NCERT Class 12 Mathematics — Application of Derivatives, Exercise 6.3. All 29 questions solved.

The longest exercise in the chapter, and the one that carries the most board marks. It has three distinct parts, and they use different methods:

Local maxima and minima (questions 1 to 4). Find the critical points where $f'(x) = 0$, then classify each one:

• First derivative test. If $f’$ changes from $+$ to $-$ at $c$, it is a local maximum; from $-$ to $+$, a local minimum; no change, a point of inflexion.
• Second derivative test. If $f'(c) = 0$ and $f”(c) < 0$, local maximum; if $f”(c) > 0$, local minimum. Inconclusive when $f”(c) = 0$.

Absolute maxima and minima on a closed interval (questions 5 to 12). Evaluate $f$ at every critical point and at both endpoints, then compare the list.

Optimisation word problems (questions 13 to 26). Build a function of one variable from the geometry, using the given constraint to eliminate the second variable, and then minimise or maximise it.

Key insight. The single most common lost mark in this exercise is forgetting the endpoints in the second group. A continuous function on a closed interval always attains its extremes, but they need not be at critical points — question 5(i) has its absolute maximum and minimum both at endpoints, because $x^3$ has no local extremum at all.

In the third group, the whole difficulty is the setup. Once the constraint has been used to reduce everything to one variable, the calculus is a single derivative.

Question 1

Find the maximum and minimum values, if any, of the following functions given by

(i) $f(x) = (2x-1)^2 + 3$   (ii) $f(x) = 9x^2 + 12x + 2$   (iii) $f(x) = -(x-1)^2 + 10$   (iv) $g(x) = x^3 + 1$

Solution. These are all answerable by inspection, without calculus — which is the point of the question.

(i) $(2x-1)^2 \ge 0$ with equality at $x = \tfrac12$, so the smallest value of $f$ is $0 + 3 = 3$. It has no maximum: as $|x|$ grows, $f$ grows without bound.

(ii) Complete the square: $9x^2 + 12x + 2 = (3x+2)^2 – 2$. The square is non-negative, so the minimum is $-2$, attained at $x = -\tfrac23$. No maximum.

(iii) Here the square is subtracted, so the largest value is $10$, at $x = 1$. No minimum.

(iv) $x^3 + 1$ takes every real value; it has neither a maximum nor a minimum.

Question 2

Find the maximum and minimum values, if any, of the following functions given by

(i) $f(x) = |x+2| – 1$   (ii) $g(x) = -|x+1| + 3$   (iii) $h(x) = \sin(2x) + 5$   (iv) $f(x) = |\sin 4x + 3|$   (v) $h(x) = x + 1$, $x \in (-1, 1)$

Solution.

(i) $|x+2| \ge 0$, so the minimum is $-1$ at $x = -2$. Unbounded above, so no maximum.

(ii) $-|x+1| \le 0$, so the maximum is $3$ at $x = -1$. Unbounded below, so no minimum.

(iii) $\sin 2x$ ranges over $[-1, 1]$, so $h$ ranges over $[4, 6]$: minimum $4$, maximum $6$.

(iv) $\sin 4x + 3$ ranges over $[2, 4]$, which is entirely positive, so the modulus changes nothing: minimum $2$, maximum $4$.

(v) On the open interval $(-1, 1)$ the function $x + 1$ gets arbitrarily close to $0$ and to $2$ without ever reaching either. So it has neither a maximum nor a minimum — the openness of the interval is the whole content of this part.

Question 3

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

(i) $f(x) = x^2$   (ii) $g(x) = x^3 – 3x$   (iii) $h(x) = \sin x + \cos x$, $0 < x < \tfrac{\pi}{2}$   (iv) $f(x) = \sin x – \cos x$, $0 < x < 2\pi$   (v) $f(x) = x^3 – 6x^2 + 9x + 15$   (vi) $g(x) = \tfrac{x}{2} + \tfrac{2}{x}$, $x > 0$   (vii) $g(x) = \tfrac{1}{x^2+2}$   (viii) $f(x) = x\sqrt{1-x}$, $0 < x < 1$

Solution.

