Application of Derivatives • Topic 3 of 3

Maxima & Minima

Optimisation — finding the largest or smallest value a quantity can take — is the headline application of derivatives, and it is the part of this chapter that appears most often in real problems and in exams.

Local extrema and critical points

A function $f$ has a local maximum at $c$ if $f(c)\ge f(x)$ for all $x$ near $c$, and a local minimum if $f(c)\le f(x)$ nearby. At such a point of a differentiable function the tangent is horizontal, so $f'(c)=0$. Points where $f'(c)=0$ or $f'$ is undefined are critical points — the only candidates for local extrema. A critical point need not be an extremum (it could be a point of inflection, like $x=0$ for $x^3$), so every candidate must be tested.

First-derivative test

Examine how $f'$ changes sign as $x$ passes through a critical point $c$:

  • $f'$ changes from $+$ to $-$: local maximum at $c$.
  • $f'$ changes from $-$ to $+$: local minimum at $c$.
  • $f'$ does not change sign: $c$ is neither (a point of inflection).

Second-derivative test

At a critical point $c$ where $f'(c)=0$, the curvature settles the matter quickly:

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

When $f''(c)=0$ the test gives no information and you must fall back on the first-derivative test. The two tests always agree where both apply; the second-derivative test is usually faster, the first-derivative test more general.

Absolute (global) extrema on a closed interval

A function continuous on a closed interval $[a,b]$ always attains an absolute maximum and an absolute minimum there (the Extreme Value Theorem). These global extreme values occur either at a critical point inside $(a,b)$ or at an endpoint. The procedure:

  • Find all critical points of $f$ in $(a,b)$.
  • Evaluate $f$ at each critical point and at both endpoints $a$ and $b$.
  • The largest of these values is the absolute maximum; the smallest is the absolute minimum.

Note the distinction: a local extremum compares $f(c)$ only to nearby values, whereas an absolute extremum is the overall biggest or smallest value on the whole interval. An endpoint can be an absolute extremum without being a local one.

Optimisation word problems

Real-world maxima/minima questions follow a fixed pattern:

  • Identify the quantity to optimise and call it $Q$.
  • Express $Q$ in terms of one variable, using a constraint (a fixed perimeter, a given volume, a geometric relation) to eliminate the others.
  • State the domain of that variable from the physical situation.
  • Differentiate, set $Q'=0$, solve for the critical value, and confirm it is a max or min with the second-derivative test.
  • Answer the question asked — give the dimensions, the maximum area, the minimum cost, etc.

Classic cases include the largest rectangle of given perimeter (a square), the open box of greatest volume from a cut sheet, the cylinder of maximum volume inscribed in a sphere or cone, the least material for a given volume, and the shortest distance from a point to a curve.

Deeper Insight — optimisation is monotonicity read at the turning points: a maximum is precisely where the function stops increasing and starts decreasing, a minimum where it stops decreasing and starts increasing — so the entire topic is the previous topic's sign chart, read at the moments $f'$ flips sign. The second-derivative test is just a shortcut for "which way is it flipping": positive curvature cups upward into a valley, negative curvature caps over into a peak. The hardest part of an optimisation problem is almost never the calculus — it is the modelling step of writing the target quantity as a function of one variable using the constraint. Get the equation right, and finding the extremum is mechanical.

A curve with a local maximum at a peak and a local minimum at a valley, each with a horizontal tangent where f prime equals zero, and a point of inflection marked between them. Local maximum, local minimum and point of inflection x y y = f(x) local max f′=0 inflection f″=0 local min f′=0 Horizontal tangents (f′=0) at the peak and the valley
Solved Examples
1
Worked Example
Find the local extrema of $f(x)=x^2-6x+5$.
Solution

$f'(x)=2x-6=0\Rightarrow x=3$. $f''(x)=2>0$, so $x=3$ is a local minimum; $f(3)=9-18+5=-4$.

Answer: Local minimum value $-4$ at $x=3$.

2
Worked Example
Find the local maximum and minimum of $f(x)=x^3-3x$.
Solution

$f'(x)=3x^2-3=0\Rightarrow x=\pm1$. $f''(x)=6x$: at $x=-1,\ f''<0$ (max, $f(-1)=2$); at $x=1,\ f''>0$ (min, $f(1)=-2$).

Answer: Local maximum $2$ at $x=-1$; local minimum $-2$ at $x=1$.

