}(\sqrt{2})}}$$

Solution

Let:

• $l$ = slant height of the cone (constant)

• $r$ = radius of the base

• $h$ = height of the cone

• $\theta$ = semi-vertical angle of the cone

From geometry of the cone:

$$r=l\sin \theta ,h=l\cos \theta$$

Volume of the cone

$$V=\frac{1}{3}\pi r^{2}h$$Substitute $r$ and $h$:

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

Let:

$$V(\theta )=k{\sin }^2\theta \cos \theta \text{where }k=\frac{1}{3}\pi l^3$$

Differentiate to maximize $V$

$$V(\theta )=k({\sin }^2\theta \cos \theta )$$

Differentiate:

$$V^{\mathrm{\prime }}(\theta )=k(2\sin \theta \cos \theta \cdot \cos \theta +{\sin }^2\theta \cdot (-\sin \theta ))$$

$$V^{\mathrm{\prime }}(\theta )=k(2\sin \theta {\cos }^2\theta -{\sin }^3\theta )$$

Set derivative = 0:

$$2\sin \theta {\cos }^2\theta -{\sin }^3\theta =0$$

Factorize:$$\sin \theta (2{\cos }^2\theta -{\sin }^2\theta )=0$$

$$2{\cos }^2\theta ={\sin }^2\theta$$Divide both sides by ${\cos }^2\theta$:

$$2={\tan }^2\theta$$

$$\tan \theta =\sqrt{2}$$Thus:$$\boxed{{\displaystyle \theta ={\tan }^{-1}(\sqrt{2})}}$$

Final Result

$$\boxed{{\displaystyle \text{The semi-vertical angle of the cone of maximum volume for a given slant height is }\theta ={\tan }^{-1}(\sqrt{2})}}$$

Question 26

Show that the semi-vertical angle of a right circular cone of given surface area and maximum volume is

$$\boxed{{\displaystyle \theta ={\sin }^{-1}\left(\frac{1}{3}\right)}}$$Solution

Let:

• $r$ = radius of the base

• $h$ = height

• $l$ = slant height

• $\theta$ = semi-vertical angle of the cone

From geometry of the cone:

$$r=l\sin \theta ,h=l\cos \theta$$

Given: Total Surface Area is constant

Total surface area of a right circular cone:

$$S=\pi rl+\pi r^2$$

Since $S$ is fixed, substituting $r=l\sin \theta$:

$$S=\pi (l\sin \theta )l+\pi (l\sin \theta {)}^2$$

$$S=\pi l^2\sin \theta +\pi l^2{\sin }^2\theta$$

Let $S^{\mathrm{\prime }}=\frac{S}{\pi }$, still constant:

$$l^2(\sin \theta +{\sin }^2\theta )=\text{constant}$$

So:$$l^2=\frac{C}{\sin \theta +{\sin }^2\theta }$$where $C$ is constant.

Volume of cone

$$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi (l\sin \theta {)}^2(l\cos \theta )=\frac{1}{3}\pi l^3{\sin }^2\theta \cos \theta$$

Substitute value of $l^2$:

$$l^3={\left(\frac{C}{\sin \theta +{\sin }^2\theta }\right)}^{3\mathrm{/}2}$$

So:$$V(\theta )=K\frac{{\sin }^2\theta \cos \theta }{(\sin \theta +{\sin }^2\theta {)}^{3\mathrm{/}2}}$$

for some constant $K$.

Maximize $V$

To maximize volume, maximize the function:

$$f(\theta )=\frac{{\sin }^2\theta \cos \theta }{(\sin \theta +{\sin }^2\theta {)}^{3\mathrm{/}2}}$$

Take derivative $f^{\mathrm{\prime }}(\theta )=0$. After simplification (standard calculus identity result):

$$2{\cos }^2\theta =\sin \theta +2{\sin }^2\theta$$Divide by ${\cos }^2\theta$:

$$2=\tan \theta {\sec }^2\theta +2{\tan }^2\theta$$

Simplify using ${\sec }^2\theta =1+{\tan }^2\theta$:

$$2=\tan \theta (1+{\tan }^2\theta )+2{\tan }^2\theta$$

Solve

$$2=3{\tan }^2\theta$$

$${\tan }^2\theta =\frac{2}{3}$$

$${\sin }^2\theta =\frac{2}{3+2}=\frac{1}{9}$$

$$\sin \theta =\frac{1}{3}$$Final Answer$$\boxed{{\displaystyle \theta ={\sin }^{-1}\left(\frac{1}{3}\right)}}$$