(i) $f'(x) = 2x = 0$ at $x = 0$, and $f”(x) = 2 > 0$. Local minimum at $x = 0$, value $0$.

(ii) $g'(x) = 3x^2 – 3 = 3(x-1)(x+1)$, zero at $x = \pm1$. With $g”(x) = 6x$: at $x = 1$, $g” = 6 > 0$, a local minimum of value $1 – 3 = -2$; at $x = -1$, $g” = -6 < 0$, a local maximum of value $-1 + 3 = 2$.

(iii) $h'(x) = \cos x – \sin x = 0$ gives $\tan x = 1$, so $x = \tfrac{\pi}{4}$ in the given range. $h”(x) = -\sin x – \cos x < 0$ there, so it is a local maximum of value $\tfrac{1}{\sqrt2} + \tfrac{1}{\sqrt2} = \sqrt2$.

(iv) $f'(x) = \cos x + \sin x = 0$ gives $\tan x = -1$, so $x = \tfrac{3\pi}{4}$ or $x = \tfrac{7\pi}{4}$ in $(0, 2\pi)$. Differentiating again, $f”(x) = -\sin x + \cos x$.

At $x = \tfrac{3\pi}{4}$: $\sin x = \tfrac{1}{\sqrt2}$ and $\cos x = -\tfrac{1}{\sqrt2}$, so $f” = -\tfrac{1}{\sqrt2} – \tfrac{1}{\sqrt2} = -\sqrt2 < 0$ — a local maximum, of value $\tfrac{1}{\sqrt2} + \tfrac{1}{\sqrt2} = \sqrt2$.

At $x = \tfrac{7\pi}{4}$: $\sin x = -\tfrac{1}{\sqrt2}$ and $\cos x = \tfrac{1}{\sqrt2}$, so $f” = \tfrac{1}{\sqrt2} + \tfrac{1}{\sqrt2} = \sqrt2 > 0$ — a local minimum, of value $-\tfrac{1}{\sqrt2} – \tfrac{1}{\sqrt2} = -\sqrt2$.

(v) $f'(x) = 3x^2 – 12x + 9 = 3(x-1)(x-3)$, zero at $x = 1$ and $x = 3$. $f”(x) = 6x – 12$: negative at $x = 1$ (local maximum, value $1 – 6 + 9 + 15 = 19$), positive at $x = 3$ (local minimum, value $27 – 54 + 27 + 15 = 15$).

(vi) $g'(x) = \tfrac12 – \tfrac{2}{x^2} = 0$ gives $x^2 = 4$, so $x = 2$ in the domain $x > 0$. $g”(x) = \tfrac{4}{x^3} > 0$ there: local minimum of value $1 + 1 = 2$.

(vii) $g'(x) = \dfrac{-2x}{\left(x^2+2\right)^2}$, zero only at $x = 0$. The derivative is positive for $x < 0$ and negative for $x > 0$, so $x = 0$ is a local maximum of value $\tfrac12$.

(viii) $f'(x) = \sqrt{1-x} – \dfrac{x}{2\sqrt{1-x}} = \dfrac{2(1-x) – x}{2\sqrt{1-x}} = \dfrac{2 – 3x}{2\sqrt{1-x}}$, zero at $x = \tfrac23$. The numerator changes from $+$ to $-$ there, so it is a local maximum, of value $\tfrac23\sqrt{\tfrac13} = \dfrac{2}{3\sqrt3} = \dfrac{2\sqrt3}{9}$.

Question 4

Prove that the following functions do not have maxima or minima:

(i) $f(x) = e^x$   (ii) $g(x) = \log x$   (iii) $h(x) = x^3 + x^2 + x + 1$

Solution. A local extremum of a differentiable function can only occur where $f’ = 0$. So it is enough to show the derivative never vanishes.

(i) $f'(x) = e^x$, which is strictly positive for every real $x$ and so never zero.