3
Worked Example
Find the absolute maximum of $f(x)=x^3-3x$ on $[0,2]$.
Solution

Critical point in $[0,2]$: $x=1$ ($f(1)=-2$). Endpoints: $f(0)=0,\ f(2)=8-6=2$. The largest value is $f(2)=2$, so the absolute maximum is $2$ at $x=2$.

Answer: Absolute maximum $2$ at $x=2$.

4
Worked Example
Divide $20$ into two parts whose product is maximum.
Solution

Let the parts be $x$ and $20-x$; product $P=x(20-x)=20x-x^2$. $P'=20-2x=0\Rightarrow x=10$. $P''=-2<0$, a maximum. The parts are $10$ and $10$, product $100$.

Answer: The parts are $10$ and $10$, giving maximum product $100$.

5
Worked Example
Find all local maxima and minima of $f(x)=x^3-6x^2+9x+15$.
Solution

$f'(x)=3x^2-12x+9=3(x-1)(x-3)=0\Rightarrow x=1,3$. $f''(x)=6x-12$. At $x=1$: $f''=-6<0$, local maximum, $f(1)=1-6+9+15=19$. At $x=3$: $f''=6>0$, local minimum, $f(3)=27-54+27+15=15$.

Answer: Local maximum $19$ at $x=1$; local minimum $15$ at $x=3$.

6
Worked Example
Find the absolute maximum and minimum values of $f(x)=2x^3-15x^2+36x+1$ on $[1,5]$.
Solution

$f'(x)=6x^2-30x+36=6(x^2-5x+6)=6(x-2)(x-3)=0\Rightarrow x=2,3$ (both in $[1,5]$). Evaluate: $f(1)=2-15+36+1=24$, $f(2)=16-60+72+1=29$, $f(3)=54-135+108+1=28$, $f(5)=250-375+180+1=56$.

Answer: Absolute maximum $56$ at $x=5$; absolute minimum $24$ at $x=1$.

7
Worked Example
Show that $f(x)=x+\dfrac{1}{x}$ has a local minimum at $x=1$ and a local maximum at $x=-1$.
Solution

$f'(x)=1-\dfrac{1}{x^2}=0\Rightarrow x^2=1\Rightarrow x=\pm1$. $f''(x)=\dfrac{2}{x^3}$. At $x=1$: $f''=2>0$, local minimum $f(1)=2$. At $x=-1$: $f''=-2<0$, local maximum $f(-1)=-2$.

Answer: Local minimum $2$ at $x=1$; local maximum $-2$ at $x=-1$.

8
Worked Example
A square sheet of tin of side $18$ cm has equal squares of side $x$ cut from each corner; the flaps are folded up to make an open box. Find $x$ for maximum volume and the maximum volume.
Solution

The box has base $(18-2x)$ by $(18-2x)$ and height $x$, so $V=x(18-2x)^2$. $\dfrac{dV}{dx}=(18-2x)^2+x\cdot 2(18-2x)(-2)=(18-2x)(18-6x)$. Setting $=0$ gives $x=9$ (rejected, no base) or $x=3$. $V''<0$ at $x=3$, so it is a maximum. $V=3(12)^2=432$ cm$^3$.

Answer: $x=3$ cm gives maximum volume $432$ cm$^3$.

9
Worked Example
Find two positive numbers whose sum is $24$ and whose product is as large as possible.
Solution

Let the numbers be $x$ and $24-x$; product $P=x(24-x)=24x-x^2$. $P'=24-2x=0\Rightarrow x=12$. $P''=-2<0$, a maximum. The numbers are $12$ and $12$.

Answer: Both numbers are $12$, giving maximum product $144$.

10
Worked Example
Show that a rectangle of given perimeter $P$ has the largest area when it is a square.
Solution

Let the sides be $x$ and $y$ with $2(x+y)=P$, so $y=\dfrac{P}{2}-x$. Area $A=x\left(\dfrac{P}{2}-x\right)=\dfrac{P}{2}x-x^2$. $\dfrac{dA}{dx}=\dfrac{P}{2}-2x=0\Rightarrow x=\dfrac{P}{4}$, and then $y=\dfrac{P}{4}$. Since $A''=-2<0$ this is a maximum, and $x=y$.

Answer: The area is greatest when $x=y=\dfrac{P}{4}$, i.e. the rectangle is a square.

