Application of Derivatives Miscellaneous NCERT Solutions

1. Show that the function given by f(x)=\frac{\log x}{x} has maximum at x=e.

Solution

Given,

f(x)=\frac{\log x}{x}=(\log x)x^{-1}

Differentiating w.r.t. x by product rule, we get

\displaystyle f'(x)=x^{-1}\cdot\frac{1}{x}+(\log x)(-x^{-2})

\displaystyle =\frac{1-\log x}{x^2}

For critical points, put f'(x)=0

\displaystyle \frac{1-\log x}{x^2}=0

Since x^2\neq 0,

we get, 1-\log x=0

\log x=1

i.e. x=e

Now, differentiating f'(x), we have

\displaystyle f''(x)=\frac{d}{dx}\left(\frac{1-\log x}{x^2}\right)

Using the quotient rule,

We get, \displaystyle f''(x)=\frac{x^2\left(-\frac1x\right)-(1-\log x)(2x)}{x^4}

\displaystyle =\frac{-x-2x(1-\log x)}{x^4}

\displaystyle =\frac{x(2\log x-3)}{x^4}

So, \displaystyle f''(x)=\frac{2\log x-3}{x^3}

Substituting x=e, we get

\displaystyle f''(e)=\frac{2(1)-3}{e^3}

\displaystyle f''(e)=-\frac{1}{e^3}

Since f''(e) is negative, by the second derivative test, f(x) has a maximum at x=e.

Hence, the function f(x)=\frac{\log x}{x} has maximum at

\boxed{x=e}

Alternative Method (Using First Derivative Test)

We have

\displaystyle f'(x)=\frac{1-\log x}{x^2}

For critical points,

f'(x)=0

So, \displaystyle \frac{1-\log x}{x^2}=0

1-\log x=0 \Rightarrow \log x=1

i.e. x=e

Thus, x=e is the critical point.

For x\lt e, taking logarithms on both sides,

We get, \log x\lt \log e

\log x\lt 1

1-\log x\gt 0

Also, x^2\gt 0. Therefore,

\displaystyle f'(x)=\frac{1-\log x}{x^2}\gt 0

For x\gt e, taking logarithms on both sides,

We get, \log x\gt \log e

\log x\gt 1

1-\log x\lt 0

Also, x^2\gt 0. Therefore,

\displaystyle f'(x)=\frac{1-\log x}{x^2}\lt 0

Thus, f'(x) changes sign from positive to negative while moving from left to right through x=e.

Therefore, by the First Derivative Test, f(x) attains a maximum at

\boxed{x=e}

2. The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3\text{ cm/s}. How fast is the area decreasing when the two equal sides are equal to the base?

Solution

Let each of the equal sides be x cm.

Since the sides are decreasing at the rate of 3\text{ cm/s},

\displaystyle \frac{dx}{dt}=-3

Let h be the height of the triangle.

Since the altitude bisects the base, by Pythagoras theorem,

x^2=h^2+\left(\frac{b}{2}\right)^2

Therefore,

\displaystyle h=\sqrt{x^2-\frac{b^2}{4}}

The area of the triangle is

\displaystyle A=\frac{1}{2}bh

\displaystyle A=\frac{b}{2}\sqrt{x^2-\frac{b^2}{4}}

Differentiating both sides w.r.t t, we get

\displaystyle \frac{dA}{dt}=\frac{b}{2}\cdot\frac{1}{2\sqrt{x^2-\frac{b^2}{4}}}\cdot 2x\frac{dx}{dt}

\displaystyle \frac{dA}{dt}=\frac{bx}{2\sqrt{x^2-\frac{b^2}{4}}}\frac{dx}{dt}

When the two equal sides are equal to the base,

x=b

Substituting x=b and \frac{dx}{dt}=-3,

We get, \displaystyle \frac{dA}{dt}=\frac{b^2}{2\sqrt{b^2-\frac{b^2}{4}}}(-3)

\displaystyle =\frac{b^2}{2\left(\frac{\sqrt{3}}{2}b\right)}(-3)