(ii) $g'(x) = \tfrac1x$, which is never zero on the domain $x > 0$.

(iii) $h'(x) = 3x^2 + 2x + 1$. Its discriminant is $2^2 – 4(3)(1) = 4 – 12 = -8 < 0$, so the quadratic has no real roots and, having a positive leading coefficient, is strictly positive everywhere.

In each case $f’$ has no zero, so there is no critical point and hence no maximum or minimum.

Question 5

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i) $f(x) = x^3$, $x \in [-2, 2]$   (ii) $f(x) = \sin x + \cos x$, $x \in [0, \pi]$   (iii) $f(x) = 4x – \tfrac12 x^2$, $x \in \left[-2, \tfrac92\right]$   (iv) $f(x) = (x-1)^2 + 3$, $x \in [-3, 1]$

Solution. In each case: list the critical points inside the interval, add the two endpoints, evaluate, compare.

(i) $f'(x) = 3x^2 = 0$ only at $x = 0$, where $f = 0$. Endpoints: $f(-2) = -8$ and $f(2) = 8$. Both extremes are at endpoints.

(ii) $f'(x) = \cos x – \sin x = 0$ at $x = \tfrac{\pi}{4}$, where $f = \sqrt2$. Endpoints: $f(0) = 1$ and $f(\pi) = -1$.

(iii) $f'(x) = 4 – x = 0$ at $x = 4$, where $f = 16 – 8 = 8$. Endpoints: $f(-2) = -8 – 2 = -10$ and $f\left(\tfrac92\right) = 18 – \tfrac{81}{8} = 7.875$.

(iv) $f'(x) = 2(x-1) = 0$ at $x = 1$, which is also the right endpoint; $f(1) = 3$. The other endpoint gives $f(-3) = 16 + 3 = 19$. So the smallest value is $3$ and the largest is $19$.

A note on the printed answer. For part (iv) the NCERT key gives “absolute minimum value $= 19$, absolute maximum value $= 3$”. The two labels are interchanged. $f(x) = (x-1)^2 + 3$ is an upward parabola with vertex $(1, 3)$, so $3$ is the smallest value it can take anywhere, and $19$ is what it reaches at the far endpoint $x = -3$. The values above are the correct way round.

Question 6

Find the maximum profit that a company can make, if the profit function is given by $p(x) = 41 – 72x – 18x^2$.

Solution. $p'(x) = -72 – 36x = 0$ gives $x = -2$, and $p”(x) = -36 < 0$, confirming a maximum.

$$p(-2) = 41 + 144 – 72 = 113$$

Question 7

Find both the maximum value and the minimum value of $3x^4 – 8x^3 + 12x^2 – 48x + 25$ on the interval $[0, 3]$.

Solution.

$$f'(x) = 12x^3 – 24x^2 + 24x – 48 = 12\left(x^3 – 2x^2 + 2x – 4\right)$$

Factor by grouping: $x^2(x-2) + 2(x-2) = (x-2)\left(x^2+2\right)$. Since $x^2 + 2$ is never zero, the only critical point is $x = 2$.

$$f(0) = 25, \qquad f(2) = 48 – 64 + 48 – 96 + 25 = -39, \qquad f(3) = 243 – 216 + 108 – 144 + 25 = 16$$

Question 8

At what points in the interval $[0, 2\pi]$ does the function $\sin 2x$ attain its maximum value?

Solution. $\sin 2x$ has maximum value $1$, attained when $2x = \tfrac{\pi}{2}$ or $2x = \tfrac{5\pi}{2}$ — the two odd multiples of $\tfrac{\pi}{2}$ that fall in $[0, 4\pi]$, which is the range of $2x$.

$$x = \frac{\pi}{4} \quad\text{and}\quad x = \frac{5\pi}{4}$$

Doubling the interval is what produces two answers rather than one; a solution that reports only $\tfrac{\pi}{4}$ has forgotten that $2x$ runs twice round.

Question 9

What is the maximum value of the function $\sin x + \cos x$?

