Solved Examples

Important Theorems and Working Rules

Before proceeding with the solved examples, here is a quick recap of the fundamental theorems and working rules used in applications of derivatives.

Theorem 1: Increasing and Decreasing Functions

Let $f$ be continuous on $[a, b]$ and differentiable on the open interval $(a, b)$. Then:

1. $f$ is strictly increasing on $[a, b]$ if $f'(x) > 0$ for every $x \in (a, b)$.
2. $f$ is strictly decreasing on $[a, b]$ if $f'(x) < 0$ for every $x \in (a, b)$.
3. $f$ is a constant function on $[a, b]$ if $f'(x) = 0$ for every $x \in (a, b)$.

Theorem 2: First Derivative Test

Let $f$ be continuous at a critical point $c$ in an open interval $I$.

1. If $f'(x)$ changes sign from positive to negative as $x$ increases through $c$, then $c$ is a point of local maximum.
2. If $f'(x)$ changes sign from negative to positive as $x$ increases through $c$, then $c$ is a point of local minimum.
3. If $f'(x)$ does not change sign as $x$ passes through $c$, then $c$ is neither a point of local maximum nor a point of local minimum. In many standard textbook cases, such a point is a point of inflection.

Theorem 3: Second Derivative Test

Let $f$ be twice differentiable at $c$.

1. $x = c$ is a point of local maximum if $f'(c) = 0$ and $f''(c) < 0$.
2. $x = c$ is a point of local minimum if $f'(c) = 0$ and $f''(c) > 0$.
3. The test fails if $f'(c) = 0$ and $f''(c) = 0$. In that case, use the First Derivative Test.

Working Rule for Absolute Maxima and Minima

To find the absolute maximum and minimum of a continuous function on a closed interval $[a, b]$:

1. Find all critical points in $(a, b)$.
2. Evaluate $f(x)$ at these critical points and at the endpoints $a$ and $b$.
3. The greatest of these values is the absolute maximum value, and the smallest is the absolute minimum value.

Example 1: Rate of Change (Area of a Circle)

Find the rate of change of the area of a circle with respect to its radius $r$ when $r = 6$ cm.

Solution: Step 1: The area of a circle is $$A = \pi r^2.$$

Step 2: Differentiate with respect to $r$. $$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r.$$

Step 3: Evaluate at $r = 6$. $$\left.\frac{dA}{dr}\right|_{r=6} = 2\pi(6) = 12\pi.$$

Answer: $12\pi \text{ cm}^2/\text{cm}$.

Example 2: Rate of Change (Volume of a Cube)

An edge of a variable cube is increasing at the rate of $3$ cm/s. How fast is the volume of the cube increasing when the edge is $10$ cm long?

Solution: Step 1: Let $x$ be the edge of the cube and $V$ its volume. Given: $$\frac{dx}{dt} = 3 \text{ cm/s}, \qquad V = x^3.$$

Step 2: Differentiate $V = x^3$ with respect to time $t$. $$\frac{dV}{dt} = 3x^2 \frac{dx}{dt}.$$

Step 3: Substitute $x = 10$ and $\frac{dx}{dt} = 3$. $$\frac{dV}{dt} = 3(10)^2(3) = 3(100)(3) = 900.$$

Answer: $900 \text{ cm}^3/\text{s}$.

Example 3: Rate of Change (The Sliding Ladder)

A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $2$ cm/s. How fast is its height on the wall decreasing when the foot of the ladder is $4$ m away from the wall?

Solution: Step 1: Let $x$ be the distance of the foot from the wall and $y$ the height of the top on the wall. The ladder length is constant: $$x^2 + y^2 = 25.$$ Given: $$\frac{dx}{dt} = 2 \text{ cm/s} = 0.02 \text{ m/s}.$$ We need $\frac{dy}{dt}$ when $x = 4$ m.

Step 2: Find $y$ when $x = 4$. $$4^2 + y^2 = 25 \implies 16 + y^2 = 25 \implies y^2 = 9 \implies y = 3 \text{ m}.$$

Step 3: Differentiate the relation $x^2 + y^2 = 25$ with respect to $t$. $$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0.$$ Dividing by 2, $$x\frac{dx}{dt} + y\frac{dy}{dt} = 0.$$

Step 4: Substitute the known values. $$4(0.02) + 3\frac{dy}{dt} = 0.$$ $$0.08 + 3\frac{dy}{dt} = 0.$$ $$\frac{dy}{dt} = -\frac{0.08}{3} = -\frac{2}{75} \text{ m/s}.$$

Step 5: Convert to cm/s. $$-\frac{2}{75} \text{ m/s} = -\frac{200}{75} \text{ cm/s} = -\frac{8}{3} \text{ cm/s}.$$ The negative sign shows the height is decreasing.