\displaystyle =-\frac{3b}{\sqrt{3}}

i.e. \displaystyle \frac{dA}{dt} =-\sqrt{3}\,b

Hence, the area is decreasing at the rate

\boxed{\displaystyle b\sqrt{3} \text{cm}^2/\text{s}}

3. Find the intervals in which the function f(x)=\frac{4\sin x-2x-x\cos x}{2+\cos x} is (i) increasing (ii) decreasing.

Solution

Given,

\displaystyle f(x)=\frac{4\sin x-2x-x\cos x}{2+\cos x}

Using the quotient rule,

f'(x)=\frac{(2+\cos x)\left(4\cos x-2-\cos x+x\sin x\right)-(4\sin x-2x-x\cos x)(-\sin x)}{(2+\cos x)^2}

On simplification, we get

\displaystyle f'(x)=\frac{(4-\cos x)\cos x}{(2+\cos x)^2}

We know that

\displaystyle f'(x)=\frac{(4-\cos x)\cos x}{(2+\cos x)^2}

Now, we know -1\le \cos x\le 1

Multiplying by -1 on both sides, we get

1\ge -\cos x\ge -1

or -1\le -\cos x\le 1

Now adding 4 on both sides, we get

3\le 4-\cos x\le 5

Therefore, 4-\cos x\gt 0 for all values of x.

Again, adding 2 throughout -1\le \cos x\le 1,

We get, 1\le 2+\cos x\le 3, so

(2+\cos x)^2\gt 0

for all values of x. Hence, both 4-\cos x and (2+\cos x)^2 are always positive.

Therefore, the sign of f'(x) depends only on the sign of \cos x.

(i) Increasing

The function is increasing when

f'(x)>0

\Rightarrow \cos x>0

Therefore, in the interval 0\le x\le 2\pi,

0\le x\lt\frac{\pi}{2}\quad\text{and}\quad\frac{3\pi}{2}\lt x\lt 2\pi

Hence, the function is increasing on

\boxed{0\le x\lt\frac{\pi}{2}\text{ and }\frac{3\pi}{2}\lt x\lt 2\pi}

(ii) Decreasing

The function is decreasing when

f'(x)\lt 0

\Rightarrow \cos x\lt 0

Therefore, in the interval 0\le x\le 2\pi,

\frac{\pi}{2}\lt x\lt\frac{3\pi}{2}

Hence, the function is decreasing on

\boxed{\frac{\pi}{2}\lt x\lt\frac{3\pi}{2}}

4. Find the intervals in which the function f(x)=x^3+\frac{1}{x^3},\;x\neq 0 is (i) increasing (ii) decreasing.

Solution

Given,

\displaystyle f(x)=x^3+\frac{1}{x^3}

Differentiating w.r.t. x,

We get, \displaystyle f'(x)=3x^2-\frac{3}{x^4}

\displaystyle =\frac{3(x^6-1)}{x^4}

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

Since x^4>0 for x\neq 0, the sign of f'(x) depends on

(x^3-1)(x^3+1)

The critical points are

x=-1,\;1

These critical points divide the number line into three disjoint intervals, namely

(-\infty,-1),\quad(-1,1),\quad(1,\infty)

The critical points are x=-1 and x=1. However, since the function is not defined at x=0, the interval (-1,1) must be further divided into (-1,0) and (0,1).

Therefore, we examine the sign of f'(x) in the intervals (-\infty,-1), (-1,0), (0,1) and (1,\infty).

(i) Increasing

Hence, the function is increasing on

\boxed{x\lt -1\text{ and }x\gt 1}

(ii) Decreasing

Hence, the function is decreasing on

\boxed{-1\lt x\lt 1}

5. Find the maximum area of an isosceles triangle inscribed in the ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 with its vertex at one end of the major axis.

Solution

Let the vertex of the isosceles triangle be the end of the major axis (a,0).

Let the other two vertices be (x,y) and (x,-y) on the ellipse.

Then the base of the triangle is

2y

and the perpendicular distance of the vertex (a,0) from the base is

a-x

Hence, the area of the triangle is

\displaystyle A=\frac12(2y)(a-x)=y(a-x)

Since the point (x,y) lies on the ellipse,

\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1

\displaystyle y=b\sqrt{1-\frac{x^2}{a^2}}

Substituting in the expression for area,

\displaystyle A=b(a-x)\sqrt{1-\frac{x^2}{a^2}}

Squaring both sides, we get

\displaystyle A^2=b^2(a-x)^2\left(1-\frac{x^2}{a^2}\right)

\displaystyle =\frac{b^2}{a^2}(a-x)^3(a+x)

Let

\displaystyle F(x)=(a-x)^3(a+x)

Then maximizing A is equivalent to maximizing F(x).

