How to Find the Largest Box You Can Fold From a Square Sheet

Businesses want the greatest profit, engineers the least material, and travellers the shortest route. At a highest or lowest point of a smooth graph the tangent is flat, so derivatives locate the candidates — and further tests decide which is a maximum and which a minimum.

This lesson covers critical and extreme points, the first derivative test, the second derivative test, and optimisation on closed intervals.

**A critical point is a point of the domain where $f^{\prime }(x)=0$ or $f^{\prime }(x)$ does not exist, a stationary point is one where $f^{\prime }(x)=0$, and an extreme point is where f has a local maximum or minimum — every local extreme occurs at a critical point, but not every critical point is an extreme.

Definitions:

- Local maximum at c** — $f(c)\ge f(x)$ for all x near c
- Local minimum at c — $f(c)\le f(x)$ for all x near c
- Point of inflection — where the graph changes the way it bends, as $x^3$ does at 0

Worked example. For $f(x)=x^3-3x$, $f^{\prime }(x)=3x^2-3=0$ gives the stationary points $x=-1$ and $x=1$, with $f(-1)=2$ and $f(1)=-2$.

Worked example 2. For $f(x)=|x-2|$, $f^{\prime }(2)$ does not exist, so $x=2$ is a critical point — and it gives the minimum value 0.

An everyday example. The highest point reached by a ball thrown straight up is a stationary point: for an instant its vertical velocity, the derivative of its height, is zero.

The substance. **$f^{\prime }(c)=0$ is necessary, not sufficient** — $x^3$ has a stationary point at 0 that is neither a maximum nor a minimum.

**At a critical point c, if $f^{\prime }(x)$ changes from positive to negative as x increases through c, f has a local maximum there; if it changes from negative to positive, a local minimum; and if its sign does not change, c is neither.

Worked example.** Classify the critical points of $f(x)=x^3-6x^2+9x+1$.

- $f^{\prime }(x)=3x^2-12x+9=3(x-1)(x-3)$
- Near $x=1$: $f^{\prime }(0)=9>0$ and $f^{\prime }(2)=-3<0$, so a local maximum, $f(1)=5$
- Near $x=3$: $f^{\prime }(2)=-3<0$ and $f^{\prime }(4)=9>0$, so a local minimum, $f(3)=1$

When this test is essential. For $f(x)=x^{2/3}$, $f^{\prime }(x)=\frac{2}{3x^{1/3}}$ is undefined at 0, negative for $x<0$ and positive for $x>0$, so 0 is a local minimum — a case the second derivative test cannot handle.

An everyday example. A trekker crossing a hill in the Western Ghats climbs, reaches the top and descends — the slope changes from positive to negative at the summit, a local maximum.

The substance. A local maximum need not be the greatest value overall — $f(1)=5$ is a local maximum, yet $f(x)$ exceeds 5 for large x.

**At a stationary point c, where $f^{\prime }(c)=0$, f has a local maximum if $f^{\prime \prime }(c)<0$ and a local minimum if $f^{\prime \prime }(c)>0$; if $f^{\prime \prime }(c)=0$, the test gives no conclusion and the first derivative test must be used.

Worked example.** For $f(x)=x^3-6x^2+9x+1$, $f^{\prime \prime }(x)=6x-12$.

- $f^{\prime \prime }(1)=-6<0$, so $x=1$ gives the local maximum 5
- $f^{\prime \prime }(3)=6>0$, so $x=3$ gives the local minimum 1

Worked example 2, where the test fails. For $f(x)=x^4$, $f^{\prime }(0)=0$ and $f^{\prime \prime }(0)=0$, yet the first derivative test shows a minimum at 0; for $x^3$, $f^{\prime \prime }(0)=0$ as well, but 0 is neither a maximum nor a minimum.

An everyday example. The dip at the bottom of a skate-park bowl bends upward, so its lowest point has a positive second derivative — a minimum.

The substance. The second derivative test is quicker but less powerful — it is silent when $f^{\prime \prime }(c)=0$ and unusable when $f^{\prime }(c)$ does not exist.

**On a closed interval $[a,b]$, a continuous function reaches its absolute maximum and minimum either at a critical point inside the interval or at an end point, so evaluate f at all of these points and pick the largest and smallest values.

Worked example (closed interval).** Find the absolute extremes of $f(x)=x^3-6x^2+9x+1$ on $[0,5]$. The candidates are $f(0)=1$, $f(1)=5$, $f(3)=1$ and $f(5)=21$. The absolute maximum is 21 at $x=5$, and the absolute minimum is 1, at $x=0$ and $x=3$.

Worked example (box). An open box is made from an 18 cm square sheet by cutting a square of side x from each corner. Its volume is $V=x(18-2x{)}^2$ for $0<x<9$.

$$V^{\prime }=(18-2x)(18-6x)=0\Rightarrow x=3$$

Since $V^{\prime \prime }=24x-144$ gives $V^{\prime \prime }(3)=-72<0$, the volume is greatest at $x=3$: $V=3\times 12^2=432$ cm³.

An everyday example. A sweet shop owner choosing the price that gives the greatest daily profit is solving exactly this kind of problem.

The substance. End points can beat critical points — on $[0,5]$ the local maximum 5 is not the absolute maximum, which occurs at the end point.