Stationary points and optimisation

What you’ll learn

• Locate and classify stationary points.
• Use derivative signs to identify increasing and decreasing intervals.
• Model a practical maximum or minimum.

Understand the idea

A stationary point has zero first derivative. Solve $f'(x)=0$, then substitute into the original function for its coordinates. A positive second derivative indicates a local minimum and a negative second derivative a local maximum. If the second derivative is zero, examine how the first derivative changes sign instead.

A positive derivative means the function is increasing and a negative derivative means it is decreasing. In an optimisation problem, first express the quantity in one variable using the constraint. State the feasible domain, classify stationary candidates and compare endpoints when they are part of that domain.

Key formulae

$$f'(a)=0\quad\text{at a stationary point}$$

$$f''(a)>0\Rightarrow\text{local minimum},\qquad f''(a)<0\Rightarrow\text{local maximum}$$

$$f'(x)>0\Rightarrow f\text{ increasing},\qquad f'(x)<0\Rightarrow f\text{ decreasing}$$

Worked example

A rectangle has perimeter 28 cm. Find its greatest possible area.

1. Let one side be $x$ cm; the other is $14-x$ cm, with $0<x<14$.
2. Area $A=x(14-x)=14x-x^2$, so $A'=14-2x$.
3. Set $A'=0$ to get $x=7$. Since $A''=-2<0$, this gives a maximum.

Answer: The maximum area is $49\text{ cm}^2$, for a $7$ cm square.