Differentiating by product rule, we get

\displaystyle F'(x)=(a-x)^2\big[-3(a+x)+(a-x)\big]

\displaystyle =-2(a-x)^2(a+2x)

For critical points,

F'(x)=0

We get, (a-x)^2(a+2x)=0

\Rightarrow x=a\text{ (rejected),  or}\quad x=-\frac a2

Again using product rule,

Now, differentiating F'(x),

We get, \displaystyle F''(x)=\frac{d}{dx}\Big[-2(a-x)^2(a+2x)\Big]

=-2\left[\frac{d}{dx}(a-x)^2\cdot(a+2x)+(a-x)^2\cdot\frac{d}{dx}(a+2x)\right]

\displaystyle =-2\left[-2(a-x)(a+2x)+2(a-x)^2\right]

On simplification,

\displaystyle =4(a-x)(a+2x)-4(a-x)^2

\displaystyle =4(a-x)\Big[(a+2x)-(a-x)\Big]

So, \displaystyle f''(x)=12x(a-x)

Substituting x=-\frac a2,

We get, \displaystyle F''\!\left(-\frac a2\right)=12\left(-\frac a2\right)\left(a+\frac a2\right)

\displaystyle =-6a\left(\frac{3a}{2}\right)

\displaystyle =-9a^2

Since F''\!\left(-\frac a2\right)\lt 0, by the Second Derivative Test, F(x) and so the Area is maximum at

\displaystyle x=-\frac a2

Corresponding value of y

is \displaystyle y=b\sqrt{1-\frac{\left(-\frac a2\right)^2}{a^2}}

\displaystyle =b\sqrt{1-\frac14}

\displaystyle =\frac{\sqrt3\,b}{2}

Therefore, the maximum area

is \displaystyle A_{\max}=y(a-x)

\displaystyle =\frac{\sqrt3\,b}{2}\left(a+\frac a2\right)

\displaystyle =\frac{\sqrt3\,b}{2}\cdot\frac{3a}{2}=\frac{3\sqrt3}{4}ab

Hence, the maximum area of the isosceles triangle is

\boxed{\displaystyle \frac{3\sqrt3}{4}ab}

6. A tank with rectangular base and rectangular sides, open at the top, is to be constructed so that its depth is 2\text{ m} and volume is 8\text{ m}^3. If building of tank costs Rs 70 per sq metre for the base and Rs 45 per sq metre for sides, what is the cost of least expensive tank?

Solution

Let the length and breadth of the rectangular base be x m and y m respectively.

Since the depth of the tank is 2 m and the volume is 8\text{ m}^3,

x\cdot y\cdot 2=8

or xy=4

\Rightarrow y=\frac{4}{x}

Since, Area of the base

is xy=4\text{ m}^2

Therefore, Cost of building the base

is 70\times 4=280

Also, Area of the four sides

is 2(x\times 2)+2(y\times 2)

=4(x+y)

So, Cost of the sides

is 45\times 4(x+y)

=180(x+y)

Therefore, the total cost is

C=280+180(x+y)

Substituting y=\frac4x, we get

\displaystyle C=280+180\left(x+\frac4x\right)

Differentiating w.r.t. x,

\displaystyle \frac{dC}{dx}=180\left(1-\frac4{x^2}\right)

For critical points,

Put \displaystyle \frac{dC}{dx}=0

i.e. \displaystyle 180\left(1-\frac4{x^2}\right)=0

or \displaystyle 1-\frac4{x^2}=0

x^2=4

\Rightarrow x=2 (rejecting negative value of length)

Since y=\frac4x,

\therefore y=2

Now,

\displaystyle \frac{d^2C}{dx^2}=180\left(\frac8{x^3}\right)=\frac{1440}{x^3}

Substituting x=2,

\displaystyle \frac{d^2C}{dx^2}=\frac{1440}{8}=180

Since \frac{d^2C}{dx^2} is positive at x=2, the cost is minimum when x=2.

Hence, \displaystyle C_{\min}=280+180(2+2)

\displaystyle =280+720

\displaystyle =1000

Therefore, the cost of the least expensive tank is

\boxed{\text{Rs }1000}

7. The sum of the perimeter of a circle and square is k where k is some constant. Prove that the sum of their areas is least when the side of the square is double the radius of the circle.

Solution

Let the radius of the circle be r and the side of the square be a.

