Solved Examples

How to Use This Section

This is your practice engine for the whole chapter: 32 fully worked problems arranged topic-wise and in rising difficulty — rate of change (Examples 1-8), increasing/decreasing functions (Examples 9-16), and maxima-minima with optimisation (Examples 17-32). The NCERT miscellaneous examples and the toughest exercise problems are all here.

Attempt each problem yourself first — set a 5-minute timer — then compare with the worked solution. If your method differs but your answer matches, read the solution anyway: the presented method is usually the one examiners award full marks fastest.

[JEE Tip] Examples marked with harder geometry (26-32) are exactly the level of JEE Main optimisation questions. Master their setups — the differentiation itself is rarely the hard part.

Solved Examples

Rate of Change (Examples 1-8)

Example 1: Circle area rate (warm-up)

Find the rate of change of the area of a circle with respect to its radius $r$ when $r = 4$ cm.

Solution:

1. $A = \pi r^2 \Rightarrow \frac{dA}{dr} = 2\pi r$.
2. At $r = 4$: $\frac{dA}{dr} = 8\pi$.

Final Answer: $8\pi$ cm²/cm.

Example 2: Balloon volume vs radius

A balloon, which always remains spherical, has a variable radius. Find the rate at which its volume is increasing with respect to the radius when the radius is 10 cm.

Solution:

1. $V = \frac{4}{3}\pi r^3 \Rightarrow \frac{dV}{dr} = 4\pi r^2$.
2. At $r = 10$: $\frac{dV}{dr} = 4\pi (100) = 400\pi$.

Final Answer: $400\pi$ cm³/cm.

Takeaway: $\frac{dV}{dr}$ equals the sphere's surface area — volume grows by "adding a skin".

Example 3: The stopping car

A car starts from a point P at time $t = 0$ and stops at point Q. The distance covered in $t$ seconds is $x = t^2\left(2 - \frac{t}{3}\right)$ metres. Find the time taken to reach Q and the distance PQ.

Solution:

1. Velocity: $v = \frac{dx}{dt} = \frac{d}{dt}\left(2t^2 - \frac{t^3}{3}\right) = 4t - t^2 = t(4 - t)$.
2. The car stops when $v = 0$: $t = 0$ (start, at P) or $t = 4$ (at Q).
3. Distance: $x(4) = 16\left(2 - \frac{4}{3}\right) = 16 \cdot \frac{2}{3} = \frac{32}{3}$ m.

Final Answer: The car reaches Q after 4 seconds; PQ $= \frac{32}{3}$ m.

Takeaway: "Starts" and "stops" are physics words for $v = 0$ — translate them before differentiating.

Example 4: The conical water tank

A water tank is an inverted right circular cone with semi-vertical angle $\tan^{-1}(0.5)$. Water is poured in at 5 m³/h. Find the rate at which the water level rises when the depth is 4 m.

Solution:

1. Use the angle: $\tan\alpha = \frac{r}{h} = 0.5 \Rightarrow r = \frac{h}{2}$.
2. One-variable volume: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \frac{h^2}{4} h = \frac{\pi h^3}{12}$.
3. Differentiate: $\frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}$.
4. Substitute: $5 = \frac{\pi (16)}{4}\frac{dh}{dt} = 4\pi\frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{5}{4\pi} = \frac{35}{88}$ m/h (taking $\pi = \frac{22}{7}$).

Final Answer: The water level rises at $\frac{35}{88}$ m/h.

Takeaway: The semi-vertical angle is the constraint in disguise — use it to eliminate $r$ before differentiating.

Example 5: The walking man's shadow

A man 2 m tall walks at 5 km/h away from a lamp post 6 m high. Find the rate at which the length of his shadow increases.

Solution:

1. Set up: man at distance $l$ from the post, shadow length $s$. The lamp, the man's head and the shadow tip form similar triangles: $\frac{s}{s + l} = \frac{2}{6}$.
2. Simplify: $6s = 2s + 2l \Rightarrow l = 2s$.
3. Differentiate: $\frac{dl}{dt} = 2\frac{ds}{dt}$, and $\frac{dl}{dt} = 5$ km/h.