Solution. Write it as a single sine using the standard amplitude form:

$$\sin x + \cos x = \sqrt2\left(\frac{1}{\sqrt2}\sin x + \frac{1}{\sqrt2}\cos x\right) = \sqrt2\sin\left(x + \frac{\pi}{4}\right)$$

Since the sine never exceeds $1$, the maximum is $\sqrt2$.

Question 10

Find the maximum value of $2x^3 – 24x + 107$ in the interval $[1, 3]$. Find the maximum value of the same function in $[-3, -1]$.

Solution. $f'(x) = 6x^2 – 24 = 0$ gives $x = \pm2$.

On $[1, 3]$, the critical point $x = 2$ lies inside:

$$f(1) = 85, \qquad f(2) = 75, \qquad f(3) = 89$$

On $[-3, -1]$, the critical point $x = -2$ lies inside:

$$f(-3) = 125, \qquad f(-2) = 139, \qquad f(-1) = 129$$

The same function, the same derivative, and completely different answers — because the interval decides which critical point is relevant.

Question 11

It is given that at $x = 1$, the function $x^4 – 62x^2 + ax + 9$ attains its maximum value on the interval $[0, 2]$. Find the value of $a$.

Solution. If the maximum on $[0, 2]$ occurs at the interior point $x = 1$, then $x = 1$ must be a critical point:

$$f'(x) = 4x^3 – 124x + a, \qquad f'(1) = 4 – 124 + a = 0$$

Question 12

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

Solution. $f'(x) = 1 + 2\cos 2x = 0$ gives $\cos 2x = -\tfrac12$, so $2x \in \left\{\tfrac{2\pi}{3}, \tfrac{4\pi}{3}, \tfrac{8\pi}{3}, \tfrac{10\pi}{3}\right\}$ and hence

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

Evaluating $f$ at these four points gives roughly $1.91$, $1.23$, $5.06$ and $4.37$. The endpoints give $f(0) = 0$ and $f(2\pi) = 2\pi \approx 6.28$.

Neither extreme is at a critical point: $f’$ is never negative for long enough to undo the steady $+x$, so the function drifts upward overall.

Question 13

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

Solution. Let the numbers be $x$ and $24 – x$, and maximise $P = x(24-x) = 24x – x^2$.

$$\frac{dP}{dx} = 24 – 2x = 0 \quad\Longrightarrow\quad x = 12$$

and $\tfrac{d^2P}{dx^2} = -2 < 0$, confirming a maximum.

Question 14

Find two positive numbers $x$ and $y$ such that $x + y = 60$ and $xy^3$ is maximum.

Solution. Substitute $y = 60 – x$ and maximise $f(x) = x(60-x)^3$:

$$f'(x) = (60-x)^3 – 3x(60-x)^2 = (60-x)^2\big[(60-x) – 3x\big] = (60-x)^2(60 – 4x)$$

The squared factor never changes sign, so the only sign change is at $x = 15$, where $f’$ goes from $+$ to $-$: a maximum. Then $y = 45$.

Question 15

Find two positive numbers $x$ and $y$ such that their sum is $35$ and the product $x^2y^5$ is a maximum.

Solution. With $y = 35 – x$, maximise $f(x) = x^2(35-x)^5$:

$$f'(x) = 2x(35-x)^5 – 5x^2(35-x)^4 = x(35-x)^4\big[2(35-x) – 5x\big] = x(35-x)^4(70 – 7x)$$

The sign change that matters is at $x = 10$, giving $y = 25$.

Note the pattern across questions 14 and 15: for $x^my^n$ with $x + y = s$, the maximum is at $x = \tfrac{m}{m+n}s$. Here $\tfrac{2}{7}(35) = 10$.

Question 16

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

Solution. Minimise $f(x) = x^3 + (16-x)^3$:

$$f'(x) = 3x^2 – 3(16-x)^2 = 3\big[x – (16-x)\big]\big[x + (16-x)\big] = 48(2x – 16)$$

using the difference of squares, since the second bracket is just $16$. This vanishes at $x = 8$, and $f”(x) = 96 > 0$, confirming a minimum.