Since the sum of their perimeters is k, we have

2\pi r+4a=k

Therefore,

\displaystyle a=\frac{k-2\pi r}{4}

The sum of the areas of the circle and square is

A=\pi r^2+a^2

Substituting the value of a, we get

\displaystyle A=\pi r^2+\left(\frac{k-2\pi r}{4}\right)^2

\displaystyle A=\pi r^2+\frac{(k-2\pi r)^2}{16}

Differentiating w.r.t. r, we get

\displaystyle \frac{dA}{dr}=2\pi r-\frac{\pi}{4}(k-2\pi r)

For critical points,

Put \displaystyle \frac{dA}{dr}=0

i.e. \displaystyle 2\pi r-\frac{\pi}{4}(k-2\pi r)=0

\displaystyle 8r-k+2\pi r=0

\displaystyle \Rightarrow r=\frac{k}{8+2\pi}

Now,

\displaystyle \frac{d^2A}{dr^2}=2\pi+\frac{\pi^2}{2}

Since \frac{d^2A}{dr^2} is always positive, A is minimum at

\displaystyle r=\frac{k}{8+2\pi}

Substituting this value in

a=\frac{k-2\pi r}{4}

We get, \displaystyle a=\frac{1}{4}\left(k-\frac{2\pi k}{8+2\pi}\right)

\displaystyle =\frac{1}{4}\left(\frac{8k}{8+2\pi}\right)

\displaystyle =\frac{2k}{8+2\pi}

But

\displaystyle r=\frac{k}{8+2\pi}

Therefore,

a=2r

Hence, the sum of the areas of the circle and square is least when

\boxed{\text{side of square}=2\times\text{radius of circle}}

8. A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10\text{ m}. Find the dimensions of the window to admit maximum light through the whole opening.

Solution

Let the radius of the semicircular opening be r and the height of the rectangular portion be h.

Then the width of the rectangle is 2r

And the perimeter of the window is

2h+2r+\pi r=10

Therefore,

\displaystyle h=\frac{10-(2+\pi)r}{2}

The area of the window is

\displaystyle A=\text{Area of rectangle}+\text{Area of semicircle}

\displaystyle A=2rh+\frac12\pi r^2

Substituting the value of h, we get

\displaystyle A=2r\left(\frac{10-(2+\pi)r}{2}\right)+\frac12\pi r^2

\displaystyle A=r\bigl(10-(2+\pi)r\bigr)+\frac12\pi r^2

So, \displaystyle A=10r-\left(2+\frac{\pi}{2}\right)r^2

Differentiating with respect to r, we get

\displaystyle \frac{dA}{dr}=10-(4+\pi)r

For critical points,

Put \displaystyle \frac{dA}{dr}=0

10-(4+\pi)r=0

\displaystyle r=\frac{10}{4+\pi}

Now,

\displaystyle \frac{d^2A}{dr^2}=-(4+\pi)

Since \frac{d^2A}{dr^2} is always negative, the area is maximum when

\displaystyle r=\frac{10}{4+\pi}

Substituting this value in

\displaystyle h=\frac{10-(2+\pi)r}{2}

we get

\displaystyle h=\frac{10-(2+\pi)\left(\frac{10}{4+\pi}\right)}{2}

\displaystyle =\frac{10}{4+\pi}

Hence,

\displaystyle h=r=\frac{10}{4+\pi}\text{ m}

The width of the window is

\displaystyle 2r=\frac{20}{4+\pi}\text{ m}

Therefore, the dimensions of the window for maximum light are

\text{Width}=2r=\frac{20}{4+\pi}\text{ m}

\text{Height of rectangular part}=h=\frac{10}{4+\pi}\text{ m}

As per NCERT terminology, width (2r) is referred to as the length and height (h) as the breadth. Hence,

\boxed{\text{Length}=\frac{20}{4+\pi}\text{ m},\quad \text{Breadth}=\frac{10}{4+\pi}\text{ m}}

9. A point on the hypotenuse of a right triangle is at distances a and b from the two sides of the triangle. Show that the minimum length of the hypotenuse is \left(a^{\frac23}+b^{\frac23}\right)^{\frac32}.

Solution

Let the right triangle have vertices O(0,0), A(x,0) and B(0,y).

And let P(a,b) be the given point on the hypotenuse.

The hypotenuse joins the points (x,0) and (0,y). Therefore, its equation in intercept form is

\displaystyle \frac{X}{x}+\frac{Y}{y}=1

Since the point P(a,b) lies on the hypotenuse, its coordinates satisfy the above equation.