Final Answer: The shadow lengthens at $\frac{5}{2}$ km/h.

Takeaway: The shadow rate is constant — independent of where the man is. Similar triangles often linearise these problems completely.

Example 6: The heated disc

A circular disc of radius 3 cm is heated and its radius expands at 0.05 cm/s. Find the rate at which its area increases when the radius is 3.2 cm.

Solution:

1. $A = \pi r^2 \Rightarrow \frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
2. At $r = 3.2$ with $\frac{dr}{dt} = 0.05$: $\frac{dA}{dt} = 2\pi(3.2)(0.05) = 0.320\pi$ cm²/s.

Final Answer: $0.320\pi$ cm²/s.

Takeaway: Use the radius at the instant asked (3.2 cm), not the initial radius (3 cm) — a deliberately planted distractor.

Example 7: Wheat in a cylindrical tank

A cylindrical tank of radius 10 m is being filled with wheat at 314 m³/h. At what rate is the depth of the wheat increasing?

Solution:

1. $V = \pi r^2 h = 100\pi h$ (radius fixed at 10).
2. $\frac{dV}{dt} = 100\pi \frac{dh}{dt} \Rightarrow 314 = 100 \times 3.14 \times \frac{dh}{dt}$.
3. $\frac{dh}{dt} = 1$ m/h.

Final Answer: 1 m/h. (NCERT Miscellaneous Exercise, Q16 — answer (A).)

Takeaway: In a cylinder the cross-section never changes, so depth rate = volume rate ÷ base area. No calculus drama needed.

Example 8: The shrinking isosceles triangle

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

Solution:

1. Area in terms of the equal side $a$: height $= \sqrt{a^2 - \frac{b^2}{4}}$, so $A = \frac{b}{4}\sqrt{4a^2 - b^2}$.
2. Differentiate: $\frac{dA}{dt} = \frac{b}{4} \cdot \frac{4a}{\sqrt{4a^2 - b^2}}\frac{da}{dt} = \frac{ab}{\sqrt{4a^2 - b^2}}\frac{da}{dt}$.
3. Substitute $a = b$, $\frac{da}{dt} = -3$: $\frac{dA}{dt} = \frac{b^2}{\sqrt{3b^2}}(-3) = -\sqrt{3}\,b$.

Final Answer: The area is decreasing at $\sqrt{3}\,b$ cm²/s.

Takeaway: Keep the fixed quantity ($b$) as a symbol throughout — the answer is supposed to contain it.

Increasing and Decreasing Functions (Examples 9-16)

Example 9: A linear warm-up

Show that $f(x) = 3x + 17$ is increasing on $\mathbb{R}$.

Solution:

1. $f'(x) = 3 > 0$ for every real $x$.
2. By the first derivative test for monotonicity, $f$ is increasing on $\mathbb{R}$.

Final Answer: Proved.

Example 10: Sine on its first half-period

Show that $f(x) = \sin x$ is (a) increasing on $\left(0, \frac{\pi}{2}\right)$, (b) decreasing on $\left(\frac{\pi}{2}, \pi\right)$, (c) neither on $(0, \pi)$.

Solution:

1. $f'(x) = \cos x$.
2. (a) On $\left(0, \frac{\pi}{2}\right)$: $\cos x > 0 \Rightarrow$ increasing.
3. (b) On $\left(\frac{\pi}{2}, \pi\right)$: $\cos x < 0 \Rightarrow$ decreasing.
4. (c) Since it rises then falls, it is neither on $(0, \pi)$.

Final Answer: As proved.

Example 11: A downward parabola

Find the intervals in which $f(x) = 6 - 9x - x^2$ is strictly increasing or decreasing.

Solution:

1. $f'(x) = -9 - 2x = 0 \Rightarrow x = -\frac{9}{2}$.
2. For $x < -\frac{9}{2}$: $f' > 0$ (increasing). For $x > -\frac{9}{2}$: $f' < 0$ (decreasing).