Answer: The height is decreasing at the rate of $\frac{8}{3} \text{ cm/s}$.

Example 4: Rate of Change (Conical Sand Pile)

Sand is pouring from a pipe at the rate of $12 \text{ cm}^3/\text{s}$. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is $4$ cm?

Solution: Step 1: Let $V$, $r$, and $h$ be the volume, radius, and height of the cone. Given: $$\frac{dV}{dt} = 12 \text{ cm}^3/\text{s}, \qquad h = \frac{r}{6} \implies r = 6h.$$

Step 2: Write the volume in terms of $h$ alone. $$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (6h)^2 h = \frac{1}{3}\pi (36h^2)h = 12\pi h^3.$$

Step 3: Differentiate with respect to time. $$\frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt}.$$

Step 4: Substitute $\frac{dV}{dt} = 12$ and $h = 4$. $$12 = 36\pi (4)^2 \frac{dh}{dt} = 36\pi(16)\frac{dh}{dt} = 576\pi \frac{dh}{dt}.$$ Hence, $$\frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi}.$$

Answer: $\frac{1}{48\pi} \text{ cm/s}$.

Example 5: Rate of Change (Marginal Revenue)

The total revenue in Rupees received from the sale of $x$ units of a product is given by $R(x) = 13x^2 + 26x + 15$. Find the marginal revenue when $x = 7$.

Solution: Step 1: Marginal revenue is the derivative of total revenue with respect to $x$. $$MR = \frac{dR}{dx} = R'(x).$$

Step 2: Differentiate $R(x)$. $$R'(x) = \frac{d}{dx}(13x^2 + 26x + 15) = 26x + 26.$$

Step 3: Evaluate at $x = 7$. $$MR = 26(7) + 26 = 182 + 26 = 208.$$

Answer: ₹ 208.

Example 6: Intervals of Monotonicity (Quadratic)

Find the intervals in which the function $f(x) = x^2 - 4x + 6$ is strictly increasing or strictly decreasing.

Solution: Step 1: Find the derivative. $$f'(x) = 2x - 4.$$

Step 2: Find the critical point. $$2x - 4 = 0 \implies x = 2.$$

Step 3: This divides the real line into two intervals: $$(-\infty, 2) \quad \text{and} \quad (2, \infty).$$

Step 4: Test the sign of $f'(x)$.

• For $x < 2$, choose $x = 0$: $$f'(0) = -4 < 0,$$ so $f(x)$ is strictly decreasing on $(-\infty, 2)$.
• For $x > 2$, choose $x = 3$: $$f'(3) = 2 > 0,$$ so $f(x)$ is strictly increasing on $(2, \infty)$.

Answer: Strictly increasing on $(2, \infty)$; strictly decreasing on $(-\infty, 2)$.

Example 7: Intervals of Monotonicity (Cubic)

Find the intervals in which $f(x) = 2x^3 - 3x^2 - 36x + 7$ is strictly increasing or strictly decreasing.

Solution: Step 1: Find the derivative. $$f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x-3)(x+2).$$

Step 2: Find the critical points. $$f'(x) = 0 \implies x = 3, -2.$$ So the intervals are: $$(-\infty, -2), \quad (-2, 3), \quad (3, \infty).$$

Step 3: Test the sign of $f'(x)$.

• In $(-\infty, -2)$, take $x=-3$: $$f'(-3)=6(-6)(-1)>0,$$ so $f$ is strictly increasing.
• In $(-2,3)$, take $x=0$: $$f'(0)=6(-3)(2)<0,$$ so $f$ is strictly decreasing.
• In $(3,\infty)$, take $x=4$: $$f'(4)=6(1)(6)>0,$$ so $f$ is strictly increasing.

Answer: Strictly increasing on $(-\infty, -2) \cup (3, \infty)$; strictly decreasing on $(-2, 3)$.

Example 8: Monotonicity of a Trigonometric Function

Find the intervals in which $f(x) = \sin x + \cos x$, $0 \le x \le 2\pi$ is strictly increasing or strictly decreasing.

