Week 10 In-Depth — Why Differentiation Solves These Problems

Read the cheatsheet first. This note explains why each technique works and walks through one substantial worked example per topic, with every step on the page.

1 — Why critical points are special

A continuous, smooth function $f$ can only hit a local maximum or minimum at:

1. an interior point where the tangent line is horizontal (i.e. $f^{\prime }(x)=0$), or
2. an interior point where $f$ is not differentiable (a corner, cusp, or vertical tangent — i.e. $f^{\prime }(x)$ undefined), or
3. an endpoint of the interval (only relevant when the domain is bounded).

The first two cases together are what we call critical points. The job of “find the max/min” reduces to “find the critical points, then sort which is which.”

Why the second derivative test works

Near a critical point $c$, Taylor’s theorem says

$$f(c+h)\approx f(c)+f^{\prime }(c)h+\frac{1}{2}{f}^{\prime \prime }(c)h^2.$$

Because $f^{\prime }(c)=0$ at a critical point, the linear term vanishes:

$$f(c+h)\approx f(c)+\frac{1}{2}{f}^{\prime \prime }(c)h^2.$$

Now $h^2\ge 0$ always, so:

• If $f^{\prime \prime }(c)>0$, then $f(c+h)\ge f(c)$ for all small $h$ — $c$ is a min.
• If $f^{\prime \prime }(c)<0$, then $f(c+h)\le f(c)$ for all small $h$ — $c$ is a max.
• If $f^{\prime \prime }(c)=0$, the quadratic term vanishes too, and the sign of $f(c+h)-f(c)$ is determined by higher-order terms — the test is inconclusive and you fall back to inspecting how $f^{\prime }$ behaves either side of $c$.

That last bullet is why you must be honest when $f^{\prime \prime }(c)=0$. It is not a saddle and it is not an inflection point by default — it is “I don’t know yet, look closer.”

Worked example with all the steps — $f(x)=3x^5-5x^3$

Differentiate:

$$f^{\prime }(x)=15x^4-15x^2=15x^2(x^2-1)=15x^2(x-1)(x+1).$$

Set to zero: $x=-1,0,1$. Three critical points.

Take $f^{\prime \prime }$:

$$f^{\prime \prime }(x)=60x^3-30x=30x(2x^2-1).$$

Now classify each:

• $x=1$: $f^{\prime \prime }(1)=30\cdot 1\cdot (2-1)=30>0$ → local min at $(1,-2)$.
• $x=-1$: $f^{\prime \prime }(-1)=30\cdot (-1)\cdot (2-1)=-30<0$ → local max at $(-1,2)$.
• $x=0$: $f^{\prime \prime }(0)=0$. Inconclusive. Check $f^{\prime }$ on either side: $f^{\prime }(-0.1)=15(-0.1{)}^2((-0.1{)}^2-1)=15(0.01)(-0.99)<0$ … wait, let me redo. Actually $15x^2$ is always $\ge 0$, and $x^2-1<0$ for $|x|<1$. So near $x=0$, $f^{\prime }(x)<0$ on both sides. Same sign → neither. So $(0,0)$ is an inflection point, not a max or min.

The takeaway: when $f^{\prime \prime }(c)=0$, slow down, write out the factored $f^{\prime }$ near $c$, and check the sign change directly.

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2 — Why optimization always reduces to one variable

Word problems give you two facts:

• A constraint linking your variables, such as $xy=192$, $2x+y=120$, or $4Lh=36$. This is a relationship with zero degrees of freedom given the constraint.
• An objective function of those variables (the thing to maximise or minimise: area, cost, length).

If you tried to apply $\frac{d}{dx}$ to a function of two variables, you’d be working in multivariable calculus. The constraint saves you — solve it for one variable in terms of the other, substitute into the objective, and now you have a function of one variable. Single-variable calculus applies.

That is the entire reason the constraint exists in the problem statement. If you find yourself stuck staring at two variables, you haven’t substituted.

Worked example — the offshore well

An offshore well sits at $W$, 6 km from the closest shore point $A$. Oil is piped to shore point $B$, 8 km from $A$ along the shoreline. Pipe runs straight from $W$ to a shore point $P$ between $A$ and $B$ (under water at \100,000$/km),thenalongtheshorefrom$P$to$B$(overlandat$$75,000$/km).Whereshould$P$ be?

Set up. Let $x$ = distance from $A$ to $P$, so $0\le x\le 8$.

Distance under water: $W$ is 6 km offshore, $P$ is $x$ km along the shore from $A$, so

$$WP=\sqrt{x^2+6^2}=\sqrt{x^2+36}.$$

Distance over land: from $P$ to $B$ is just $8-x$.

Cost objective (in dollars, dropping the per-km units):

$$C(x)=100000\sqrt{x^2+36}+75000(8-x).$$

Differentiate:

$$C^{\prime }(x)=100000\cdot \frac{x}{\sqrt{x^2+36}}-75000.$$

Set to zero:

$$\frac{100000x}{\sqrt{x^2+36}}=75000⟺\frac{x}{\sqrt{x^2+36}}=\frac{3}{4}.$$

Square both sides:

$$\frac{x^2}{x^2+36}=\frac{9}{16}⟹16x^2=9x^2+324⟹7x^2=324⟹x=\sqrt{324/7}\approx \mathrm{6.803.}$$

Classify. Either confirm with $C^{\prime \prime }(x)>0$ at $x\approx 6.8$, or note C(0) = 100\,000 \cdot 6 + 75\,000 \cdot 8 = \1,200,000$and$C(8) = 100,000 \sqrt{100} \approx $1,000,000$ — the interior critical point gives a lower value than both endpoints, so it’s the minimum.

