$$\frac{dy}{dx} = 0$$

Solve for x, then sub back into the curve (not the derivative) for y. Max, min, or inflexion — all look the same until you check either side. Reflex: Stationary = gradient zero; classify afterwards, never assume.

$$\begin{gathered}\frac{d^2y}{dx^2} > 0 \to \text{minimum}\\[2pt] \frac{d^2y}{dx^2} < 0 \to \text{maximum}\\[2pt] \frac{d^2y}{dx^2} = 0 \to \text{no conclusion}\end{gathered}$$

Zero does NOT mean inflexion — it means nothing yet. Reflex: Zero? Switch to the sign test.

$$x = a^-,\ a,\ a^+ \quad \text{sign of } \frac{dy}{dx}$$

$-,0,+$ → minimum; $+,0,-$ → maximum; same sign both sides → point of inflexion. Use when $\frac{d^2y}{dx^2}=0$ or tedious.

Reflex: Sign flips → turning point; sign keeps → inflexion.

Use the constraint to express the target in ONE variable, differentiate, set to zero, solve — then prove max/min with the second derivative (or sign) test. Reject roots that make lengths negative.

Reflex: Constraint eliminates a variable → single-variable target → dy/dx = 0 → justify the nature.

$A$ and $A^2$ peak at the SAME $x$ (squaring is increasing for positive quantities) — so ditch the root and differentiate $A^2$: cleaner derivative, same location. ($V^2 = 85.5 \Rightarrow V = 9.25$.)

Reflex: root in the target → optimise the square; square-root the final value at the end.