Solution: Step 1: Differentiate. $$f'(x) = \cos x - \sin x.$$

Step 2: Find critical points by solving $f'(x)=0$. $$\cos x - \sin x = 0 \implies \tan x = 1.$$ In $[0,2\pi]$, the solutions are $$x = \frac{\pi}{4}, \quad \frac{5\pi}{4}.$$

Step 3: The intervals are $$(0, \tfrac{\pi}{4}), \quad (\tfrac{\pi}{4}, \tfrac{5\pi}{4}), \quad (\tfrac{5\pi}{4}, 2\pi).$$

Step 4: Test the sign of $f'(x)$.

• In $(0,\pi/4)$, choose $x=0$: $$f'(0)=1>0,$$ so $f$ is strictly increasing.
• In $(\pi/4,5\pi/4)$, choose $x=\pi/2$: $$f'(\pi/2)=0-1=-1<0,$$ so $f$ is strictly decreasing.
• In $(5\pi/4,2\pi)$, choose $x=3\pi/2$: $$f'(3\pi/2)=0-(-1)=1>0,$$ so $f$ is strictly increasing.

Answer: Strictly increasing on $(0, \pi/4) \cup (5\pi/4, 2\pi)$; strictly decreasing on $(\pi/4, 5\pi/4)$.

Example 9: Proving Monotonicity Everywhere

Show that the function $f(x) = x - \sin x$ is strictly increasing for all $x \in \mathbb{R}$.

Solution: Step 1: Differentiate. $$f'(x) = 1 - \cos x.$$

Step 2: Use the fact that for all real $x$, $$-1 \le \cos x \le 1.$$ Therefore, $$1 - \cos x \ge 0.$$ So $f'(x) \ge 0$ for all real $x$.

Step 3: The equality $f'(x)=0$ occurs only when $\cos x = 1$, that is, at isolated points $x = 2n\pi$, $n \in \mathbb{Z}$. Since the derivative is never negative and is zero only at isolated points, the function never decreases. Also, for any $x_1 < x_2$, $$f(x_2)-f(x_1)=\int_{x_1}^{x_2}(1-\cos x)\,dx > 0,$$ because $1-\cos x \ge 0$ and is not identically zero on any interval. Hence $f$ is strictly increasing on $\mathbb{R}$.

Answer: Proved analytically.

Example 10: Monotonicity of a Logarithmic Function

Prove that $y = \log(1+x) - \frac{2x}{2+x}, \; x > -1$, is an increasing function throughout its domain.

Solution: Step 1: Differentiate with respect to $x$. $$y' = \frac{1}{1+x} - \frac{(2+x)(2)-2x(1)}{(2+x)^2}.$$

Step 2: Simplify the derivative. $$y' = \frac{1}{1+x} - \frac{4+2x-2x}{(2+x)^2} = \frac{1}{1+x} - \frac{4}{(2+x)^2}.$$ Taking the LCM, $$y' = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2}.$$ Expand the numerator: $$(2+x)^2 - 4(1+x) = (x^2+4x+4) - (4+4x) = x^2.$$ Hence, $$y' = \frac{x^2}{(1+x)(2+x)^2}.$$

Step 3: Analyze the sign. For $x > -1$,

• $x^2 \ge 0$,
• $1+x > 0$,
• $(2+x)^2 > 0$. So, $$y' \ge 0 \quad \text{for all } x>-1.$$ Thus the function is increasing throughout its domain.

Answer: Proved analytically.

Example 11: Local Maxima and Minima (First Derivative Test)

Find all points of local maxima and local minima of $f(x) = x^3 - 6x^2 + 9x + 15$.

Solution: Step 1: Differentiate. $$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3).$$

Step 2: Find critical points. $$f'(x)=0 \implies x=1,3.$$

Step 3: Apply the First Derivative Test.

• For $x<1$, choose $x=0$: $$f'(0)=3(-1)(-3)>0.$$ So the function is increasing before $x=1$.

• For $1<x<3$, choose $x=2$: $$f'(2)=3(1)(-1)<0.$$ So the function is decreasing after $x=1$. Hence $x=1$ is a point of local maximum.

• For $1<x<3$, $f'(x)<0$.

• For $x>3$, choose $x=4$: $$f'(4)=3(3)(1)>0.$$ So the sign changes from negative to positive at $x=3$. Hence $x=3$ is a point of local minimum.

Step 4: Find the corresponding function values. $$f(1)=1-6+9+15=19,$$ $$f(3)=27-54+27+15=15.$$

Answer: Local maximum at $(1,19)$; local minimum at $(3,15)$.

Example 12: Second Derivative Test Application

Find local maxima and minima for $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$.

