Stationary points and optimisation

The idea

At a stationary point $\frac{dy}{dx} = 0$. The second derivative tells you which kind: $\frac{d^2y}{dx^2} > 0$ is a minimum, $< 0$ a maximum. For optimisation, write the quantity in one variable, differentiate and set to zero.

1. Differentiate and solve $\frac{dy}{dx} = 0$.
2. Find $y$ at each solution.
3. Use $\frac{d^2y}{dx^2}$ to decide maximum or minimum.

Watch out If $\frac{d^2y}{dx^2} = 0$, the test fails: check the gradient on either side.