Chapter 05

Medium

Applications of Derivatives — Optimization

00 · Symbol Glossary

$\text{abs. max / abs. min}$absolute maximum / absolute minimum

The single largest (or smallest) value a function attains over an entire domain or interval — not just locally near one point, as with the local max/min of Chapter 04.

$[a,b]$closed interval

The interval from $a$ to $b$ including both endpoints, introduced in Chapter 01 for the Intermediate Value Theorem. Reused here because absolute extrema on a closed interval always exist and can occur at an endpoint.

$x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$Newton's method update rule

An iterative formula that produces a sequence of increasingly accurate approximations $x_1, x_2, x_3,\ldots$ to a root of $f$ , starting from an initial guess $x_0$ .

$MR,\ MC$marginal revenue, marginal cost

$MR = R'(x)$ and $MC = C'(x)$ — the derivatives of total revenue and total cost with respect to quantity $x$ . They measure the revenue or cost of producing one additional unit, not the total revenue or cost itself.

01 · Absolute Extrema on a Closed Interval

Chapter 04's local max/min described behavior near a point. Absolute extrema ask a different question: over an entire interval, where is the function largest or smallest overall? On a closed interval, this question always has an answer, and it is always found at one of a short, specific list of candidate points.

Definition — Extreme Value Theorem

If $f$ is continuous on a closed interval $[a,b]$ , then $f$ attains both an absolute maximum and an absolute minimum somewhere on $[a,b]$ .

Definition — Closed Interval Method

To find the absolute extrema of a continuous $f$ on $[a,b]$ :

Find all critical points of $f$ inside $(a,b)$ (Chapter 04: where $f'=0$ or $f'$ undefined).
Evaluate $f$ at every critical point found in step 1, and at both endpoints $a$ and $b$ .
The largest value from step 2 is the absolute maximum; the smallest is the absolute minimum.

Step-by-step — Find the absolute extrema of $f(x)=x^3-3x^2$ on $[-1,3]$

Differentiate:

f'(x) = 3x^2-6x

(power rule).

Find critical points: set

3x^2-6x=0 \implies 3x(x-2)=0 \implies x=0

x=2

. Both lie inside

(-1,3)

, so both are candidates.

Evaluate $f$ at every candidate — critical points first:

f(0)=0-0=0

f(2)=8-12=-4

Evaluate $f$ at both endpoints:

f(-1)=-1-3=-4

f(3)=27-27=0

Compare all four values:

\{0, -4, -4, 0\}

x=\{0,2,-1,3\}

respectively. The largest is

0

(attained at both

x=0

and

x=3

); the smallest is

-4

(attained at both

x=2

and

x=-1

Conclude: absolute maximum

0

, absolute minimum

-4

[-1,3]

❌ Only checking critical points, ignoring endpoints

On the same function, stopping after step 3 above and reporting "absolute max $0$ at $x=0$ , absolute min $-4$ at $x=2$ " happens to be numerically correct here only because the endpoint values tie the interior values. Test instead on $[-1,1]$ : critical point $x=0$ is the only one inside $(-1,1)$ ( $x=2$ is outside), giving $f(0)=0$ . But $f(-1)=-4$ and $f(1)=1-3=-2$ .

Why it breaks: on $[-1,1]$ , ignoring the endpoint $x=-1$ would miss that $f(-1)=-4$ is actually the absolute minimum on this interval — smaller than any interior critical-point value.

Consequence: the closed interval method requires checking every endpoint every time; a critical point being "the only stationary point" says nothing about whether an endpoint is more extreme.

02 · Optimization Word Problems

The closed interval method (or, when the domain is unbounded, the first/second derivative test from Chapter 04) becomes a practical tool once a real quantity is translated into a function to be maximized or minimized.

Step-by-step — Maximize the area of a rectangular region with fixed perimeter 40

Name the variables and write the constraint: let the rectangle have width

w

and height

h

. Perimeter constraint:

2w+2h=40

Solve the constraint for one variable:

h = \dfrac{40-2w}{2} = 20-w

Write the quantity to optimize as a function of the single remaining variable: area

A(w) = w\cdot h = w(20-w) = 20w-w^2

State the domain: both

w>0

and

h=20-w>0

, so

0<w<20

Differentiate and find critical points:

A'(w) = 20-2w

. Set to

0

20-2w=0 \implies w=10

Confirm it's a maximum:

A''(w)=-2<0

everywhere, so by the second derivative test (Chapter 04),

w=10

gives a local max — and since

A(w)

is a downward parabola on an open interval with only one critical point, it is also the absolute maximum.

Compute the final answer:

h=20-10=10

; the rectangle is a

10\times10

square, giving area

A(10)=20(10)-100=200-100=100

Why the domain matters as much as the function

The optimization is only over $0<w<20$ , not all real numbers — $w=0$ or $w=20$ would give a degenerate rectangle with zero area. Step 4's domain restriction is what makes the closed-interval or open-interval reasoning in step 6 valid; skipping it risks optimizing over inputs that don't correspond to an actual rectangle.

03 · Marginal Analysis

In cost/revenue models, the derivative of a total function gives the marginal version of that quantity: the approximate effect of producing one more unit. This is a direct application of the derivative's meaning as an instantaneous rate of change (Chapter 02).