Solution: Step 1: Find the first derivative. $$f'(x) = 12x^3 + 12x^2 - 24x = 12x(x^2+x-2) = 12x(x+2)(x-1).$$ So the critical points are $$x=0, -2, 1.$$

Step 2: Find the second derivative. $$f''(x)=36x^2+24x-24.$$

Step 3: Evaluate $f''(x)$ at the critical points.

• At $x=0$: $$f''(0)=-24<0,$$ so $x=0$ is a point of local maximum.
• At $x=-2$: $$f''(-2)=36(4)+24(-2)-24=144-48-24=72>0,$$ so $x=-2$ is a point of local minimum.
• At $x=1$: $$f''(1)=36+24-24=36>0,$$ so $x=1$ is a point of local minimum.

Step 4: Find the function values. $$f(0)=12,$$ $$f(-2)=3(16)+4(-8)-12(4)+12=48-32-48+12=-20,$$ $$f(1)=3+4-12+12=7.$$

Answer: Local maximum at $(0,12)$; local minima at $(-2,-20)$ and $(1,7)$.

Example 13: Local Extrema of Rational Function

Find the local maximum and minimum values of $f(x) = x + \frac{1}{x}, \; x \ne 0$.

Solution: Step 1: Differentiate. $$f'(x) = 1 - \frac{1}{x^2}.$$

Step 2: Find critical points. $$1 - \frac{1}{x^2}=0 \implies x^2=1 \implies x=1,-1.$$

Step 3: Find the second derivative. $$f''(x)=\frac{2}{x^3}.$$

Step 4: Classify the critical points.

• At $x=1$: $$f''(1)=2>0,$$ so $x=1$ is a point of local minimum. Minimum value: $$f(1)=1+1=2.$$
• At $x=-1$: $$f''(-1)=-2<0,$$ so $x=-1$ is a point of local maximum. Maximum value: $$f(-1)=-1-1=-2.$$

Answer: Local minimum value is $2$ at $x=1$; local maximum value is $-2$ at $x=-1$.

Example 14: Absolute Extrema on a Closed Interval

Find the absolute maximum and absolute minimum values of $f(x) = 2x^3 - 15x^2 + 36x + 1$ on the interval $[1, 5]$.

Solution: Step 1: Find the derivative. $$f'(x)=6x^2-30x+36=6(x-2)(x-3).$$ Critical points in $(1,5)$ are $x=2$ and $x=3$.

Step 2: Evaluate $f(x)$ at the critical points and endpoints. $$f(1)=2-15+36+1=24,$$ $$f(2)=16-60+72+1=29,$$ $$f(3)=54-135+108+1=28,$$ $$f(5)=250-375+180+1=56.$$

Step 3: Compare the values. The greatest value is $56$ and the least value is $24$.

Answer: Absolute maximum is $56$ at $x=5$; absolute minimum is $24$ at $x=1$.

Example 15: Absolute Extrema involving Fractional Powers

Find the absolute maximum and minimum values of $f(x) = 12x^{4/3} - 6x^{1/3}$ on $[-1, 1]$.

Solution: Step 1: Differentiate. $$f'(x)=16x^{1/3}-2x^{-2/3}=\frac{16x-2}{x^{2/3}}.$$

Step 2: Find critical points.

• $f'(x)=0$ when $$16x-2=0 \implies x=\frac18.$$
• $f'(x)$ is undefined when $$x=0.$$ So the critical points are $x=0$ and $x=\frac18$.

Step 3: Evaluate at $x=-1,0,\frac18,1$. $$f(-1)=12(1)-6(-1)=18,$$ $$f(0)=0,$$ $$f\left(\frac18\right)=12\left(\frac1{16}\right)-6\left(\frac12\right)=\frac34-3=-\frac94,$$ $$f(1)=12-6=6.$$

Step 4: Compare the values. The greatest value is $18$, and the least value is $-\frac94$.

Answer: Absolute maximum is $18$ at $x=-1$; absolute minimum is $-\frac94$ at $x=\frac18$.

Optimization

Example 16: The Open Box Proble

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 of the square to be cut off so that the volume of the box is the maximum possible?

Solution: Step 1: Let the side of each cut square be $x$ cm. Then the dimensions of the box are:

• length $=18-2x$
• breadth $=18-2x$
• height $=x$ So, $$V(x)=x(18-2x)^2.$$ The domain is $$0 \le x \le 9.$$

Step 2: Differentiate. $$V'(x)=(18-2x)^2 + x\cdot 2(18-2x)(-2).$$ Factorizing, $$V'(x)=(18-2x)(18-6x).$$

Step 3: Find critical points. $$V'(x)=0 \implies x=9 \quad \text{or} \quad x=3.$$

Step 4: Check the values of $V(x)$ at the endpoints and critical point. $$V(0)=0,$$ $$V(3)=3(12)^2=432,$$ $$V(9)=0.$$ Thus the maximum volume occurs at $x=3$.