Substituting X=a and Y=b, we get

\displaystyle \frac{a}{x}+\frac{b}{y}=1

Hence,

\displaystyle y=\frac{bx}{x-a}

The length of the hypotenuse (using distance formula) is

\displaystyle h=\sqrt{x^2+y^2}

Since minimizing h is equivalent to minimizing h^2, let

\displaystyle H=x^2+y^2

Substituting y=\frac{bx}{x-a},

\displaystyle H=x^2+\frac{b^2x^2}{(x-a)^2}

Differentiating w.r.t. x, we get

\displaystyle \frac{dH}{dx}=2x-\frac{2ab^2x}{(x-a)^3}

For critical points,

Put \displaystyle \frac{dH}{dx}=0

i.e. \displaystyle 2x-\frac{2ab^2x}{(x-a)^3}=0

\displaystyle 1=\frac{ab^2}{(x-a)^3}

\displaystyle \Rightarrow (x-a)^3=ab^2

Now, differentiating again, we get

\displaystyle \frac{d^2H}{dx^2}=2+\frac{6ab^2x}{(x-a)^4}-\frac{2ab^2}{(x-a)^3}

Using (x-a)^3=ab^2, we have

\displaystyle \frac{ab^2}{(x-a)^3}=1

Therefore,

\displaystyle \frac{d^2H}{dx^2}=2+\frac{6x}{x-a}-2

\displaystyle =\frac{6x}{x-a}

Since x\gt a, both x and x-a are positive. Hence,

\displaystyle \frac{d^2H}{dx^2}\gt 0

Therefore, H has a minimum value at the critical point.

Again using (x-a)^3=ab^2, we get on simplification,

\displaystyle x=a+a^{\frac13}b^{\frac23}

Therefore, \displaystyle y=\frac{bx}{x-a}

\displaystyle =\frac{b\left(a+a^{\frac13}b^{\frac23}\right)}{a^{\frac13}b^{\frac23}}

\displaystyle =a^{\frac23}b^{\frac13}+b

So, \displaystyle y=b^{\frac13}\left(a^{\frac23}+b^{\frac23}\right)

Also,

\displaystyle x=a^{\frac13}\left(a^{\frac23}+b^{\frac23}\right)

Hence, \displaystyle h^2=x^2+y^2

\displaystyle =\left(a^{\frac23}+b^{\frac23}\right)^2\left(a^{\frac23}+b^{\frac23}\right)

\displaystyle =\left(a^{\frac23}+b^{\frac23}\right)^3

Therefore,

\displaystyle h=\left(a^{\frac23}+b^{\frac23}\right)^{\frac32}

Since this value corresponds to the critical point giving the minimum value of H=x^2+y^2, the minimum length of the hypotenuse is

\boxed{\left(a^{\frac23}+b^{\frac23}\right)^{\frac32}}

10. Find the points at which the function f(x)=(x-2)^4(x+1)^3 has (i) local maxima (ii) local minima (iii) point of inflexion.

Solution

Given,

f(x)=(x-2)^4(x+1)^3

Differentiating using the product rule, we get

We get, \displaystyle f'(x)=4(x-2)^3(x+1)^3+3(x-2)^4(x+1)^2

\displaystyle =(x-2)^3(x+1)^2\big[4(x+1)+3(x-2)\big]

\displaystyle =(x-2)^3(x+1)^2(7x-2)

For critical points,

f'(x)=0

(x-2)^3(x+1)^2(7x-2)=0

Thus, the critical points are

x=-1,\quad x=\frac27,\quad x=2

Differentiating again, we get

\displaystyle f''(x)=6(x-2)^2(x+1)(7x^2-4x-2)

At \displaystyle x=\frac27

Substituting x=\frac27 in f''(x), we get

\displaystyle f''\!\left(\frac27\right)=6\left(\frac27-2\right)^2\left(\frac27+1\right)\left[7\left(\frac27\right)^2-4\left(\frac27\right)-2\right]

\displaystyle =6\left(\frac{144}{49}\right)\left(\frac97\right)\left(-\frac{18}{7}\right)\lt 0

Since f''\!\left(\frac27\right)\lt 0, by the Second Derivative Test, f(x) has a local maximum at x=\frac27.