Plug back in:

$$C(6.803)=100000\sqrt{6.803^2+36}+75000(8-6.803)\approx \$996\mathrm{862.70.}$$

So pipe should hit the shore $\approx 6.80$ km from $A$, at a total cost of \approx \996,862.70$.

Why this matters more than the answer

The recipe is always the same. Variables → constraint → objective → substitute → optimise → classify → answer with units. Once you’ve done it twice, every optimization problem is a fill-in-the-blank.

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3 — Where the chain rule comes from in related rates

Two quantities $A$ and $r$ are linked: $A=\pi r^2$. Both depend on time.

You don’t get to write $\frac{dA}{dt}$ directly. You have to use the chain rule:

$$\frac{dA}{dt}=\frac{dA}{dr}\cdot \frac{dr}{dt}.$$

That’s literally what the chain rule says: $A$ varies with $r$, $r$ varies with $t$, so $A$‘s rate of change in time is the product.

For $A=\pi r^2$: $\frac{dA}{dr}=2\pi r$. So

$$\frac{dA}{dt}=2\pi r\cdot \frac{dr}{dt}.$$

Notice that $\frac{dA}{dt}$ depends on the current value of $r$, not just on $\dot{r}$. That’s why the question always tells you “at the instant when $r=3$ mm” — without an instant, you can’t compute a number.

The “differentiate, then substitute” rule

If you substitute $r=3$ before differentiating, you get $A=9\pi$ — a constant. Constants differentiate to zero. So your final answer is $\dot{A}=0$, which is wrong.

Always:

1. Write the symbolic relationship first.
2. Differentiate symbolically with respect to $t$.
3. Then substitute the instantaneous values.

Worked example — the artery

A drug widens an artery. $\dot{r}=+0.02$ mm/month, constant. Find $\dot{A}$ when $r=3$ mm.

Relationship: $A=\pi r^2$.

Differentiate w.r.t. $t$:

$$\frac{dA}{dt}=2\pi r\cdot \frac{dr}{dt}.$$

Substitute the instant:

$$\frac{dA}{dt}{|}_{r=3}=2\pi (3)(0.02)=0.12\pi \approx 0.377{\text{ mm}}^2/\text{month}.$$

A nice consequence: because $\dot{r}$ is constant, $\dot{A}$ is linear in $r$ — and $r$ itself grows linearly in time. So the rate at which the area grows increases over time, even though the radius grows at a constant rate. That’s the geometric intuition: a bigger circle has more boundary, so each unit of radius adds more area.

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4 — How these three connect

All three techniques use $f^{\prime }(x)$, but they use it for different reasons:

Technique	What’s $f^{\prime }(x)$ doing?
Critical points	$f^{\prime }(x)=0$ is the condition for an interior extremum.
Optimization	Same — you build an objective, then critical-point it. Optimization is critical-point analysis dressed up in a real-world story.
Related rates	$f^{\prime }(x)$ is used as a converter between rates of change. The chain rule’s middle term ($dA/dr=2\pi r$) is just $f^{\prime }$ for $A(r)$.

So really, this whole week is one technique applied three ways. If you understand $f^{\prime }(x)=0$ deeply, optimization is just substitution then critical-point analysis, and related rates is just chain rule then instant values.

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5 — Exam-style sample, end-to-end

A 4 m wide rectangular tank with an open top is to be built. It must hold 36 m³. The base costs \10$/m^{2}andthesidescost$$5$/m². What dimensions minimise the cost?

Setup.

Item	LaTeX model
Variables	Length $L$, height $h$, fixed width $4$
Constraint	$4Lh=36\Rightarrow Lh=9\Rightarrow h=\frac{9}{L}$
Base area	$4L$
Side area	$2(4h)+2(Lh)=8h+2Lh$
Objective	Minimise total cost $C$

$$C=10\cdot 4L+5\cdot (8h+2Lh)=40L+40h+10Lh.$$

Sub in $h=9/L$ and $Lh=9$:

$$C(L)=40L+40\cdot \frac{9}{L}+10\cdot 9=40L+\frac{360}{L}+90.$$

Differentiate.

$$C^{\prime }(L)=40-\frac{360}{L^2}.$$

Critical point.

$$C^{\prime }(L)=0⟹L^2=9⟹L=3\text{m}$$

The positive root is the only physical solution.

Classify.

$$C^{\prime \prime }(L)=\frac{720}{L^3}>0$$

So the critical point is a minimum.

Plug back in.

$$\begin{array}{rl}h& =\frac{9}{3}=3\text{m},\\ L\times w\times h& =3\times 4\times 3\text{m},\\ C(3)& =120+120+90=\$330.\end{array}$$

Answer. The tank should be 3 m long, 4 m wide, 3 m tall, costing \330$.

That layout — Variables, Constraint, Objective, Substitute, Differentiate, Critical point, Classify, Answer — is the template for every optimisation question on the paper.

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Where to look next

• The cheatsheet for fast revision + the quiz.
• The workshop PDFs (cited in frontmatter) for more practice problems.
• Whatever you add to this folder — Lecture Atlas picks up any new Markdown note automatically.