(You may also verify by second derivative: $V''(3)<0$.)

Answer: The side of the square to be cut off should be $3$ cm.

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Example 17: Cutting a Wire

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

Solution: Step 1: Let the length used for the square be $x$ m. Then the length used for the circle is $(28-x)$ m.

Step 2: Express both areas in terms of $x$. For the square:

• perimeter $=x$
• side $=x/4$
• area $$A_1=\left(\frac{x}{4}\right)^2=\frac{x^2}{16}.$$

For the circle:

• circumference $=28-x$
• radius $$r=\frac{28-x}{2\pi}$$
• area $$A_2=\pi r^2 = \pi\left(\frac{28-x}{2\pi}\right)^2 = \frac{(28-x)^2}{4\pi}.$$

So the total area is $$A(x)=\frac{x^2}{16}+\frac{(28-x)^2}{4\pi}.$$

Step 3: Differentiate. $$A'(x)=\frac{x}{8}-\frac{28-x}{2\pi}.$$

Step 4: Set $A'(x)=0$. $$\frac{x}{8}=\frac{28-x}{2\pi}.$$ Cross-multiplying, $$2\pi x = 8(28-x)$$ $$2\pi x = 224 - 8x$$ $$x(2\pi+8)=224$$ $$x=\frac{224}{2(\pi+4)}=\frac{112}{\pi+4}.$$ So the circle gets length $$28-x = 28 - \frac{112}{\pi+4} = \frac{28\pi}{\pi+4}.$$

Step 5: Verify minimum. $$A''(x)=\frac18 + \frac{1}{2\pi}>0,$$ so the total area is indeed minimum.

Answer: Length for square $= \frac{112}{\pi+4}$ m, and length for circle $= \frac{28\pi}{\pi+4}$ m.

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Example 18: Cylinder inscribed in a Sphere

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is $\frac{2R}{\sqrt{3}}$.

Solution: Step 1: Let the cylinder have radius $r$ and height $h$. From the right triangle formed in the central cross-section of the sphere, $$r^2 + \left(\frac{h}{2}\right)^2 = R^2.$$ Thus, $$r^2 = R^2 - \frac{h^2}{4}.$$

Step 2: Write the volume as a function of $h$. $$V = \pi r^2 h = \pi\left(R^2 - \frac{h^2}{4}\right)h = \pi R^2 h - \frac{\pi}{4}h^3.$$

Step 3: Differentiate with respect to $h$. $$\frac{dV}{dh}=\pi R^2 - \frac{3\pi}{4}h^2.$$ Set this equal to zero: $$\pi R^2 = \frac{3\pi}{4}h^2.$$ Cancelling $\pi$, $$R^2 = \frac34 h^2 \implies h^2 = \frac{4R^2}{3}.$$ So, $$h = \frac{2R}{\sqrt{3}}.$$ (We take the positive value since height is positive.)

Step 4: Check for maximum. $$\frac{d^2V}{dh^2} = -\frac{3\pi}{2}h < 0$$ for $h>0$, so the volume is maximum.

Answer: The height of the cylinder of maximum volume is $\frac{2R}{\sqrt{3}}$.

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Example 19: Cone inscribed in a Sphere

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: Step 1: Let the cone have base radius $R$ and height $h$. In the axial section of the sphere, by geometry, $$R^2 + (h-r)^2 = r^2.$$ Expanding, $$R^2 + h^2 - 2hr + r^2 = r^2$$ $$R^2 = 2hr - h^2.$$

Step 2: Write the volume of the cone in terms of $h$. $$V = \frac13 \pi R^2 h = \frac13 \pi (2hr - h^2)h = \frac{\pi}{3}(2rh^2 - h^3).$$

Step 3: Differentiate with respect to $h$. $$\frac{dV}{dh} = \frac{\pi}{3}(4rh - 3h^2).$$ Set this equal to zero: $$h(4r - 3h)=0.$$ Ignoring $h=0$, we get $$h=\frac{4r}{3}.$$

Step 4: Verify it is a maximum. $$V''(h)=\frac{\pi}{3}(4r-6h).$$ At $h=\frac{4r}{3}$, $$V''\left(\frac{4r}{3}\right)=\frac{\pi}{3}(4r-8r) < 0.$$ So the volume is maximum.

