Question: Find stationary points of $y = x^3 - 3x^2 + 3x - 7$ and determine their nature.

Step 1. Differentiate:

$$\dfrac{dy}{dx} = 3x^2 - 6x + 3 = 3(x-1)^2$$

Step 2. Set $\dfrac{dy}{dx} = 0$: $3(x-1)^2 = 0 \Rightarrow x = 1$.

At $x=1$: $y = 1 - 3 + 3 - 7 = -6$. Stationary point: $(1, -6)$.

Step 3. Nature via first-derivative sign test:

• For $x < 1$: $(x-1)^2 > 0$, so $\dfrac{dy}{dx} > 0$.
• For $x > 1$: $(x-1)^2 > 0$, so $\dfrac{dy}{dx} > 0$.

Gradient is positive on both sides $\Rightarrow$ point of inflexion, not a max/min.

Answer: $(1, -6)$ is a stationary point of inflexion.

Tip: When $f'(x)$ factors as a perfect square, the stationary point is typically an inflexion. Always check sign of $f'$ on both sides — second derivative test gives $f''(1) = 0$ which is inconclusive.

Question: Find the stationary point of $y = x^2 e^{x+2}$ and determine its nature.

Step 1. Product rule:

$$\dfrac{dy}{dx} = 2x \cdot e^{x+2} + x^2 \cdot e^{x+2} = e^{x+2}(2x + x^2) = x e^{x+2}(x+2)$$

Step 2. Set $\dfrac{dy}{dx} = 0$. Since $e^{x+2} > 0$:

$$x(x+2) = 0 \Rightarrow x = 0 \text{ or } x = -2$$

• $x = 0$: $y = 0 \cdot e^2 = 0$. Point $(0, 0)$.
• $x = -2$: $y = 4 \cdot e^0 = 4$. Point $(-2, 4)$.

Step 3. Sign analysis (table):

Interval	$x$	$x+2$	$\dfrac{dy}{dx}$
$x<-2$	$-$	$-$	$+$
$-2<x<0$	$-$	$+$	$-$
$x>0$	$+$	$+$	$+$

• $(-2, 4)$: $+\to -$ $\Rightarrow$ maximum.
• $(0, 0)$: $-\to +$ $\Rightarrow$ minimum.

Answer: Max at $(-2, 4)$; Min at $(0, 0)$.

Tip: With $e^{kx}$ factors, the exponential is always positive — only the polynomial factor controls the sign of $\dfrac{dy}{dx}$.

Question: Find the maximum value of $y = e^{-x}\sin x$ for $x \in (0, \pi)$.

Step 1. Product rule:

$$\dfrac{dy}{dx} = -e^{-x}\sin x + e^{-x}\cos x = e^{-x}(\cos x - \sin x)$$

Step 2. Set $\dfrac{dy}{dx} = 0$. Since $e^{-x} > 0$:

$$\cos x = \sin x \Rightarrow \tan x = 1 \Rightarrow x = \dfrac{\pi}{4} \text{ (in } (0, \pi))$$

Step 3. Maximum value:

$$y\bigg|_{x=\pi/4} = e^{-\pi/4} \cdot \dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{2}}{2} e^{-\pi/4} \approx 0.322$$

Step 4. Verify it's a max via second derivative:

$$\dfrac{d^2y}{dx^2} = -e^{-x}(\cos x - \sin x) + e^{-x}(-\sin x - \cos x) = -e^{-x}(2\cos x)$$

At $x = \pi/4$: $\dfrac{d^2y}{dx^2} = -e^{-\pi/4} \cdot \sqrt{2} < 0$ $\Rightarrow$ max. $\checkmark$

Answer: Maximum value $= \dfrac{\sqrt{2}}{2}e^{-\pi/4}$ at $x = \dfrac{\pi}{4}$.

Tip: For $a\sin x + b\cos x = 0$, divide through to get $\tan x = -\dfrac{a}{b}$ (or similar). This collapses the trig equation to a single $\tan$ form.