Final Answer: Strictly increasing on $\left(-\infty, -\frac{9}{2}\right)$; strictly decreasing on $\left(-\frac{9}{2}, \infty\right)$.

Example 12: A quartic with three critical points

Find the intervals in which $f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11$ is (a) increasing (b) decreasing.

Solution:

1. Differentiate and factorise: $f'(x) = \frac{6}{5}x^3 - \frac{12}{5}x^2 - 6x + \frac{36}{5} = \frac{6}{5}(x - 1)(x + 2)(x - 3)$.
2. Critical values: $x = -2, 1, 3$ — four intervals.
3. Sign table: $(-\infty, -2)$: $(-)(-)(-) < 0$; $(-2, 1)$: $(-)(+)(-) > 0$; $(1, 3)$: $(+)(+)(-) < 0$; $(3, \infty)$: $(+)(+)(+) > 0$.

Final Answer: Increasing on $(-2, 1)$ and $(3, \infty)$; decreasing on $(-\infty, -2)$ and $(1, 3)$.

Takeaway: This is NCERT's Miscellaneous Example 33 — the model for handling three critical points cleanly with a sign table.

Example 13: An inverse-trig composition

Show that $f(x) = \tan^{-1}(\sin x + \cos x)$ is an increasing function on $\left(0, \frac{\pi}{4}\right)$.

Solution:

1. Chain rule: $f'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2} = \frac{\cos x - \sin x}{2 + \sin 2x}$.
2. Denominator: $2 + \sin 2x > 0$ always (since $\sin 2x \geq -1$).
3. Numerator: on $\left(0, \frac{\pi}{4}\right)$, $\cos x > \sin x$, so $\cos x - \sin x > 0$.
4. Hence $f'(x) > 0$ on the interval.

Final Answer: $f$ is increasing on $\left(0, \frac{\pi}{4}\right)$. ∎

Takeaway: $\tan^{-1}$ is itself increasing, so $f$ inherits the monotonicity of the inner function $\sin x + \cos x$ — a slick way to double-check.

Example 14: Symmetric reciprocal powers

Find the intervals in which $f(x) = x^3 + \frac{1}{x^3}$, $x \neq 0$, is (a) increasing (b) decreasing.

Solution:

1. Differentiate: $f'(x) = 3x^2 - \frac{3}{x^4} = \frac{3(x^6 - 1)}{x^4}$.
2. Sign: $x^4 > 0$ always, so the sign follows $x^6 - 1$: positive when $|x| > 1$, negative when $0 < |x| < 1$.

Final Answer: Increasing on $(-\infty, -1)$ and $(1, \infty)$; decreasing on $(-1, 0)$ and $(0, 1)$.

Takeaway: Exclude $x = 0$ from every interval you report — it's not in the domain.

Example 15: A hard rational-trig function

Find the intervals in which $f(x) = \frac{4\sin x - 2x - x\cos x}{2 + \cos x}$, $x \in [0, 2\pi]$, is (i) increasing (ii) decreasing.

Solution:

1. Rewrite: $f(x) = \frac{4\sin x}{2 + \cos x} - x$ (split the numerator: $\frac{-2x - x\cos x}{2+\cos x} = -x$).
2. Differentiate the first term (quotient rule): $f'(x) = \frac{4\cos x(2 + \cos x) - 4\sin x(-\sin x)}{(2+\cos x)^2} - 1 = \frac{8\cos x + 4}{(2+\cos x)^2} - 1$.
3. Combine: $f'(x) = \frac{8\cos x + 4 - (2 + \cos x)^2}{(2+\cos x)^2} = \frac{4\cos x - \cos^2 x}{(2+\cos x)^2} = \frac{\cos x\,(4 - \cos x)}{(2 + \cos x)^2}$.
4. Sign: $4 - \cos x > 0$ and $(2+\cos x)^2 > 0$ always, so the sign of $f'$ is the sign of $\cos x$.