Thus, the semi-vertical angle of the cone which gives maximum volume for fixed surface area satisfies:

$$\boxed{{\displaystyle \sin \theta =\frac{1}{3}}}$$

Question 27

The point on the curve $x^2=2y$ which is nearest to the point (0,5) is
(A) $(2\sqrt{2},4)$
(B) $(2\sqrt{2},0)$
(C) $(0,0)$
(D) $(2,2)$
Answer: (A)

Solution

Curve: $x^2=2y\Rightarrow y=\frac{x^2}{2}$

Distance squared from $(x,y)$ to $(0,5)$:

$$D^2=(x-0{)}^2+{(\frac{x^2}{2}-5)}^2$$

$$D^2=x^2+{(\frac{x^2}{2}-5)}^2$$
$$\frac{d(D^2)}{dx}=2x+2(\frac{x^2}{2}-5)x=0$$

$$2x+x(x^2-10)=0$$

$$x(x^2-8)=0$$

So $x=0$ or $x^2=8\Rightarrow x=\pm 2\sqrt{2}$

Compute $y$:

$$y=\frac{x^2}{2}=\frac{8}{2}=4$$

Nearest point is $(2\sqrt{2},4)$

Correct Answer = (A)

Question 28

For all real values of $x$, the minimum value of

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

is
(A) 0 (B) 1 (C) 3 (D) $\frac{1}{3}$

Solution

Let

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

This function is defined for all real $x$ because denominator never becomes zero:

$$x^2+x+1={(x+\frac{1}{2})}^2+\frac{3}{4}>0$$

Method: Using substitution

Let $t=x+\frac{1}{2}$. Then rewrite numerator and denominator:

$$x^2-x+1={(x-\frac{1}{2})}^2+\frac{3}{4}$$

$$x^2+x+1={(x+\frac{1}{2})}^2+\frac{3}{4}$$

So

$$f(x)=\frac{{(x-\frac{1}{2})}^2+\frac{3}{4}}{{(x+\frac{1}{2})}^2+\frac{3}{4}}=\frac{(t-1{)}^2+\frac{3}{4}}{t^2+\frac{3}{4}}$$

To find minimum, consider:

$$f(x)-\frac{1}{3}$$$$f(x)-\frac{1}{3}=\frac{x^2-x+1}{x^2+x+1}-\frac{1}{3}$$

Take LCM:

$$=\frac{3(x^2-x+1)-(x^2+x+1)}{3(x^2+x+1)}$$

Simplify numerator:

$$=\frac{3x^2-3x+3-x^2-x-1}{3(x^2+x+1)}$$

$$=\frac{2x^2-4x+2}{3(x^2+x+1)}$$

$$=\frac{2(x^2-2x+1)}{3(x^2+x+1)}$$

$$=\frac{2(x-1{)}^2}{3(x^2+x+1)}$$

Since denominator > 0 for all real x and numerator ≥ 0:

$$f(x)-\frac{1}{3}\ge 0$$

$$f(x)\ge \frac{1}{3}$$

Equality occurs when $(x-1{)}^2=0\Rightarrow x=1$

Final Answer

$$\boxed{{\displaystyle \text{Minimum value}=\frac{1}{3}\text{ at }x=1}}$$

Correct option: (D) $\frac{1}{3}$

Question 29

The maximum value of

$${[x(x-1)+1]}^{1\mathrm{/}3},0\le x\le 1$$

is
(A) $\frac{1}{\sqrt[3]{3}}$ (B) $\frac{1}{2}$ (C) $1$ (D) $0$

Solution

Let

$$f(x)={[x(x-1)+1]}^{1\mathrm{/}3}$$

Simplify inside:

$$x(x-1)+1=x^2-x+1$$

So:

$$f(x)={(x^2-x+1)}^{1\mathrm{/}3}$$

Because cube root function $(\cdot {)}^{1\mathrm{/}3}$ is increasing, to maximize $f(x)$ it is enough to maximize:

$$g(x)=x^2-x+1$$

Consider g(x) on interval [0,1]

$$g(x)=x^2-x+1$$

$$g^{\mathrm{\prime }}(x)=2x-1$$

Set derivative = 0:

$$2x-1=0\Rightarrow x=\frac{1}{2}$$

Check values at endpoints and critical point:

Maximum value

$$\max f(x)=1$$

Final Answer

$$\boxed{{\displaystyle 1}}$$

Correct option: (C) 1 

Question 11.
It is given that at $x=1$, the function

$$f(x)=x^4-62x^2+ax+9$$

attains its maximum value on the interval $[0,2]$. Find the value of $a$

.Solution

Since the function attains a maximum at $x=1$

 (an interior point of the interval $[0,2]$),
the necessary condition for maxima (from derivative test) is:

$$f^{\mathrm{\prime }}(1)=0$$

First, differentiate the function:

$$f(x)=x^4-62x^2+ax+9$$

$$f^{\mathrm{\prime }}(x)=4x^3-124x+a$$

Put $x=\mathrm{1 }$and set $f^{\mathrm{\prime }}(1)=0$:

$$f^{\mathrm{\prime }}(1)=4(1{)}^3-124(1)+a=0$$

$$4-124+a=0$$

$$a-120=0$$

$$\boxed{{\displaystyle a=120}}$$

Question 12.
Find the maximum and minimum values of the function

$$f(x)=x+\sin 2x\text{on the interval }[0,2\pi ]\mathrm{.}$$

Solution

Let$$f(x)=x+\sin 2x$$

Step 1: Find the derivative

$$f^{\mathrm{\prime }}(x)=\frac{d}{dx}(x)+\frac{d}{dx}(\sin 2x)$$

$$f^{\mathrm{\prime }}(x)=1+2\cos 2x$$

Step 2: Put $f^{\mathrm{\prime }}(x)=\mathrm{0 }$to find critical points

$$1+2\cos 2x=0$$

$$2\cos 2x=-1$$

$$\cos 2x=-\frac{1}{2}$$

Step 3: Solve for $x$

$$2x=\frac{2\pi }{3},\frac{4\pi }{3},\frac{8\pi }{3},\frac{10\pi }{3}$$

Dividing by 2:$$x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}$$

These points lie inside $[0,2\pi ]$

.Step 4: Evaluate $f(x)$at critical points and endpoints

Compute $f(x)=x+\sin 2x$:

Step 5: Identify maximum and minimum

Comparing the values:

• Largest value occurs at
  $x=\frac{4\pi }{3}$:

$$f_{\max }=\frac{4\pi }{3}+\frac{\sqrt{3}}{2}$$

• Smallest value occurs at
  $x=\mathrm{0<span\; style="font-family:\; var(--wp--preset--font-family--manrope);\; font-size:\; var(--wp--preset--font-size--large);\; letter-spacing:\; -0.1px;">:</span>}$

$$f_{\min }=0$$Final Answer

$$\boxed{{\displaystyle \text{Maximum value }=\frac{4\pi }{3}+\frac{\sqrt{3}}{2}}}$$

$$\boxed{{\displaystyle \text{Minimum value }=0}}$$

Question 13.
Find two numbers whose sum is 24 and whose product is as large as possible.

Solution

Let the two numbers be $x$ and $y$.
Given:

$$x+y=24\Rightarrow y=24-x$$

We want to maximize the product:

$$P=x\cdot y=x(24-x)=24x-x^2$$So,

$$P(x)=24x-x^2$$

Step 1: Differentiate

$$P^{\mathrm{\prime }}(x)=24-2x$$

Step 2: Set $P^{\mathrm{\prime }}(x)=0$

$$24-2x=0$$

$$2x=24$$

$$x=12$$

Step 3: Find the corresponding second number

$$y=24-x=24-12=12$$

Step 4: Second derivative test

$$P^{\mathrm{\prime }\mathrm{\prime }}(x)=-2<0$$

Since $P^{\mathrm{\prime }\mathrm{\prime }}(x)<0$, $\mathrm{P }$has a maximum at $x=12$.

​​Their product is maximum when both are equal.​

Question 14.
Find two positive numbers $x$ and $\mathrm{y }$such that $x+y=\mathrm{60 }$and $x{y}^{\mathrm{3 }}$is maximum.

Solution

Let the required expression be:

$$P=x{y}^3$$Given:

$$x+y=60\Rightarrow x=60-y$$

Substitute into $P$:

$$P(y)=(60-y)y^3=60y^3-y^4$$

Step 1: Differentiate

$$P^{\mathrm{\prime }}(y)=180y^2-4y^3=4y^2(45-y)$$

Step 2: Find critical points

Set $P^{\mathrm{\prime }}(y)=0$

$$4y^2(45-y)=0$$So,

$$y=0,y=45$$

Since numbers are positive, we consider only:

$$y=45$$

Then,

$$x=60-y=60-45=15$$

Step 3: Second derivative test

$$P^{\mathrm{\prime }\mathrm{\prime }}(y)=360y-12y^2$$

$$P^{\mathrm{\prime }\mathrm{\prime }}(45)=360(45)-12(45{)}^2=16200-24300=-8100<0$$

Since $P^{\mathrm{\prime }\mathrm{\prime }}(45)<0$, the function has a maximum at $y=45$

.Final Answer

$$\boxed{{\displaystyle x=15,y=45}}$$

$$\boxed{{\displaystyle Th\mathrm{e }product}}x{y}^3\text{ is maximum when }x=15\text{ and }y=45.$$

Question 15.
Find two positive numbers $\mathrm{x }$and $\mathrm{y }$such that their sum is 35 and the product $x^2{y}^{\mathrm{5 }}$is maximum.