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 to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

Solution. If a square of side $x$ is cut from each corner, the base of the box measures $(18 – 2x)$ by $(18 – 2x)$ and the height is $x$:

$$V = x(18-2x)^2, \qquad 0 < x < 9$$

$$\frac{dV}{dx} = (18-2x)^2 + x \cdot 2(18-2x)(-2) = (18-2x)\big[(18-2x) – 4x\big] = (18-2x)(18-6x)$$

The root $x = 9$ is excluded (it gives zero volume), leaving $x = 3$. Checking the sign: $V’$ is positive just below $3$ and negative just above, so it is a maximum.

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. Cutting $x$ from each corner leaves a base of $(45 – 2x)$ by $(24 – 2x)$:

$$V = x(45-2x)(24-2x) = 4x^3 – 138x^2 + 1080x$$

$$\frac{dV}{dx} = 12x^2 – 276x + 1080 = 12\left(x^2 – 23x + 90\right) = 12(x-5)(x-18)$$

Only $x = 5$ lies in the permissible range $0 < x < 12$ (the shorter side limits it), and there $V” = 24x – 276 = -156 < 0$: a maximum.

Question 19

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

Solution. Let the circle have radius $a$, and let the inscribed rectangle have sides $2a\cos\theta$ and $2a\sin\theta$ — this parametrisation automatically satisfies the constraint that the diagonal is a diameter.

$$A = 4a^2\sin\theta\cos\theta = 2a^2\sin 2\theta$$

The sine is largest when $2\theta = \tfrac{\pi}{2}$, that is $\theta = \tfrac{\pi}{4}$. Then $\cos\theta = \sin\theta$ and the two sides are equal — the rectangle is a square, of area $2a^2$.

Question 20

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

Solution. With radius $r$ and height $h$, the total surface is $S = 2\pi r^2 + 2\pi rh$, a fixed constant. Solve for $h$ and substitute into the volume:

$$h = \frac{S – 2\pi r^2}{2\pi r}, \qquad V = \pi r^2 h = \frac{r\left(S – 2\pi r^2\right)}{2} = \frac{Sr – 2\pi r^3}{2}$$

$$\frac{dV}{dr} = \frac{S – 6\pi r^2}{2} = 0 \quad\Longrightarrow\quad S = 6\pi r^2$$

Substituting back:

$$h = \frac{6\pi r^2 – 2\pi r^2}{2\pi r} = \frac{4\pi r^2}{2\pi r} = 2r$$

which is exactly the diameter. Since $\tfrac{d^2V}{dr^2} = -6\pi r < 0$, this is a maximum.

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.

Solution. The constraint is $\pi r^2 h = 100$, so $h = \dfrac{100}{\pi r^2}$. Substituting into the total surface area:

$$S = 2\pi r^2 + 2\pi rh = 2\pi r^2 + \frac{200}{r}$$

$$\frac{dS}{dr} = 4\pi r – \frac{200}{r^2} = 0 \quad\Longrightarrow\quad r^3 = \frac{50}{\pi}$$

$$\frac{d^2S}{dr^2} = 4\pi + \frac{400}{r^3} > 0$$

so this is a minimum. For the height, note that $r^3 = \tfrac{50}{\pi}$ gives $\pi r^2 = \tfrac{50}{r}$, hence

$$h = \frac{100}{\pi r^2} = \frac{100r}{50} = 2r$$

— the same “height equals diameter” relation as question 20.