Final Answer: Increasing on $\left(0, \frac{\pi}{2}\right)$ and $\left(\frac{3\pi}{2}, 2\pi\right)$; decreasing on $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$.

Takeaway: The pre-simplification in step 1 is the whole game — spotting $\frac{-x(2+\cos x)}{2+\cos x}$ saves five minutes of quotient-rule agony.

Example 16: The famous $\frac{\log x}{x}$

Show that $f(x) = \frac{\log x}{x}$ has a maximum at $x = e$.

Solution:

1. Differentiate: $f'(x) = \frac{\frac{1}{x}\cdot x - \log x}{x^2} = \frac{1 - \log x}{x^2}$.
2. Critical point: $f'(x) = 0 \Rightarrow \log x = 1 \Rightarrow x = e$.
3. Sign change: for $x < e$, $\log x < 1$ so $f' > 0$; for $x > e$, $f' < 0$. Maximum at $x = e$ with value $\frac{1}{e}$.

Final Answer: $f$ has a (local and absolute) maximum $\frac{1}{e}$ at $x = e$. ∎

Takeaway: This result powers the classic JEE comparison $e^\pi > \pi^e$: since $\frac{\log x}{x}$ peaks at $e$, $\frac{\log e}{e} > \frac{\log \pi}{\pi}$, and cross-multiplying gives it.

Maxima, Minima and Optimisation (Examples 17-32)

Example 17: A bell-shaped rational function

Find the local maximum value of $g(x) = \frac{1}{x^2 + 2}$.

Solution:

1. $g'(x) = \frac{-2x}{(x^2+2)^2} = 0 \Rightarrow x = 0$.
2. Sign of $g'$: positive for $x < 0$, negative for $x > 0$ — local maximum at $x = 0$.
3. Value: $g(0) = \frac{1}{2}$.

Final Answer: Local maximum value $\frac{1}{2}$ at $x = 0$.

Example 18: Mixed powers, three critical points

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

Solution:

1. Differentiate and factorise: $f'(x) = 4(x-2)^3(x+1)^3 + 3(x-2)^4(x+1)^2 = (x-2)^3(x+1)^2(7x - 2)$.
2. Critical points: $x = 2, -1, \frac{2}{7}$.
3. Sign analysis: $(x+1)^2 \geq 0$ never changes sign. Through $x = \frac{2}{7}$: $f'$ goes $(+) \to (-)$ — local maximum. Through $x = 2$: $f'$ goes $(-) \to (+)$ — local minimum. Through $x = -1$: no sign change — inflexion.

Final Answer: Local maximum at $x = \frac{2}{7}$; local minimum at $x = 2$; point of inflexion at $x = -1$.

Takeaway: Odd powers in $f'$ flip the sign, even powers don't — read the multiplicities instead of testing every interval.

Example 19: Trig absolute extremes on a closed interval

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

Solution:

1. Differentiate: $f'(x) = -2\cos x \sin x + \cos x = \cos x(1 - 2\sin x)$.
2. Critical points in $(0, \pi)$: $\cos x = 0 \Rightarrow x = \frac{\pi}{2}$; $\sin x = \frac{1}{2} \Rightarrow x = \frac{\pi}{6}, \frac{5\pi}{6}$.
3. Evaluate all candidates: $f(0) = 1$; $f\left(\frac{\pi}{6}\right) = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}$; $f\left(\frac{\pi}{2}\right) = 0 + 1 = 1$; $f\left(\frac{5\pi}{6}\right) = \frac{5}{4}$; $f(\pi) = 1$.

Final Answer: Absolute maximum $\frac{5}{4}$ (at $x = \frac{\pi}{6}$ and $\frac{5\pi}{6}$); absolute minimum 1 (at $x = 0, \frac{\pi}{2}, \pi$).

Takeaway: Factorise $f'$ fully — the $\cos x$ factor alone would miss two of the three critical points.

Example 20: $x + \sin 2x$ over a full period

Find the maximum and minimum values of $f(x) = x + \sin 2x$ on $[0, 2\pi]$.