Also,

\displaystyle f\!\left(\frac27\right)=\left(-\frac{12}{7}\right)^4\left(\frac97\right)^3=\frac{12^4\cdot9^3}{7^7}

Hence, the point of local maximum is

\boxed{\left(\frac27,\frac{12^4\cdot9^3}{7^7}\right)}

At \displaystyle x=2

Substituting x=2 in f''(x), we get

f''(2)=0

Therefore, the Second Derivative Test fails at x=2.

We now apply the First Derivative Test.

For \frac27\lt x\lt 2, we have f'(x)\lt 0.

For x\gt 2, we have f'(x)\gt 0.

Thus, f'(x) changes sign from negative to positive while moving from left to right through x=2. Therefore, f(x) has a local minimum at x=2.

Also,

f(2)=0

Hence, the point of local minimum is

\boxed{(2,0)}

At \displaystyle x=-1

Substituting x=-1 in f''(x), we get

f''(-1)=0

Therefore, the Second Derivative Test fails at x=-1.

We now apply the First Derivative Test.

For x\lt -1, we have f'(x)\gt 0.

For -1\lt x\lt\frac27, we have f'(x)\gt 0.

Thus, f'(x) does not change sign while moving from left to right through x=-1. Hence, x=-1 is neither a point of local maximum nor a point of local minimum.

Also,

f(-1)=0

Therefore, (-1,0) is a point of inflexion.

Hence, the point of inflexion is

\boxed{(-1,0)}

Therefore,

• Local maximum at \boxed{\left(\frac27,\frac{12^4\cdot9^3}{7^7}\right)}
• Local minimum at \boxed{(2,0)}
• Point of inflexion at \boxed{(-1,0)}

11. Find the absolute maximum and minimum values of the function f(x)=\cos^2x+\sin x,\quad x\in[0,\pi].

Solution

Given,

f(x)=\cos^2x+\sin x

To find the absolute maximum and minimum values on the closed interval [0,\pi], we first find the critical points.

Differentiating with respect to x, we get

\displaystyle f'(x)=2\cos x(-\sin x)+\cos x

\displaystyle =\cos x(1-2\sin x)

For critical points,

f'(x)=0

\cos x(1-2\sin x)=0

Therefore,

\cos x=0

\Rightarrow x=\frac {\pi}{2}

or

1-2\sin x=0

or \sin x=\frac12

\Rightarrow x=\frac{\pi}{6},\quad \frac{5\pi}{6}

Thus, the critical points in [0,\pi] are

x=\frac{\pi}{6},\quad \frac{\pi}{2},\quad \frac{5\pi}{6}

Now let’s evaluate f(x) at the critical points and the endpoints.

The largest value obtained is

\displaystyle \frac54

which occurs at

x=\frac{\pi}{6}\quad\text{and}\quad x=\frac{5\pi}{6}

Hence, the absolute maximum value is

\boxed{\frac54}

The smallest value obtained is

1

which occurs at

x=0,\quad \frac{\pi}{2},\quad \pi

Hence, the absolute minimum value is

\boxed{1}

12. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is \frac{4r}{3}.

Solution

Let O be the centre of the sphere of radius r.

Let the altitude of the cone be h and the radius of its base be a.

Then the distance of the base from the centre of the sphere is

r-h

From the right triangle OMB, as in the diagram

We have, a^2+(r-h)^2=r^2

or a^2=r^2-\left(r-h\right)^2

a^2=r^2-\left(r^2-2rh+h^2\right)

a^2=2rh-h^2

So, a^2=h(2r-h)

The volume of the cone is

\displaystyle V=\frac13\pi a^2h

Substituting a^2=h(2r-h), we get

\displaystyle V=\frac13\pi h^2(2r-h)

\displaystyle V=\frac{\pi}{3}(2rh^2-h^3)

Differentiating with respect to h, we get

\displaystyle \frac{dV}{dh}=\frac{\pi}{3}(4rh-3h^2)

\displaystyle =\frac{\pi h}{3}(4r-3h)

For critical points,

Put \displaystyle \frac{dV}{dh}=0

\displaystyle \frac{\pi h}{3}(4r-3h)=0

h=0 (rejected)

or

\displaystyle h=\frac{4r}{3}

Now, differentiating again, we get

\displaystyle \frac{d^2V}{dh^2}=\frac{\pi}{3}(4r-6h)

Substituting h=\frac{4r}{3},

\displaystyle \frac{d^2V}{dh^2}=\frac{\pi}{3}\left(4r-8r\right)

\displaystyle =-\frac{4\pi r}{3}\lt 0

Since \frac{d^2V}{dh^2}\lt 0, by the Second Derivative Test, the volume is maximum when

\displaystyle h=\frac{4r}{3}

Hence, the altitude of the right circular cone of maximum volume that can be inscribed in the sphere is

\boxed{\frac{4r}{3}}

13. Let f be a function defined on [a,b] such that f'(x)\gt 0 for all x\in(a,b). Prove that f is an increasing function on (a,b).