11
Worked Example
A wire of length $28$ cm is cut into two pieces; one is bent into a square and the other into a circle. Find the lengths of the two pieces so that the combined area is a minimum.
Solution

Let the circle's piece have length $x$, so its radius is $r=\dfrac{x}{2\pi}$ and its area $\pi r^2=\dfrac{x^2}{4\pi}$. The square's piece is $28-x$ with side $\dfrac{28-x}{4}$ and area $\dfrac{(28-x)^2}{16}$. Total $A=\dfrac{x^2}{4\pi}+\dfrac{(28-x)^2}{16}$. $\dfrac{dA}{dx}=\dfrac{x}{2\pi}-\dfrac{28-x}{8}=0\Rightarrow x=\dfrac{28\pi}{\pi+4}$. As $A''>0$, this gives the minimum; the square's piece is $\dfrac{112}{\pi+4}$.

Answer: Circle piece $=\dfrac{28\pi}{\pi+4}$ cm, square piece $=\dfrac{112}{\pi+4}$ cm.

12
Worked Example
Find the point on the curve $y=x^2$ that is nearest to the point $(0,3)$.
Solution

For a point $(x,x^2)$ on the curve, the squared distance is $D=x^2+(x^2-3)^2$. $\dfrac{dD}{dx}=2x+2(x^2-3)(2x)=2x(2x^2-5)=0\Rightarrow x=0$ or $x^2=\dfrac{5}{2}$. At $x=0$, $D=9$; at $x^2=\dfrac{5}{2}$, $D=\dfrac{5}{2}+\left(\dfrac{5}{2}-3\right)^2=\dfrac{5}{2}+\dfrac{1}{4}=\dfrac{11}{4}$, which is smaller.

Answer: The nearest points are $\left(\pm\sqrt{\dfrac{5}{2}},\,\dfrac{5}{2}\right)$, at distance $\dfrac{\sqrt{11}}{2}$ from $(0,3)$.

13
Worked Example
Show that the height of a closed right circular cylinder of given volume and least total surface area equals its diameter.
Solution

For volume $V=\pi r^2 h$, the total surface area is $S=2\pi r^2+2\pi r h$. Substitute $h=\dfrac{V}{\pi r^2}$: $S=2\pi r^2+\dfrac{2V}{r}$. $\dfrac{dS}{dr}=4\pi r-\dfrac{2V}{r^2}=0\Rightarrow V=2\pi r^3$, so $h=\dfrac{V}{\pi r^2}=\dfrac{2\pi r^3}{\pi r^2}=2r$. Since $\dfrac{d^2S}{dr^2}=4\pi+\dfrac{4V}{r^3}>0$, this is a minimum.

Answer: $S$ is least when $h=2r$, i.e. the height equals the diameter.

Key Points

  • Local extrema of a differentiable function occur only at critical points, where $f'(c)=0$ or $f'$ is undefined; not every critical point is an extremum.
  • First-derivative test: $f'$ changing $+\to-$ gives a maximum, $-\to+$ gives a minimum, no sign change gives a point of inflection.
  • Second-derivative test: $f''(c)>0 \Rightarrow$ minimum, $f''(c)<0 \Rightarrow$ maximum, $f''(c)=0 \Rightarrow$ inconclusive (use the first-derivative test).
  • Absolute extrema on $[a,b]$ occur at an interior critical point or an endpoint; evaluate $f$ at all of them and pick the largest and smallest.
  • Local vs absolute: a local extremum beats only nearby values, an absolute extremum beats all values on the interval.
  • Optimisation recipe: name the quantity, reduce it to one variable using the constraint, state the domain, set the derivative to zero, then confirm with $f''$.
  • Standard results: maximum area for a fixed perimeter is a square; least-surface closed cylinder of given volume has height $=$ diameter.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.At a critical point with $f''(c)>0$, $f$ has a local:
Explanation: Positive second derivative $\Rightarrow$ local minimum.
Q2.$f(x)=x^2-6x+5$ has a minimum at:
Explanation: $f'=2x-6=0\Rightarrow x=3$, and $f''=2>0$.
Q3.Two parts of $20$ with maximum product are:
Explanation: Product is maximised at $x=10$, giving $10\times10=100$.
Q4.Absolute extrema on $[a,b]$ are found among:
Explanation: Check interior critical points and both endpoints.
Practice more MCQs → 📝 Assignment · 35 marks