Answer: The altitude of the cone of maximum volume is $\frac{4r}{3}$.

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Example 20: Semi-vertical angle of a Cone

Show that the semi-vertical angle of the cone of maximum volume and given slant height is $\tan^{-1} \sqrt{2}$.

Solution: Step 1: Let $\theta$ be the semi-vertical angle, $l$ the fixed slant height, $r$ the base radius, and $h$ the height. Then, $$r = l\sin\theta, \qquad h = l\cos\theta.$$

Step 2: Write the volume of the cone in terms of $\theta$. $$V = \frac13 \pi r^2 h = \frac13 \pi (l\sin\theta)^2 (l\cos\theta) = \frac{\pi l^3}{3} \sin^2\theta \cos\theta.$$

Step 3: Differentiate with respect to $\theta$. $$\frac{dV}{d\theta} = \frac{\pi l^3}{3}\left[2\sin\theta\cos\theta\cdot\cos\theta + \sin^2\theta(-\sin\theta)\right].$$ Factor out $\sin\theta$: $$\frac{dV}{d\theta} = \frac{\pi l^3}{3} \sin\theta (2\cos^2\theta - \sin^2\theta).$$

Step 4: Set derivative equal to zero. Ignoring $\sin\theta=0$ (which gives degenerate cone), we get $$2\cos^2\theta - \sin^2\theta = 0.$$ Thus, $$\sin^2\theta = 2\cos^2\theta \implies \tan^2\theta = 2.$$ Since $\theta$ is acute, $$\tan\theta = \sqrt{2}.$$ Hence, $$\theta = \tan^{-1}(\sqrt{2}).$$

Answer: The semi-vertical angle is $\tan^{-1}(\sqrt{2})$.

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Example 21: Window Shape

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

Solution: Step 1: Let the rectangle have width $2x$ and height $y$. Then the semicircle has radius $x$.

Step 2: Use the perimeter condition. Perimeter consists of the base $2x$, the two vertical sides $2y$, and the semicircular arc $\pi x$. So, $$2x + 2y + \pi x = 10.$$ Hence, $$2y = 10 - 2x - \pi x$$ $$y = 5 - x - \frac{\pi x}{2}.$$

Step 3: Write the total area. Area of rectangle: $$A_1 = 2xy.$$ Area of semicircle: $$A_2 = \frac12 \pi x^2.$$ So, $$A = 2xy + \frac12 \pi x^2.$$ Substitute $y$: $$A = 2x\left(5 - x - \frac{\pi x}{2}\right) + \frac12 \pi x^2.$$ Simplify: $$A = 10x - 2x^2 - \pi x^2 + \frac12 \pi x^2 = 10x - 2x^2 - \frac{\pi}{2}x^2.$$

Step 4: Differentiate. $$A'(x)=10-4x-\pi x.$$ Set equal to zero: $$10-(4+\pi)x=0 \implies x=\frac{10}{4+\pi}.$$

Step 5: Verify maximum. $$A''(x)=-4-\pi<0,$$ so the area is maximum.

Step 6: Find the dimensions. Width of rectangle: $$2x=\frac{20}{4+\pi} \text{ m}.$$ Height of rectangle: $$y = 5 - \frac{10}{4+\pi} - \frac{5\pi}{4+\pi} = \frac{10}{4+\pi} \text{ m}.$$ Thus the semicircle radius is also $$x=\frac{10}{4+\pi} \text{ m}.$$

Answer: Width of the rectangular part $= \frac{20}{4+\pi}$ m, height of the rectangular part $= \frac{10}{4+\pi}$ m.

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Example 22: Right Circular Cylinder

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: Step 1: Let $r$ be the radius and $h$ the height of the cylinder. Let the fixed total surface area be $S$. Then, $$S = 2\pi r^2 + 2\pi rh.$$ So, $$h = \frac{S - 2\pi r^2}{2\pi r}.$$

Step 2: Write the volume in terms of $r$. $$V = \pi r^2 h = \pi r^2 \left(\frac{S - 2\pi r^2}{2\pi r}\right) = \frac12 r(S - 2\pi r^2) = \frac{S}{2}r - \pi r^3.$$

Step 3: Differentiate with respect to $r$. $$\frac{dV}{dr} = \frac{S}{2} - 3\pi r^2.$$ Set this equal to zero: $$\frac{S}{2} = 3\pi r^2 \implies S = 6\pi r^2.$$

Step 4: Substitute this into the surface-area relation for $h$. $$h = \frac{6\pi r^2 - 2\pi r^2}{2\pi r} = \frac{4\pi r^2}{2\pi r} = 2r.$$ Thus the height equals the diameter of the base.