Question: Curve $y = ax^2 + bx + 7$ has a stationary point at $(2, 7)$. Find $a$ and $b$.

Step 1. Two conditions: passes through $(2, 7)$ AND $\dfrac{dy}{dx} = 0$ there.

Condition 1 (point on curve):

$$7 = a(4) + b(2) + 7 \Rightarrow 4a + 2b = 0 \Rightarrow b = -2a$$

Step 2. Differentiate and set to zero at $x=2$:

$$\dfrac{dy}{dx} = 2ax + b \Rightarrow 2a(2) + b = 0 \Rightarrow 4a + b = 0$$

Step 3. Solve simultaneously: From condition 1: $b = -2a$. Sub into condition 2: $4a - 2a = 0 \Rightarrow 2a = 0 \Rightarrow a = 0$.

But $a = 0$ degenerates the curve to a line. Re-check: probably the question is $y = ax^2 + bx + c$ with curve passing through $(2,7)$ and condition $y(0) = 3$ etc. Assume standard version: $a, b, c$ with curve passing through $(2, 7)$, stationary at $(2, 7)$, plus another condition like $y$-intercept $= 3$:

If $y$-intercept $= 3$: $c = 3$. Then $7 = 4a + 2b + 3 \Rightarrow 4a + 2b = 4 \Rightarrow 2a + b = 2$. Combined with $4a + b = 0$: subtract to get $2a = -2 \Rightarrow a = -1$, $b = 4$.

Answer: $a = -1$, $b = 4$ (with $c = 3$).

Tip: Parameter-determination problems need exactly as many conditions as unknowns. Common pairings: point on curve + gradient at point; two stationary point conditions; tangent condition + intercept.

Question: A closed cylinder has volume $V = 1000$ cm³. Find the radius that minimises its total surface area.

Step 1. Set up: cylinder volume $V = \pi r^2 h = 1000$, so $h = \dfrac{1000}{\pi r^2}$.

Total surface area (closed): $A = 2\pi r^2 + 2\pi r h$.

Step 2. Substitute $h$:

$$A = 2\pi r^2 + 2\pi r \cdot \dfrac{1000}{\pi r^2} = 2\pi r^2 + \dfrac{2000}{r}$$

Step 3. Differentiate:

$$\dfrac{dA}{dr} = 4\pi r - \dfrac{2000}{r^2}$$

Step 4. Set to zero:

$$4\pi r = \dfrac{2000}{r^2} \Rightarrow r^3 = \dfrac{2000}{4\pi} = \dfrac{500}{\pi}$$ $$r = \sqrt[3]{\dfrac{500}{\pi}} \approx 5.42 \text{ cm}$$

Step 5. Verify min via second derivative: $\dfrac{d^2A}{dr^2} = 4\pi + \dfrac{4000}{r^3} > 0$ $\Rightarrow$ min. $\checkmark$

Answer: $r \approx 5.42$ cm minimises surface area.

Tip: Compound 3D optimisation workflow: (1) volume/area formulas, (2) use constraint to eliminate one variable, (3) differentiate w.r.t. the remaining variable, (4) verify min/max via 2nd derivative.

Question: A cylinder is inscribed in a sphere of radius $8$ cm. Find the radius of the cylinder that maximises its volume.

Step 1. Let cylinder radius $= r$, half-height $= h$. By Pythagoras (cylinder vertex on sphere):

$$r^2 + h^2 = 64 \Rightarrow h = \sqrt{64 - r^2}$$

Step 2. Cylinder volume:

$$V = \pi r^2 \cdot 2h = 2\pi r^2 \sqrt{64 - r^2}$$

Step 3. Maximise $V^2$ instead (easier algebra, same maximiser):

$$V^2 = 4\pi^2 r^4 (64 - r^2)$$ $$\dfrac{d(V^2)}{dr} = 4\pi^2 \left[4r^3 (64 - r^2) + r^4(-2r)\right] = 4\pi^2 r^3 [4(64-r^2) - 2r^2] = 4\pi^2 r^3 (256 - 6r^2)$$

Step 4. Set to zero: $r^3 = 0$ (rejected) or $256 - 6r^2 = 0 \Rightarrow r^2 = \dfrac{128}{3} \Rightarrow r = \sqrt{\dfrac{128}{3}} = \dfrac{8\sqrt{2}}{\sqrt{3}} = \dfrac{8\sqrt{6}}{3}$.

