How to Fold a Square Sheet Into the Biggest Possible Open Box

Engineers want the strongest beam from the least material; a shopkeeper wants the price that brings the most profit. These questions ask for a maximum or minimum.

This part covers local and absolute extrema with critical points, the first derivative test, the second derivative test, and optimisation problems.

**A local maximum or minimum is the highest or lowest value near a point, an absolute maximum or minimum is the greatest or least value over the whole domain or interval, and critical points — where $f^{\prime }(c)=0$ or $f^{\prime }(c)$ does not exist — are the only interior points where local extrema can occur.

Critical points.** If $f$ has a local extremum at an interior point $c$ and $f^{\prime }(c)$ exists, then $f^{\prime }(c)=0$. So the candidates are points where $f^{\prime }(c)=0$ or $f^{\prime }(c)$ does not exist.

Worked example — not an extremum. $f(x)=x^3$ has $f^{\prime }(0)=0$, but $f$ keeps increasing through $0$, so $x=0$ is neither a maximum nor a minimum — it is a point of inflection.

An everyday example. A hilly trek passes many local peaks, but only one of them is the highest point of the whole route — the absolute maximum.

The substance. A local maximum can be lower than a local minimum elsewhere — local means best only nearby.

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

Geometrically, the graph rises before a peak and falls after it.

Worked example.** $f(x)=x^3-3x^2-9x+5$ has $f^{\prime }(x)=3(x-3)(x+1)$, with critical points $x=-1$ and $x=3$.

- Through $x=-1$, $f^{\prime }$ changes from $+$ to $-$ — a local maximum, with $f(-1)=-1-3+9+5=10$
- Through $x=3$, $f^{\prime }$ changes from $-$ to $+$ — a local minimum, with $f(3)=27-27-27+5=-22$

An everyday example. A cyclist riding over a flyover feels the road change from uphill to downhill exactly at the top — the sign change of the slope.

The substance. **The first derivative test also works where $f^{\prime }$ does not exist**, such as $|x|$ at $0$, where the second derivative test cannot be used.

**If $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)<0$, $f$ has a local maximum at $c$; if $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)>0$, a local minimum; and if $f^{\prime \prime }(c)=0$, the test fails and another method is needed.

Why it works.** $f^{\prime \prime }(c)<0$ means $f^{\prime }$ is decreasing near $c$; since $f^{\prime }(c)=0$, $f^{\prime }$ goes from positive to negative, giving a peak. $f^{\prime \prime }(c)>0$ gives a valley.

Worked example. For $f(x)=x^3-3x^2-9x+5$, $f^{\prime \prime }(x)=6x-6$.

$$f^{\prime \prime }(-1)=-12<0\Rightarrow \text{local maximum at }x=-1$$

$$f^{\prime \prime }(3)=12>0\Rightarrow \text{local minimum at }x=3$$

An everyday example. A bowl sitting upright curves up and holds water at its lowest point, while an upturned bowl curves down with its highest point on top — like $f^{\prime \prime }>0$ at a minimum and $f^{\prime \prime }<0$ at a maximum.

The substance. **$f^{\prime \prime }(c)=0$ does not mean there is no extremum** — it only means this test cannot decide.

**Express the quantity to be optimised as a function of one variable, find its critical points and apply a derivative test; on a closed interval $[a,b]$, compare the values at the critical points and at both endpoints to find the absolute maximum and minimum.

Worked example 1 — the open box.** Equal squares of side $x$ cm are cut from the corners of an $18$ cm square sheet, and the sides are folded up.

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

$V^{\prime }(x)=0$ at $x=3$, since $x=9$ leaves no box. With $V^{\prime \prime }(x)=24x-144$, $V^{\prime \prime }(3)=-72<0$, a maximum:

$$V(3)=3(12{)}^2=432{\text{ cm}}^3$$

Worked example 2 — a closed interval. Find the absolute extrema of $f(x)=2x^3-15x^2+36x+1$ on $[1,5]$.

$$f^{\prime }(x)=6(x-2)(x-3)f(1)=24,f(2)=29,f(3)=28,f(5)=56$$

The absolute maximum is $56$ at $x=5$, and the absolute minimum is $24$ at $x=1$.

An everyday example. A farmer fencing a rectangular plot beside a canal with a fixed length of wire uses exactly this method to enclose the largest area.

The substance. Absolute extrema on a closed interval often sit at the endpoints, which is why the endpoints must always be checked.