Step 5: Verify maximum. $$\frac{d^2V}{dr^2} = -6\pi r < 0$$ for $r>0$, so the volume is maximum.

Answer: Proved analytically: for maximum volume, $h = 2r$.

JEE Main Standard

Example 23: Monotonicity with Parameters

Find the value of $K$ such that the function $f(x) = Kx^3 - 9x^2 + 9x + 3$ is monotonically increasing on $\mathbb{R}$.

Solution: Step 1: For $f(x)$ to be monotonically increasing on $\mathbb{R}$, we need $$f'(x) \ge 0 \quad \text{for all } x \in \mathbb{R}.$$ Differentiate: $$f'(x) = 3Kx^2 - 18x + 9.$$

Step 2: This is a quadratic in $x$. A quadratic $ax^2+bx+c$ is non-negative for all real $x$ if

1. $a>0$, and
2. its discriminant $D \le 0$.

Here, $$a = 3K.$$ So first we need $$3K > 0 \implies K > 0.$$

Step 3: Compute the discriminant. $$D = (-18)^2 - 4(3K)(9) = 324 - 108K.$$ For non-negativity everywhere, $$324 - 108K \le 0$$ $$108K \ge 324$$ $$K \ge 3.$$

Step 4: Combine the conditions. The condition $K \ge 3$ already ensures $K>0$.

Answer: $K \in [3, \infty)$.

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Example 24: Maxima and Minima of Logarithmic Polynomial

Find the points of local maxima and local minima of the function $f(x) = x^2 \ln x$ for $x > 0$.

Solution: Step 1: Differentiate using the product rule. $$f'(x)=2x\ln x + x^2\cdot \frac{1}{x}=2x\ln x + x = x(2\ln x + 1).$$

Step 2: Find critical points. Since $x>0$, $x \neq 0$, so $$2\ln x + 1 = 0$$ $$\ln x = -\frac12$$ $$x = e^{-1/2} = \frac{1}{\sqrt e}.$$

Step 3: Find the second derivative. $$f''(x)=\frac{d}{dx}(2x\ln x + x)=2\ln x + 2 + 1 = 2\ln x + 3.$$

Step 4: Evaluate at the critical point. $$f''\left(e^{-1/2}\right)=2\left(-\frac12\right)+3 = -1+3 = 2 > 0.$$ Therefore, $x = e^{-1/2}$ is a point of local minimum.

Step 5: Find the minimum value. $$f\left(e^{-1/2}\right)=\left(e^{-1/2}\right)^2\ln\left(e^{-1/2}\right)=e^{-1}\left(-\frac12\right)=-\frac{1}{2e}.$$

There is no local maximum in the domain $x>0$.

Answer: Local minimum at $x = \frac{1}{\sqrt e}$ with value $-\frac{1}{2e}$. No local maximum.

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Example 25: Shortest Distance Optimization

Find the shortest distance from the point $(0, c)$ to the parabola $y = x^2$, where $0 \le c \le 5$.

Solution: Step 1: Let $P(x, x^2)$ be any point on the parabola. The square of the distance from $(0,c)$ to $P$ is $$D = (x-0)^2 + (x^2-c)^2.$$ So, $$D = x^2 + x^4 - 2cx^2 + c^2 = x^4 + (1-2c)x^2 + c^2.$$ We minimize $D$ instead of the distance itself.

Step 2: Differentiate with respect to $x$. $$D' = 4x^3 + 2(1-2c)x = 2x(2x^2 + 1 - 2c).$$ Set $D' = 0$: $$2x(2x^2 + 1 - 2c)=0.$$ Hence, $$x=0 \quad \text{or} \quad 2x^2 + 1 - 2c = 0.$$ So, $$x^2 = c - \frac12.$$

Step 3: Analyze cases.

• Case 1: $0 \le c < \frac12$. Then $c - \frac12 < 0$, so $x^2 = c - \frac12$ has no real solution. The only critical point is $x=0$. Then $$D = c^2,$$ so the shortest distance is $$c.$$

• Case 2: $c \ge \frac12$. Then we have critical points $$x=0, \quad x=\pm\sqrt{c-\frac12}.$$ Now, $$D'' = 12x^2 + 2(1-2c).$$ At $x=0$, $$D''(0)=2(1-2c) \le 0,$$ so $x=0$ does not give the minimum when $c \ge \frac12$. At $x^2 = c-\frac12$, $$D'' = 12\left(c-\frac12\right)+2(1-2c)=8c-4 \ge 0,$$ so these points give the minimum.