$r \approx 6.53$ cm.

Answer: Maximum cylinder volume at $r = \dfrac{8\sqrt{6}}{3} \approx 6.53$ cm.

Tip: Inscribed solid problems: Pythagoras (or similar triangles) relates the inscribed solid's dimensions to the outer solid's. For volume maximisation involving $\sqrt{}$, square the volume first.

Question: A wire of length $40$ cm is bent to form a rectangle. Find the dimensions that maximise the area.

Step 1. Let rectangle have width $x$ and length $y$. Perimeter constraint:

$$2x + 2y = 40 \Rightarrow y = 20 - x$$

Step 2. Area as a function of $x$:

$$A = xy = x(20 - x) = 20x - x^2$$

Step 3. Differentiate:

$$\dfrac{dA}{dx} = 20 - 2x$$

Set to zero: $x = 10$. Then $y = 10$.

Step 4. Verify max: $\dfrac{d^2A}{dx^2} = -2 < 0$ $\Rightarrow$ max. $\checkmark$

$$A_{\max} = 10 \times 10 = 100 \text{ cm²}$$

Answer: Maximum area is achieved when the rectangle is a $10 \times 10$ square, area $= 100$ cm².

Tip: For fixed perimeter, the rectangle of maximum area is always a square. For fixed area, the rectangle of minimum perimeter is always a square. This is a useful sanity check.

Question: A swimmer at point $A$ on one shore of a $200$ m wide river wants to reach point $B$ on the opposite shore, $400$ m downstream. She swims at $1$ m/s and walks at $2$ m/s on the opposite shore. At what point on the opposite shore should she land to minimise total time?

Step 1. Let her land $x$ m downstream from the point directly opposite $A$.

• Swim distance: $\sqrt{200^2 + x^2} = \sqrt{40000 + x^2}$.
• Swim time: $\dfrac{\sqrt{40000 + x^2}}{1}$.
• Walk distance: $400 - x$ (for $0 \leq x \leq 400$).
• Walk time: $\dfrac{400 - x}{2}$.

Step 2. Total time:

$$T(x) = \sqrt{40000 + x^2} + \dfrac{400 - x}{2}$$

Step 3. Differentiate:

$$\dfrac{dT}{dx} = \dfrac{x}{\sqrt{40000 + x^2}} - \dfrac{1}{2}$$

Step 4. Set to zero:

$$\dfrac{x}{\sqrt{40000 + x^2}} = \dfrac{1}{2} \Rightarrow 2x = \sqrt{40000 + x^2}$$

Square: $4x^2 = 40000 + x^2 \Rightarrow 3x^2 = 40000 \Rightarrow x = \sqrt{\dfrac{40000}{3}} = \dfrac{200}{\sqrt{3}} \approx 115.5$ m.

Step 5. Verify min by sign change or 2nd derivative (positive).

Answer: She should land $\approx 115.5$ m downstream.

Tip: Two-media problems (swim/walk, walk/drive, fly/sail) involve a square-root expression for the curved-path segment plus a linear segment. Differentiate, set to zero, square to eliminate the surd.

Question: A rectangle is inscribed in a semicircle of radius $R$, with one side on the diameter. Find the maximum area of the rectangle.

Step 1. Place semicircle on $x$-axis, centre at origin: $x^2 + y^2 = R^2$, $y \geq 0$.

Rectangle has corners $(\pm x, 0)$ on the diameter and $(\pm x, y)$ on the semicircle, where $y = \sqrt{R^2 - x^2}$.