Solution:

1. Critical points: $f'(x) = 1 + 2\cos 2x = 0 \Rightarrow \cos 2x = -\frac{1}{2} \Rightarrow 2x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}$, i.e. $x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$.
2. Evaluate candidates: $f(0) = 0$; $f\left(\frac{\pi}{3}\right) = \frac{\pi}{3} + \frac{\sqrt 3}{2}$; $f\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} - \frac{\sqrt 3}{2}$; $f\left(\frac{4\pi}{3}\right) = \frac{4\pi}{3} + \frac{\sqrt 3}{2}$; $f\left(\frac{5\pi}{3}\right) = \frac{5\pi}{3} - \frac{\sqrt 3}{2}$; $f(2\pi) = 2\pi$.
3. Compare: the largest is $2\pi$ (endpoint), the smallest is 0 (endpoint).

Final Answer: Maximum value $2\pi$ at $x = 2\pi$; minimum value 0 at $x = 0$.

Takeaway: Four interior critical points, yet both champions are endpoints — the rising trend $x$ dominates the bounded wiggle $\sin 2x$.

Example 21: Minimum sum of cubes

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

Solution:

1. Setup: numbers $x$ and $16 - x$; $S(x) = x^3 + (16 - x)^3$, $0 < x < 16$.
2. Differentiate: $S'(x) = 3x^2 - 3(16 - x)^2 = 3\left[x^2 - (16-x)^2\right]$. By difference of squares, $x^2 - (16-x)^2 = (x - (16-x))(x + (16-x)) = (2x - 16)(16)$, so $S'(x) = 48(2x - 16)$.
3. Critical point: $x = 8$; $S''(x) = 96 > 0$: minimum.

Final Answer: The numbers are 8 and 8.

Takeaway: Use the difference-of-squares factoring $a^2 - b^2 = (a-b)(a+b)$ on $S'$ — it avoids expanding cubes entirely.

Example 22: Box from a rectangular sheet

A rectangular sheet of tin 45 cm by 24 cm is made into an open box by cutting equal squares of side $x$ from the corners. Find $x$ so that the volume is maximum.

Solution:

1. Volume: $V(x) = x(45 - 2x)(24 - 2x)$, $0 < x < 12$.
2. Expand and differentiate: $V(x) = 4x^3 - 138x^2 + 1080x$, so $V'(x) = 12x^2 - 276x + 1080 = 12(x - 5)(x - 18)$.
3. Critical points: $x = 5$ or $x = 18$ (rejected: outside domain).
4. Confirm: $V''(x) = 24x - 276$, $V''(5) = -156 < 0$: maximum. $V(5) = 5 \times 35 \times 14 = 2450$ cm³.

Final Answer: $x = 5$ cm (maximum volume 2450 cm³).

Example 23: The best rectangle in a circle (proof)

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

Solution:

1. Setup: circle of radius $R$; a rectangle with sides $x, y$ inscribed in it has diagonal $= 2R$, so $x^2 + y^2 = 4R^2$.
2. Target: maximise $A = xy$, or equivalently $A^2 = x^2(4R^2 - x^2)$. Let $t = x^2$: maximise $g(t) = t(4R^2 - t)$, $0 < t < 4R^2$.
3. Optimise: $g'(t) = 4R^2 - 2t = 0 \Rightarrow t = 2R^2$; $g''(t) = -2 < 0$: maximum.
4. Conclude: $x^2 = 2R^2 \Rightarrow x = R\sqrt{2}$ and $y = \sqrt{4R^2 - 2R^2} = R\sqrt{2} = x$ — a square.

Final Answer: The inscribed rectangle of maximum area is the square (side $R\sqrt{2}$). ∎

Takeaway: Maximising $A^2$ instead of $A$, and substituting $t = x^2$, turns the whole proof into a parabola-vertex problem.