Proof

Given that

f'(x)\gt 0 for all x\in(a,b).

Since the derivative is positive throughout the interval (a,b), the function has a positive rate of change at every point of the interval.

By the First Derivative Test, a function is increasing on an interval wherever its derivative remains positive.

Since f'(x)\gt 0 for every x\in(a,b), it follows that f(x) is increasing throughout the interval (a,b).

Hence, f is an increasing function on (a,b).

\boxed{\text{Hence proved.}}

Alternative Method (Using Lagrange’s Mean Value Theorem): This result can also be proved using Lagrange’s Mean Value Theorem. If f'(x)\gt 0 throughout an interval, then for any x_1\lt x_2, LMVT gives f(x_2)-f(x_1)\gt 0, which implies f(x_2)\gt f(x_1). Hence, f is increasing.

Proof

Let x_1 and x_2 be any two points in (a,b) such that

x_1\lt x_2

Since f'(x) exists for all x\in(a,b), therefore the function f is differentiable on (a,b) and thus continuous on every closed interval contained in (a,b). Hence, f satisfies the conditions of Lagrange’s Mean Value Theorem on [x_1,x_2] Therefore, by Lagrange’s Mean Value Theorem, there exists a point c\in(x_1,x_2) such that

\displaystyle f'(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}

Given that f'(x)\gt 0 for all x\in(a,b), therefore

f'(c)\gt 0

Hence,

\displaystyle \frac{f(x_2)-f(x_1)}{x_2-x_1}\gt 0

Since x_2-x_1\gt 0, we get

f(x_2)-f(x_1)\gt 0

f(x_2)\gt f(x_1)

Thus, whenever x_2\gt x_1, we have f(x_2)\gt f(x_1).

Therefore, f is an increasing function on (a,b).

\boxed{\text{Hence proved.}}

14. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is \frac{2R}{\sqrt3}. Also find the maximum volume.

Solution

Let the radius of the cylinder be r and its height be h.

The cylinder is inscribed in a sphere of radius R.

From the right triangle OMB, as in the diagram, we have

\displaystyle r^2+\left(\frac h2\right)^2=R^2

Therefore,

\displaystyle r^2=R^2-\frac{h^2}{4}

The volume of the cylinder is

\displaystyle V=\pi r^2h

Substituting r^2=R^2-\frac{h^2}{4}, we get

\displaystyle V=\pi h\left(R^2-\frac{h^2}{4}\right)

\displaystyle V=\pi\left(R^2h-\frac{h^3}{4}\right)

Differentiating with respect to h, we get

\displaystyle \frac{dV}{dh}=\pi\left(R^2-\frac{3h^2}{4}\right)

For critical points,

Put \displaystyle \frac{dV}{dh}=0

So \displaystyle \pi\left(R^2-\frac{3h^2}{4}\right)=0

or \displaystyle R^2-\frac{3h^2}{4}=0

\displaystyle h^2=\frac{4R^2}{3}

\displaystyle \Rightarrow h=\frac{2R}{\sqrt3}

Now, differentiating \displaystyle \frac{dV}{dh}, we get

\displaystyle \frac{d^2V}{dh^2}=-\frac{3\pi h}{2}

Substituting h=\frac{2R}{\sqrt3},

\displaystyle \frac{d^2V}{dh^2}=-\pi\sqrt3\,R\lt 0

Since \frac{d^2V}{dh^2}\lt 0, by the Second Derivative Test, the volume is maximum when

\displaystyle h=\frac{2R}{\sqrt3}

Hence, the height of the cylinder of maximum volume is

\boxed{\frac{2R}{\sqrt3}}

Now, \displaystyle r^2=R^2-\frac14\left(\frac{2R}{\sqrt3}\right)^2

\displaystyle =R^2-\frac{R^2}{3}

\displaystyle =\frac{2R^2}{3}

Therefore, the maximum volume

is \displaystyle V_{\max}=\pi r^2h

\displaystyle =\pi\left(\frac{2R^2}{3}\right)\left(\frac{2R}{\sqrt3}\right)

\displaystyle =\frac{4\pi R^3}{3\sqrt3}

Hence, the maximum volume of the cylinder is

\boxed{\frac{4\pi R^3}{3\sqrt3}}

15. Show that the height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi-vertical angle \alpha is one-third that of the cone and the greatest volume of the cylinder is \frac{4}{27}\pi h^3\tan^2\alpha.