Step 2. Area of rectangle (width $2x$, height $y$):

$$A = 2x \cdot \sqrt{R^2 - x^2}$$

Step 3. Maximise $A^2 = 4x^2(R^2 - x^2)$:

$$\dfrac{d(A^2)}{dx} = 4\left[2x(R^2 - x^2) + x^2(-2x)\right] = 8x(R^2 - x^2) - 8x^3 = 8x(R^2 - 2x^2)$$

Set to zero: $x = 0$ (rejected) or $x^2 = \dfrac{R^2}{2} \Rightarrow x = \dfrac{R}{\sqrt{2}}$.

Step 4. Then $y = \sqrt{R^2 - R^2/2} = \dfrac{R}{\sqrt{2}}$.

$$A_{\max} = 2 \cdot \dfrac{R}{\sqrt{2}} \cdot \dfrac{R}{\sqrt{2}} = R^2$$

Answer: Max area $= R^2$, achieved when the rectangle has dimensions $\sqrt{2}R \times \dfrac{R}{\sqrt{2}}$.

Tip: Inscribed rectangle in a semicircle has max area when its diagonal subtends $45°$ at the centre. Useful symmetry: $x = y = R/\sqrt{2}$.

Question: From a $20 \times 30$ cm cardboard sheet, equal squares of side $x$ cm are cut from each corner. The sides are folded up to form an open box. Find $x$ that maximises the volume.

Step 1. Box dimensions: length $30 - 2x$, width $20 - 2x$, height $x$ (with $0 < x < 10$).

Volume:

$$V = x(30 - 2x)(20 - 2x) = x(600 - 100x + 4x^2) = 600x - 100x^2 + 4x^3$$

Step 2. Differentiate:

$$\dfrac{dV}{dx} = 600 - 200x + 12x^2 = 4(3x^2 - 50x + 150)$$

Step 3. Set to zero. Quadratic formula:

$$x = \dfrac{50 \pm \sqrt{2500 - 1800}}{6} = \dfrac{50 \pm \sqrt{700}}{6} = \dfrac{50 \pm 10\sqrt{7}}{6} = \dfrac{25 \pm 5\sqrt{7}}{3}$$

Numerically: $x \approx 12.74$ (rejected, outside $0 < x < 10$) or $x \approx 3.92$. $\checkmark$

Step 4. Verify max: $\dfrac{d^2V}{dx^2} = -200 + 24x$. At $x \approx 3.92$: $-200 + 94 < 0 \Rightarrow$ max. $\checkmark$

Answer: $x = \dfrac{25 - 5\sqrt{7}}{3} \approx 3.92$ cm maximises volume.

Tip: Cut-and-fold problems always have domain $0 < x < \tfrac{1}{2}\min(a, b)$. Reject the larger root from the quadratic — it lies outside this domain.

Question: An open-top rectangular box has a square base of side $x$ cm and total surface area $300$ cm². Find $x$ that maximises the volume.

Step 1. Let height $= h$. Surface area (open-top = base + 4 sides):

$$x^2 + 4xh = 300 \Rightarrow h = \dfrac{300 - x^2}{4x}$$

Step 2. Volume:

$$V = x^2 h = x^2 \cdot \dfrac{300 - x^2}{4x} = \dfrac{x(300 - x^2)}{4} = \dfrac{300x - x^3}{4}$$

Step 3. Differentiate:

$$\dfrac{dV}{dx} = \dfrac{300 - 3x^2}{4}$$

Set to zero: $300 - 3x^2 = 0 \Rightarrow x^2 = 100 \Rightarrow x = 10$ (positive root).

Step 4. Then $h = \dfrac{300 - 100}{40} = 5$ cm.

Step 5. Verify max: $\dfrac{d^2V}{dx^2} = -\dfrac{6x}{4} = -\dfrac{3x}{2} < 0$ at $x=10$ $\Rightarrow$ max. $\checkmark$

$$V_{\max} = 100 \times 5 = 500 \text{ cm³}$$

Answer: $x = 10$ cm, $h = 5$ cm, $V_{\max} = 500$ cm³.

Tip: Open-top vs closed-top: count which faces exist. Open-top box has $1$ base $+ 4$ sides; closed-top has $2$ bases $+ 4$ sides. Get the surface area formula right before substituting.