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 square’s piece have length $a$, so its side is $\tfrac{a}{4}$ and its area is $\tfrac{a^2}{16}$. The circle’s piece has length $28 – a$, so its radius is $\tfrac{28-a}{2\pi}$ and its area is $\pi\left(\tfrac{28-a}{2\pi}\right)^2 = \tfrac{(28-a)^2}{4\pi}$.

$$A(a) = \frac{a^2}{16} + \frac{(28-a)^2}{4\pi}$$

$$\frac{dA}{da} = \frac{a}{8} – \frac{28-a}{2\pi} = 0 \quad\Longrightarrow\quad 2\pi a = 8(28 – a)$$

$$a(2\pi + 8) = 224 \quad\Longrightarrow\quad a = \frac{224}{2\pi + 8} = \frac{112}{\pi + 4}$$

The other piece is $28 – \tfrac{112}{\pi+4} = \tfrac{28\pi}{\pi+4}$. The second derivative is positive, so this is the minimum.

Question 23

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

Solution. Let the cone have height $h$ measured from the apex, and base radius $\rho$. The geometry of a chord in a circle gives the relation

$$\rho^2 = h(2\mathrm{R} – h)$$

so the volume is a function of $h$ alone:

$$V = \frac13\pi\rho^2 h = \frac13\pi h^2(2\mathrm{R} – h)$$

$$\frac{dV}{dh} = \frac13\pi\left(4\mathrm{R}h – 3h^2\right) = \frac{\pi h}{3}(4\mathrm{R} – 3h) = 0 \quad\Longrightarrow\quad h = \frac{4\mathrm{R}}{3}$$

Substituting:

$$V = \frac13\pi \cdot \frac{16\mathrm{R}^2}{9} \cdot \frac{2\mathrm{R}}{3} = \frac{32\pi\mathrm{R}^3}{81}$$

The sphere’s volume is $\tfrac43\pi\mathrm{R}^3$, and

$$\frac{32/81}{4/3} = \frac{32}{81} \cdot \frac{3}{4} = \frac{8}{27}$$

Question 24

Show that the right circular cone of least curved surface and given volume has an altitude equal to $\sqrt2$ times the radius of the base.

Solution. The curved surface is $S = \pi r\ell$ with $\ell = \sqrt{r^2 + h^2}$, and the volume constraint is $\tfrac13\pi r^2h = V$, so $h = \dfrac{3V}{\pi r^2}$.

Minimising $S$ is the same as minimising $S^2$, which avoids the square root:

$$S^2 = \pi^2 r^2\left(r^2 + h^2\right) = \pi^2 r^4 + \pi^2 r^2 h^2 = \pi^2 r^4 + \frac{9V^2}{r^2}$$

$$\frac{d\left(S^2\right)}{dr} = 4\pi^2 r^3 – \frac{18V^2}{r^3} = 0 \quad\Longrightarrow\quad r^6 = \frac{9V^2}{2\pi^2}$$

Now $h^2 = \dfrac{9V^2}{\pi^2 r^4}$, so

$$\frac{h^2}{r^2} = \frac{9V^2}{\pi^2 r^6} = \frac{9V^2}{\pi^2} \cdot \frac{2\pi^2}{9V^2} = 2$$

giving $h = \sqrt2\,r$.

Question 25

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan^{-1}\sqrt2$.

Solution. With slant height $\ell$ fixed and semi-vertical angle $\alpha$, the radius is $r = \ell\sin\alpha$ and the height is $h = \ell\cos\alpha$:

$$V = \frac13\pi\ell^3\sin^2\alpha\cos\alpha$$

$$\frac{dV}{d\alpha} = \frac13\pi\ell^3\left(2\sin\alpha\cos^2\alpha – \sin^3\alpha\right) = \frac13\pi\ell^3\sin\alpha\left(2\cos^2\alpha – \sin^2\alpha\right)$$

Setting the bracket to zero gives $\tan^2\alpha = 2$, so $\alpha = \tan^{-1}\sqrt2$.

Question 26

Show that the semi-vertical angle of a right circular cone of given surface area and maximum volume is $\sin^{-1}\left(\tfrac13\right)$.