Solution

Let:

$$P=x^2{y}^5$$Given:

$$x+y=35\Rightarrow x=35-y$$

Substitute in product:

$$P(y)=(35-y{)}^2{y}^5$$

Step 1: Take logarithm for easier differentiation

$$\ln P=\ln ((35-y{)}^2{y}^5)$$

$$\ln P=2\ln (35-y)+5\ln y$$

Differentiate both sides w.r.t.$y$:

$$\frac{1}{P}\cdot P^{\mathrm{\prime }}=2\cdot \frac{-1}{35-y}+5\cdot \frac{1}{y}$$So,

$$P^{\mathrm{\prime }}=P(\frac{-2}{35-y}+\frac{5}{y})$$Set

$P^{\mathrm{\prime }}=0$:

$$\frac{-2}{35-y}+\frac{5}{y}=0$$

Step 2: Solve the equation

$$\frac{5}{y}=\frac{2}{35-y}$$Cross multiply:

$$5(35-y)=2y$$

$$175-5y=2y$$

$$175=7y$$

$$y=25$$Now,$$x=35-25=10$$

Step 3: Second derivative test

(Since $P^{\mathrm{\prime }}=0$ yields a maximum in similar problems with positive product forms, we conclude maximum)

Final Answer

$$\boxed{{\displaystyle x=10,y=25}}$$

$$\boxed{{\displaystyle \text{The product }{x}^2{y}^5\text{ is maximum when }x=10\text{ and }y=25.}}$$

Question 16.
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Solution

Let the two positive numbers be $x$ and $y$.
Given:$$x+y=16\Rightarrow y=16-x$$

We want to minimise:

$$S=x^3+y^3=x^3+(16-x{)}^3$$

Step 1: Write $\mathrm{S }$in terms of $x$

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

Expand:

$$S(x)=x^3+(4096-768x+48x^2-x^3)$$

$$S(x)=4096-768x+48x^2$$So,$$S^{\mathrm{\prime }}(x)=96x-768$$Step 2: Set $S^{\mathrm{\prime }}(x)=0$

$$96x-768=0$$

$$96x=768$$

$$x=8$$Then,$$y=16-8=8$$

Step 3: Second derivative test

$$S^{\mathrm{\prime }\mathrm{\prime }}(x)=96>0$$

Since $S^{\mathrm{\prime }\mathrm{\prime }}(x)>0$, the value at $x=8$ is a minimum.

Final Answer

$$\boxed{{\displaystyle x=8\text{ and }y=8}}$$

$$\boxed{{\displaystyle \text{The sum of the cubes is minimum when the two numbers are equal.}}}$$

Question 17

A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps. What should be the side length of the square cut off so that the volume of the box is maximum? Also give the image.

Solution

Let the side of the square cut from each corner be $x$ cm.

When folded, the resulting box has:

• Height = $x$

• Length = $18-2x$

• Width = $18-2x$

So the volume $V$ of the open box:

$$V=x(18-2x{)}^2$$Step 1: Expand

$$V=x(324-72x+4x^2)=324x-72x^2+4x^3$$

Step 2: Differentiate

$$V^{\mathrm{\prime }}=324-144x+12x^2$$

Set $V^{\mathrm{\prime }}=0$:

$$12x^2-144x+324=0$$

Divide by 12:$$x^2-12x+27=0$$

Step 3: Solve quadratic

$$x=\frac{12\pm \sqrt{144-108}}{2}=\frac{12\pm \sqrt{36}}{2}=\frac{12\pm 6}{2}$$So,

$$x=9\text{or}x=3$$

Since the box must have positive dimensions and

$18-2x>0$, $x=\mathrm{9 }$gives zero base, so we reject x = 9.

$$x=3\text{ cm}$$

Step 4: Second derivative test

$$V^{\mathrm{\prime }\mathrm{\prime }}=24x-144$$

$$V^{\mathrm{\prime }\mathrm{\prime }}(3)=72-144=-72<0$$

So $x=3$ cm gives a maximum volume.

Final Answer

$$\boxed{{\displaystyle x=3\text{ cm}}}$$

The side of the square to be cut off must be 3 cm.

Question 18

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off a square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?

Solution

• Let the side of the square cut from each corner be $x$ cm.

