To find the maximum or minimum value of $f(x)$ on an interval $[a,b]$:
Example 1: A rectangle has perimeter 40 m. Maximise the area.
Let width $= x$, so length $= 20 - x$. Area $A = x(20-x) = 20x-x^2$.
$A’(x) = 20-2x = 0 \Rightarrow x = 10$. $A’‘(10) = -2 < 0$: maximum.
Maximum area $= 100$ m$^2$ (square).
The derivative gives the instantaneous rate of change. Related rates problems involve two or more quantities changing with time.
Example 2: A spherical balloon is inflated at 50 cm$^3$/s. At what rate is the radius increasing when $r = 5$ cm?
$V = \dfrac{4}{3}\pi r^3 \Rightarrow \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}$.
$$\frac{dr}{dt} = \frac{1}{4\pi r^2}\frac{dV}{dt} = \frac{50}{4\pi(25)} = \frac{50}{100\pi} = \frac{1}{2\pi} \approx 0.159 \text{ cm/s}$$
Area enclosed by $y = f(x)$, $y = g(x)$, $x = a$, $x = b$ (with $f \geq g$):
$$A = \int_a^b [f(x) - g(x)]\,dx$$
If $f$ and $g$ cross, split at the intersection points and sum the absolute values.
Example 3: Find the area enclosed by $y = x^2$ and $y = 2x - x^2$.
Intersections: $x^2 = 2x-x^2 \Rightarrow 2x^2-2x=0 \Rightarrow x=0$ or $x=1$.
On $[0,1]$: $2x-x^2 \geq x^2 \Rightarrow$ upper curve is $2x-x^2$.
$$A = \int_0^1 [(2x-x^2)-x^2]\,dx = \int_0^1 (2x-2x^2)\,dx = \left[x^2 - \frac{2x^3}{3}\right]_0^1 = 1 - \frac{2}{3} = \frac{1}{3}$$
Rotating $y = f(x)$ about the $x$-axis from $a$ to $b$:
$$V = \pi\int_a^b [f(x)]^2\,dx$$
Rotating about the $y$-axis (using the shell method):
$$V = 2\pi\int_a^b x f(x)\,dx$$
Example 4: Volume when $y = \sqrt{x}$ ($x \in [0,4]$) is rotated about the $x$-axis:
$$V = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi \approx 25.1 \text{ units}^3$$
Length of curve $y = f(x)$ from $a$ to $b$:
$$L = \int_a^b \sqrt{1 + [f’(x)]^2}\,dx$$
Example 5: Arc length of $y = \ln\sec x$ from $0$ to $\pi/4$.
$f’(x) = \tan x$, so $1+(f’)^2 = \sec^2 x$.
$$L = \int_0^{\pi/4}\sec x\,dx = [\ln|\sec x + \tan x|]_0^{\pi/4} = \ln(\sqrt{2}+1)$$
KEY TAKEAWAY: Calculus quantifies change, area, and volume. Optimisation requires solving $f’=0$ and classifying; area requires setting up the correct integral including correct limits; volume requires squaring the radius.
EXAM TIP: In optimisation word problems, define your variable clearly, express the quantity to optimise as a function of one variable, and always verify the nature of the critical point.
COMMON MISTAKE: In volume calculations, forgetting to square $f(x)$ before integrating: $V = \pi\int [f(x)]^2\,dx$, not $\pi[\int f(x)\,dx]^2$.