Example 24: The manufacturer's profit

A manufacturer can sell $x$ items at a price of ₹$\left(5 - \frac{x}{100}\right)$ each. The cost price of $x$ items is ₹$\left(\frac{x}{5} + 500\right)$. Find the number of items he should sell to earn maximum profit.

Solution:

1. Revenue: $S(x) = x\left(5 - \frac{x}{100}\right) = 5x - \frac{x^2}{100}$.
2. Profit: $P(x) = S(x) - C(x) = 5x - \frac{x^2}{100} - \frac{x}{5} - 500 = \frac{24}{5}x - \frac{x^2}{100} - 500$.
3. Optimise: $P'(x) = \frac{24}{5} - \frac{x}{50} = 0 \Rightarrow x = 240$; $P''(x) = -\frac{1}{50} < 0$: maximum.

Final Answer: Selling 240 items gives maximum profit.

Takeaway: Profit = revenue − cost, always. (The maximum profit itself is $P(240) = 576 - 500 = 76$, i.e. ₹76 — worth computing if asked.)

Example 25: The 3 m × 8 m aluminium sheet

An open-topped box is made by removing equal squares from each corner of a 3 m by 8 m rectangular sheet and folding up the sides. Find the volume of the largest such box.

Solution:

1. Volume: $V(x) = x(3 - 2x)(8 - 2x) = 4x^3 - 22x^2 + 24x$, $0 < x < \frac{3}{2}$.
2. Differentiate: $V'(x) = 12x^2 - 44x + 24 = 4(x - 3)(3x - 2)$.
3. Critical points: $x = 3$ (rejected — exceeds $\frac{3}{2}$) or $x = \frac{2}{3}$.
4. Confirm: $V''\left(\frac{2}{3}\right) = 24 \cdot \frac{2}{3} - 44 = -28 < 0$: maximum.
5. Value: $V\left(\frac{2}{3}\right) = \frac{2}{3}\left(3 - \frac{4}{3}\right)\left(8 - \frac{4}{3}\right) = \frac{2}{3} \cdot \frac{5}{3} \cdot \frac{20}{3} = \frac{200}{27}$ m³.

Final Answer: Largest volume $= \frac{200}{27}$ m³ (cut squares of side $\frac{2}{3}$ m).

Takeaway: State the domain first — it instantly disqualifies the fake root $x = 3$ with the one-word reason "breadth would be negative".

Example 26: The cheapest tank

An open-topped tank with a rectangular base and rectangular sides is to have depth 2 m and volume 8 m³. Building costs are ₹70/m² for the base and ₹45/m² for the sides. What is the cost of the least expensive tank?

Solution:

1. Constraint: base $x \times y$ with depth 2: volume $2xy = 8 \Rightarrow xy = 4$.
2. Cost: $C = 70xy + 45 \times 2 \times 2(x + y) = 280 + 180(x + y)$.
3. Minimise $x + y$ given $xy = 4$: $x + \frac{4}{x}$ has derivative $1 - \frac{4}{x^2} = 0 \Rightarrow x = 2$, so $y = 2$ (second derivative $\frac{8}{x^3} > 0$: minimum).
4. Cost: $C = 280 + 180(4) = 1000$.

Final Answer: The least expensive tank costs ₹1000.

Takeaway: Constants (the ₹280 base cost) ride along untouched — optimise only the variable part.

Example 27: Circle plus square, fixed total perimeter (proof)

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

Solution:

1. Variables: circle radius $r$, square side $a$: $2\pi r + 4a = k$, so $a = \frac{k - 2\pi r}{4}$.
2. Target: $A(r) = \pi r^2 + a^2 = \pi r^2 + \frac{(k - 2\pi r)^2}{16}$.
3. Differentiate: $A'(r) = 2\pi r + \frac{2(k - 2\pi r)(-2\pi)}{16} = 2\pi r - \frac{\pi(k - 2\pi r)}{4}$.
4. Solve: $A'(r) = 0 \Rightarrow 8r = k - 2\pi r \Rightarrow r = \frac{k}{8 + 2\pi}$; then $a = \frac{k - 2\pi r}{4} = \frac{8r}{4} = 2r$.
5. Confirm: $A''(r) = 2\pi + \frac{\pi^2}{2} > 0$: minimum.