Step 4: Find the minimum value. Substitute $x^2 = c - \frac12$ into $$D = x^4 + (1-2c)x^2 + c^2.$$ Using $x^2 = c - \frac12$ and $x^4 = \left(c - \frac12\right)^2$, $$D = \left(c - \frac12\right)^2 + (1-2c)\left(c - \frac12\right) + c^2 = c - \frac14.$$ Hence the shortest distance is $$\sqrt{c - \frac14}.$$

Answer:

• If $0 \le c < \frac12$, the shortest distance is $c$.
• If $c \ge \frac12$, the shortest distance is $\sqrt{c - \frac14}$.

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Example 26: Piecewise Function Absolute Extrema

Find the absolute maximum and minimum values of $f(x) = x^2 - 5|x| + 6$ on the interval $[-4, 4]$.

Solution: Step 1: Write the function piecewise. For $x < 0$, $|x| = -x$, so $$f(x) = x^2 + 5x + 6.$$ For $x \ge 0$, $|x| = x$, so $$f(x) = x^2 - 5x + 6.$$ Also, $f(x)$ is an even function, so the graph is symmetric about the $y$-axis.

Step 2: Find critical points. For $x>0$, $$f'(x)=2x-5=0 \implies x=\frac52.$$ By symmetry, there is also a corresponding critical point at $$x=-\frac52.$$ At $x=0$, the derivative is not defined because of $|x|$, so $x=0$ is also a critical point.

Step 3: Evaluate the function at all critical points and endpoints. $$f(0)=6,$$ $$f\left(\frac52\right)=\left(\frac52\right)^2 - 5\left(\frac52\right) + 6 = \frac{25}{4} - \frac{25}{2} + 6 = -\frac14,$$ $$f\left(-\frac52\right)=-\frac14,$$ $$f(4)=16-20+6=2,$$ $$f(-4)=2.$$

Step 4: Compare the values. The greatest value is $6$ and the least value is $-\frac14$.

Answer: Absolute maximum is $6$ at $x=0$; absolute minimum is $-\frac14$ at $x=\pm \frac52$.

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Example 27: Number of Local Maxima/Minima involving Integrals

Find the number of local maxima and local minima for the function $f(x) = \int_0^x (t^2 - 5t + 6) \, dt$.

Solution: Step 1: By the Fundamental Theorem of Calculus, $$f'(x)=x^2-5x+6.$$

Step 2: Find critical points. $$x^2-5x+6=0 \implies (x-2)(x-3)=0.$$ So the critical points are $$x=2,3.$$

Step 3: Use the First Derivative Test.

• For $x<2$, choose $x=0$: $$f'(0)=6>0.$$

• For $2<x<3$, choose $x=2.5$: $$f'(2.5)=(0.5)(-0.5)<0.$$ So the sign changes from positive to negative at $x=2$, giving a local maximum.

• For $2<x<3$, $f'(x)<0$.

• For $x>3$, choose $x=4$: $$f'(4)=2>0.$$ So the sign changes from negative to positive at $x=3$, giving a local minimum.

Answer: The function has 1 local maximum and 1 local minimum.

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Example 28: Minimum Distance between a Curve and a Line

Find the shortest distance between the curve $y = e^x$ and the line $y = x$.

Solution: Step 1: A general point on the curve is $P(x, e^x)$. The line $y=x$ can be written as $$x - y = 0.$$ The perpendicular distance from $P(x,e^x)$ to this line is $$d = \frac{|x - e^x|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - e^x|}{\sqrt{2}}.$$ Since $e^x > x$ for all real $x$, this becomes $$d = \frac{e^x - x}{\sqrt{2}}.$$ So it is enough to minimize $$\phi(x)=e^x - x.$$

Step 2: Differentiate. $$\phi'(x)=e^x - 1.$$ Set equal to zero: $$e^x - 1 = 0 \implies e^x = 1 \implies x=0.$$

Step 3: Verify minimum. $$\phi''(x)=e^x > 0$$ for all $x$, so $x=0$ gives the minimum.

Step 4: Find the corresponding distance. At $x=0$, the point on the curve is $(0,1)$, and $$d = \frac{e^0 - 0}{\sqrt2} = \frac{1}{\sqrt2}.$$

Answer: The shortest distance is $\frac{1}{\sqrt2}$ units.