Solution. The total surface area $S = \pi r^2 + \pi r\ell$ is fixed, so $\ell = \dfrac{S – \pi r^2}{\pi r}$ and

$$h^2 = \ell^2 – r^2 = \frac{\left(S – \pi r^2\right)^2}{\pi^2 r^2} – r^2 = \frac{S^2 – 2\pi r^2 S}{\pi^2 r^2}$$

Maximise $V^2 = \tfrac19\pi^2 r^4 h^2 = \tfrac19 r^2\left(S^2 – 2\pi r^2 S\right)$:

$$\frac{d\left(V^2\right)}{dr} = \frac19\left(2rS^2 – 8\pi r^3 S\right) = 0 \quad\Longrightarrow\quad S = 4\pi r^2$$

Substituting back, $\ell = \dfrac{4\pi r^2 – \pi r^2}{\pi r} = 3r$, so

$$\sin\alpha = \frac{r}{\ell} = \frac13$$

Choose the correct answer in Questions 27 to 29.

Question 27

The point on the curve $x^2 = 2y$ which is nearest to the point $(0, 5)$ is

Solution. Minimise the squared distance — the square root is monotonic, so the minimising point is the same and the algebra is far cleaner:

$$D = x^2 + (y-5)^2 = 2y + (y-5)^2 = y^2 – 8y + 25$$

using the curve to replace $x^2$ by $2y$. Then $\tfrac{dD}{dy} = 2y – 8 = 0$ gives $y = 4$, and $x^2 = 8$, so $x = \pm2\sqrt2$.

Option (B) is not even on the curve. Option (D), $(2, 2)$, is on the curve, but its squared distance from $(0, 5)$ is $4 + 9 = 13$, against $8 + 1 = 9$ for the answer.

Question 28

For all real values of $x$, the minimum value of $\dfrac{1 – x + x^2}{1 + x + x^2}$ is

Solution. Differentiate with the quotient rule. The numerator of $f’$ works out to

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

so $f'(x) = \dfrac{2\left(x^2-1\right)}{\left(1+x+x^2\right)^2}$, zero at $x = \pm1$. Evaluating:

$$f(1) = \frac{1}{3}, \qquad f(-1) = \frac{3}{1} = 3$$

The denominator $1 + x + x^2$ is never zero (its discriminant is $-3$), so $f$ is defined and continuous everywhere; $\tfrac13$ is the minimum and $3$ the maximum.

Question 29

The maximum value of $\big[x(x-1) + 1\big]^{1/3}$, $0 \le x \le 1$, is

Solution. The cube root is increasing, so it is enough to maximise the inside:

$$g(x) = x^2 – x + 1 = \left(x – \tfrac12\right)^2 + \tfrac34$$

On $[0, 1]$ this has minimum $\tfrac34$ at $x = \tfrac12$ and maximum $1$ at both endpoints. Taking the cube root of $1$ gives $1$.

Option (A) is $\left(\tfrac34\right)^{1/3}$ mis-remembered, or the cube root of the minimum rather than the maximum.

Common mistakes

• Forgetting the endpoints. For an absolute extremum on a closed interval the endpoints are candidates too. Question 5(i), 10 and 12 all have their answers at endpoints.
• Using the second derivative test where it fails. When $f”(c) = 0$ the test is silent and you must fall back on the first derivative test — as in question 3(iv).
• Treating every zero of $f’$ as an extremum. In questions 14, 15 and 17 the derivative has a repeated factor whose zero is not a sign change and therefore not an extremum.
• Not using the constraint to eliminate a variable. In every word problem from 13 to 26, the constraint must be substituted before differentiating. Trying to differentiate a two-variable expression is what makes these questions feel hard.
• Minimising a square root. In questions 24 and 27, minimise the square instead — the minimising point is identical and the algebra is much shorter.
• Missing a second answer. Question 8 has two points and question 10 has two separate intervals with different answers.
• Reporting a length that the geometry forbids. In question 18, $x = 18$ solves the equation but would need a cut wider than the sheet’s shorter side.

Practise next

• Miscellaneous Exercise on Chapter 6 — the same optimisation methods on harder geometry, including a window, a tank and an inscribed cylinder.
• Exercise 6.2 — the sign of $f’$, which is what the first derivative test reads.