Final Answer: The combined area is least when $a = 2r$ — side of square = diameter of circle. ∎

Example 28: The Norman window

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

Solution:

1. Variables: semicircle radius $r$ (so rectangle width $2r$), rectangle height $h$. Perimeter: $2r + 2h + \pi r = 10$.
2. Target (area): $A = 2rh + \frac{\pi r^2}{2}$. From the perimeter, $h = \frac{10 - 2r - \pi r}{2}$, so $A(r) = r(10 - 2r - \pi r) + \frac{\pi r^2}{2} = 10r - 2r^2 - \frac{\pi r^2}{2}$.
3. Differentiate: $A'(r) = 10 - 4r - \pi r = 0 \Rightarrow r = \frac{10}{\pi + 4}$; $A''(r) = -4 - \pi < 0$: maximum.
4. Height: $h = \frac{10 - (2 + \pi)r}{2} = \frac{10}{\pi + 4}$.

Final Answer: Width $= 2r = \frac{20}{\pi + 4}$ m, rectangle height $= \frac{10}{\pi + 4}$ m (equal to the radius).

Takeaway: Answer in the requested form — "dimensions" means the width AND the height, not just $r$.

Example 29: Minimum hypotenuse (a JEE-grade proof)

A point on the hypotenuse of a right triangle is at distances $a$ and $b$ from the two sides. Show that the minimum length of the hypotenuse is $\left(a^{2/3} + b^{2/3}\right)^{3/2}$.

Solution:

1. Parametrise by the angle $\theta$ the hypotenuse makes with one side. The hypotenuse through the point splits into two segments: $L(\theta) = \frac{a}{\sin\theta} + \frac{b}{\cos\theta}$, $0 < \theta < \frac{\pi}{2}$.
2. Differentiate: $L'(\theta) = -\frac{a\cos\theta}{\sin^2\theta} + \frac{b\sin\theta}{\cos^2\theta}$.
3. Solve $L'(\theta) = 0$: $b\sin^3\theta = a\cos^3\theta \Rightarrow \tan^3\theta = \frac{a}{b} \Rightarrow \tan\theta = \left(\frac{a}{b}\right)^{1/3}$.
4. Substitute back: with $\tan\theta = \frac{a^{1/3}}{b^{1/3}}$, we get $\sin\theta = \frac{a^{1/3}}{\sqrt{a^{2/3} + b^{2/3}}}$ and $\cos\theta = \frac{b^{1/3}}{\sqrt{a^{2/3} + b^{2/3}}}$. Then $$L = a \cdot \frac{\sqrt{a^{2/3}+b^{2/3}}}{a^{1/3}} + b \cdot \frac{\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}} = \left(a^{2/3} + b^{2/3}\right)\sqrt{a^{2/3} + b^{2/3}} = \left(a^{2/3} + b^{2/3}\right)^{3/2}$$
5. Nature: $L \to \infty$ at both ends of the domain, so the single interior critical point is the minimum.

Final Answer: Minimum hypotenuse length $= \left(a^{2/3} + b^{2/3}\right)^{3/2}$. ∎

Takeaway: When lengths blow up at the domain's ends, a lone interior critical point must be the minimum — say so instead of computing a messy second derivative.

Example 30: Largest cone in a sphere

Show that the altitude of the right circular cone of maximum volume inscribed in a sphere of radius $r$ is $\frac{4r}{3}$, and that its volume is $\frac{8}{27}$ of the sphere's volume.