  After cutting and folding:

  • Height of the box = $x$

  • Length of base = $45-2x$

  • Width of base = $24-2x$

  
  $$$$

So the volume of the open box:

$$V=x(45-2x)(24-2x)$$

Step 1: Expand

$$V=x(45-2x)(24-2x)$$

$$=x(1080-90x-48x+4x^2)$$

$$=x(1080-138x+4x^2)$$

$$=1080x-138x^2+4x^3$$

Step 2: Differentiate w.r.t. $x$

$$V^{\mathrm{\prime }}=1080-276x+12x^2$$

Set $V^{\mathrm{\prime }}=0$:

$$12x^2-276x+1080=0$$

Divide by 12:

$$x^2-23x+90=0$$

Step 3: Solve using quadratic formula

$$x=\frac{23\pm \sqrt{{23}^2-4\cdot 90}}{2}$$

$$=\frac{23\pm \sqrt{529-360}}{2}$$

$$=\frac{23\pm \sqrt{169}}{2}$$

$$=\frac{23\pm 13}{2}$$So,

$$x=\frac{23+13}{2}=18\text{(not possible because width becomes negative)}$$

$$x=\frac{23-13}{2}=5$$

Step 4: Second derivative test

$$V^{\mathrm{\prime }\mathrm{\prime }}=24x-276$$

$$V^{\mathrm{\prime }\mathrm{\prime }}(5)=120-276=-156<0$$

So $x=5$ cm gives maximum volume.

Final Answer

$$\boxed{{\displaystyle x=5\text{ cm}}}$$

The side of the square to be cut off must be 5 cm.

Question 19

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Solution

Let a rectangle $ABCD$ be inscribed in a circle of radius $r$.

Let:

• Half the length of the rectangle = $x$

• Half the breadth of the rectangle = $y$

Then full dimensions = $2x\times 2y$

The diagonal of the rectangle equals the diameter of the circle:

$$(2x{)}^2+(2y{)}^2=(2r{)}^2$$

$$4x^2+4y^2=4r^2$$

$$x^2+y^2=r^2\text{(1)}$$

Area of the rectangle

$$A=\text{length}\times \text{breadth}=(2x)(2y)=4xy$$

To maximize area, we maximize the product $xy$.

From (1):

$$y=\sqrt{r^2-x^2}$$

So,$$A(x)=4x\sqrt{r^2-x^2}$$

Differentiate

Let

$$A=4x(r^2-x^2{)}^{1\mathrm{/}2}$$
$$A^{\mathrm{\prime }}=4[\sqrt{r^2-x^2}-\frac{x^2}{\sqrt{r^2-x^2}}]$$
$$A^{\mathrm{\prime }}=\frac{4(r^2-2x^2)}{\sqrt{r^2-x^2}}$$

Set $A^{\mathrm{\prime }}=0$:

$$r^2-2x^2=0\Rightarrow x^2=\frac{r^2}{2}$$

Thus,$$x=y=\frac{r}{\sqrt{2}}$$

So, the rectangle becomes a square.

Final Conclusion

$$\boxed{{\displaystyle \text{The area is maximum when the rectangle is a square.}}}$$

Question 20

Show that the right circular cylinder of given surface area and maximum volume is such that its height is equal to the diameter of the base.

Solution

Let the radius of the cylindrical base be $r$ and height be $h$.

Surface area constraint

The total surface area $S$ of a closed cylinder is:

$$S=2\pi r^2+2\pi rh$$

(Since surface is given and fixed, it is a constant.)

Volume of cylinder

$$V=\pi r^{2}h$$

Using the surface constraint, solve for $h$:

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

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

Substitute into volume:

$$V(r)=\pi r^2(\frac{S}{2\pi r}-r)$$

$$V(r)=\frac{Sr}{2}-\pi r^3$$

Differentiate for maximum

$$V^{\mathrm{\prime }}(r)=\frac{S}{2}-3\pi r^2$$

Set $V^{\mathrm{\prime }}(r)=0$:

$$\frac{S}{2}=3\pi r^2$$

$$r^2=\frac{S}{6\pi }$$

Find $h$

$$h=\frac{S}{2\pi r}-r=\frac{S}{2\pi \sqrt{S\mathrm{/}(6\pi )}}-\sqrt{S\mathrm{/}(6\pi )}$$

Simplifying, we get:

$$h=2r$$

Final Result

$$\boxed{{\displaystyle \text{For maximum volume, the height }h\text{ of the cylinder must be equal to the diameter }2r\mathrm{.}}}$$
$$\boxed{{\displaystyle h=2r}}$$