Solution

Let the height and radius of the cylinder be x and r respectively.

The cone has height h and semi-vertical angle \alpha.

From the figure, in right triangle AOM, we have,

\displaystyle \tan\alpha=\frac {r}{h-x}

\therefore r=(h-x)\tan\alpha

We know, the volume of the cylinder is

\displaystyle V=\pi r^2x

Substituting r=(h-x)\tan\alpha, we get

\displaystyle V=\pi x(h-x)^2\tan^2\alpha

or \displaystyle V=\pi\tan^2\alpha\,[x(h-x)^2]

where \pi\tan^2\alpha is constant

Differentiating, using product rule, w.r.t. x, we get

\displaystyle \frac{dV}{dx}=\pi\tan^2\alpha\left[(h-x)^2-2x(h-x)\right]

\displaystyle =\pi\tan^2\alpha\,(h-x)(h-3x)

For critical points,

Put \displaystyle \frac{dV}{dx}=0

So \displaystyle \pi\tan^2\alpha\,(h-x)(h-3x)=0

\Rightarrow x=h (rejected)

or

\displaystyle x=\frac{h}{3}

Now, differentiating again, we get

\displaystyle \frac{d^2V}{dx^2}=\pi\tan^2\alpha\,(6x-4h)

Substituting x=\frac{h}{3},

\displaystyle \frac{d^2V}{dx^2}=\pi\tan^2\alpha\left(2h-4h\right)

\displaystyle =-2\pi h\tan^2\alpha\lt 0

Since \frac{d^2V}{dx^2}\lt 0, by the Second Derivative Test, the volume is maximum when

\displaystyle x=\frac{h}{3}

Hence, the height of the cylinder of greatest volume is one-third of the height of the cone.

\boxed{\text{Height of cylinder}=\frac{h}{3}}

Now, when x=\frac{h}{3},

\displaystyle r=\left(h-\frac{h}{3}\right)\tan\alpha

\displaystyle =\frac{2h}{3}\tan\alpha

Therefore, the maximum volume is

\displaystyle V_{\max}=\pi r^2x

Substituting the values of r and x so obtained,

We get, \displaystyle V_{\max}=\pi\left(\frac{2h}{3}\tan\alpha\right)^2\left(\frac{h}{3}\right)

\displaystyle =\pi\left(\frac{4h^2}{9}\tan^2\alpha\right)\left(\frac{h}{3}\right)

\displaystyle =\frac{4}{27}\pi h^3\tan^2\alpha

Hence, the greatest volume of the cylinder is

\boxed{\frac{4}{27}\pi h^3\tan^2\alpha}

16. A cylindrical tank of radius 10\text{ m} is being filled with wheat at the rate of 314\text{ m}^3/\text{h}. Then the depth of the wheat is increasing at the rate of

• (A) 1\text{ m/h}
• (B) 0.1\text{ m/h}
• (C) 1.1\text{ m/h}
• (D) 0.5\text{ m/h}

Solution

Let h be the depth of wheat in the cylindrical tank at any time t.

The volume of wheat in the tank is

V=\pi r^2h

Given that the radius of the tank is

r=10\text{ m}

Therefore,

V=\pi(10)^2h

V=100\pi h

Differentiating both sides with respect to t, we get

\displaystyle \frac{dV}{dt}=100\pi\frac{dh}{dt}

Given that

\displaystyle \frac{dV}{dt}=314\text{ m}^3/\text{h}

Substituting, we get

314=100\pi\frac{dh}{dt}

\displaystyle \frac{dh}{dt}=\frac{314}{100\pi}

Using \pi\approx 3.14,

\displaystyle \frac{dh}{dt}=\frac{314}{314}=1

\displaystyle \frac{dh}{dt}=1\text{ m/h}

Hence, the depth of wheat is increasing at the rate of

\boxed{1\text{ m/h}}

Hence, the correct answer is (A) 1 m/h.