Solution:

1. Setup: let the cone's altitude be $h$; its base circle sits at distance $h - r$ from the centre, so base radius $R_c$ satisfies $R_c^2 = r^2 - (h - r)^2 = 2rh - h^2$.
2. Volume: $V(h) = \frac{\pi}{3}(2rh - h^2)h = \frac{\pi}{3}(2rh^2 - h^3)$, $0 < h < 2r$.
3. Differentiate: $V'(h) = \frac{\pi}{3}(4rh - 3h^2) = \frac{\pi h}{3}(4r - 3h) = 0 \Rightarrow h = \frac{4r}{3}$.
4. Confirm: $V''(h) = \frac{\pi}{3}(4r - 6h)$; at $h = \frac{4r}{3}$: $V'' = \frac{\pi}{3}(4r - 8r) < 0$: maximum.
5. Compare volumes: $V\left(\frac{4r}{3}\right) = \frac{\pi}{3}\left(2r \cdot \frac{16r^2}{9} - \frac{64r^3}{27}\right) = \frac{\pi}{3} \cdot \frac{32r^3}{27} = \frac{8}{27} \cdot \frac{4}{3}\pi r^3$.

Final Answer: Altitude $\frac{4r}{3}$; maximum cone volume $= \frac{8}{27}$ of the sphere's volume. ∎

Example 31: Largest cylinder in a sphere

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

Solution:

1. Setup: cylinder of height $h$; half the height and the cylinder radius $x$ make a right triangle with the sphere radius: $x^2 = R^2 - \frac{h^2}{4}$.
2. Volume: $V(h) = \pi x^2 h = \pi h\left(R^2 - \frac{h^2}{4}\right) = \pi R^2 h - \frac{\pi h^3}{4}$, $0 < h < 2R$.
3. Differentiate: $V'(h) = \pi R^2 - \frac{3\pi h^2}{4} = 0 \Rightarrow h^2 = \frac{4R^2}{3} \Rightarrow h = \frac{2R}{\sqrt{3}}$.
4. Confirm: $V''(h) = -\frac{3\pi h}{2} < 0$: maximum.
5. Maximum volume: $V = \pi \cdot \frac{2R}{\sqrt 3}\left(R^2 - \frac{R^2}{3}\right) = \pi \cdot \frac{2R}{\sqrt 3} \cdot \frac{2R^2}{3} = \frac{4\pi R^3}{3\sqrt{3}}$.

Final Answer: Height $= \frac{2R}{\sqrt 3}$; maximum volume $= \frac{4\pi R^3}{3\sqrt 3}$. ∎

Takeaway: Draw the axial cross-section: the sphere becomes a circle, the cylinder a rectangle, and Pythagoras hands you the constraint.

Example 32: Largest cylinder in a cone

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

Solution:

1. Setup: cone base radius $= h\tan\alpha$. For a cylinder of radius $x$, similar triangles give cylinder height $H = \frac{h(h\tan\alpha - x)}{h\tan\alpha} = h - \frac{x}{\tan\alpha}$.
2. Volume: $V(x) = \pi x^2\left(h - \frac{x}{\tan\alpha}\right) = \pi h x^2 - \frac{\pi x^3}{\tan\alpha}$.
3. Differentiate: $V'(x) = 2\pi h x - \frac{3\pi x^2}{\tan\alpha} = \pi x\left(2h - \frac{3x}{\tan\alpha}\right) = 0 \Rightarrow x = \frac{2h\tan\alpha}{3}$.
4. Cylinder height at this $x$: $H = h - \frac{2h}{3} = \frac{h}{3}$ — one-third of the cone's height.
5. Confirm and evaluate: $V''\left(\frac{2h\tan\alpha}{3}\right) = 2\pi h - \frac{6\pi}{\tan\alpha}\cdot\frac{2h\tan\alpha}{3} = -2\pi h < 0$ (maximum), and $V = \pi\left(\frac{2h\tan\alpha}{3}\right)^2 \cdot \frac{h}{3} = \frac{4}{27}\pi h^3\tan^2\alpha$.

Final Answer: Cylinder height $= \frac{h}{3}$; greatest volume $= \frac{4}{27}\pi h^3 \tan^2\alpha$. ∎

Takeaway: Compare with the curved-surface version (Section 4): greatest surface needs $x = \frac{r}{2}$, greatest volume needs $x = \frac{2r}{3}$. Read which quantity is being maximised!