Definition — Marginal Cost, Revenue, Profit

For total cost $C(x)$ , total revenue $R(x)$ , and profit $\Pi(x) = R(x)-C(x)$ , all as functions of quantity $x$ :

MC(x) = C'(x) \qquad MR(x) = R'(x) \qquad \Pi'(x) = R'(x)-C'(x) = MR(x)-MC(x)

Profit is maximized (Section 01–02's method) where $\Pi'(x)=0$ , i.e. where $MR(x)=MC(x)$ — marginal revenue equals marginal cost.

Step-by-step — Maximize profit given $R(x)=50x-0.02x^2$ and $C(x)=10x+2000$

Form the profit function:

\Pi(x) = R(x)-C(x) = (50x-0.02x^2) - (10x+2000) = 40x - 0.02x^2 - 2000

Differentiate:

\Pi'(x) = 40 - 0.04x

Set $\Pi'(x)=0$ and solve:

40-0.04x=0 \implies 0.04x=40 \implies x=1000

Confirm a maximum:

\Pi''(x)=-0.04<0

everywhere, so

x=1000

is the unique maximum (concave-down profit function, Chapter 04 second derivative test).

Compute the maximum profit:

\Pi(1000) = 40(1000)-0.02(1000)^2-2000 = 40{,}000 - 20{,}000 - 2000 = 18{,}000

04 · Newton's Method

Optimization and root-finding are closely linked: setting $f'(x)=0$ to find critical points is itself a root-finding problem, and not every such equation can be solved by factoring. Newton's method approximates a root numerically using the tangent line.

Definition — Newton's Method

Starting from an initial guess $x_0$ near a root of $f$ , generate successive approximations:

x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

Geometrically: draw the tangent line to $f$ at $x_n$ , and let $x_{n+1}$ be where that tangent line crosses the $x$ -axis. Repeating this typically converges rapidly toward an actual root.

Step-by-step — Approximate a root of $f(x)=x^2-5$ starting from $x_0=2$

Differentiate:

f'(x)=2x

Compute the first iterate:

x_1 = x_0 - \dfrac{f(x_0)}{f'(x_0)} = 2 - \dfrac{2^2-5}{2(2)} = 2 - \dfrac{-1}{4} = 2+0.25=2.25

Compute the second iterate:

x_2 = x_1 - \dfrac{f(x_1)}{f'(x_1)} = 2.25 - \dfrac{2.25^2-5}{2(2.25)} = 2.25 - \dfrac{0.0625}{4.5} \approx 2.25 - 0.0139 \approx 2.2361

Compare to the exact value:

\sqrt5 \approx 2.2360679...

— after just two iterations, the approximation is accurate to four decimal places. This is why Newton's method is the standard numerical tool wherever an equation cannot be solved in closed form.

❌ Newton's method fails at a critical point of f itself

Applying Newton's method to $f(x)=x^{1/3}$ starting from $x_0=1$ : $f'(x) = \frac13 x^{-2/3}$ , so $f'(1)=\frac13$ , giving $x_1 = 1 - \dfrac{1}{1/3} = 1-3=-2$ . Next: $f'(-2)=\frac13(-2)^{-2/3}$ is still defined (a negative number raised to a fractional power with even-looking denominator behaves subtly, but the iterates in fact grow in magnitude here rather than converging: $\lvert x_1\rvert=2>\lvert x_0\rvert=1$ ).

Why it breaks: more directly, whenever some iterate lands where $f'(x_n)=0$ , the update formula divides by zero and the method is undefined at that step — the tangent line at a point with zero slope never crosses the $x$ -axis (it's horizontal), so there is no next iterate to compute.

Consequence: Newton's method has no universal guarantee of convergence; a poor initial guess or a landing point with $f'(x_n)=0$ can make it fail outright or diverge instead of approaching a root.

05 · Exercises

EXERCISE 5.1

Differentiate, find critical points inside the interval, then evaluate the function at those points and at both endpoints.

$f(x)=x^3-6x^2+5$ on $[0,5]$ . $f'(x)=3x^2-12x=3x(x-4)$ . Critical points: $x=0,4$ ; only $x=4$ is strictly inside $(0,5)$ ( $x=0$ is an endpoint here too). $f(0)=5$ . $f(4)=64-96+5=-27$ . $f(5)=125-150+5=-20$ . Comparing $\{5,-27,-20\}$ : absolute max $5$ at $x=0$ , absolute min $-27$ at $x=4$ .

Find the absolute maximum and minimum of $f(x)=x^3-6x^2+5$ on $[0,5]$ .

EXERCISE 5.2

Set up a constraint equation, solve for one variable, substitute into the quantity to optimize, then apply calculus to the resulting single-variable function.

Perimeter $2w+2h=60 \implies h=30-w$ . Area $A(w)=w(30-w)=30w-w^2$ , domain $0<w<30$ . $A'(w)=30-2w=0 \implies w=15$ . $A''(w)=-2<0$ , confirming a maximum. $h=30-15=15$ ; max area $=15\times15=225$ .

A rectangular region has fixed perimeter $60$ . Find the dimensions that maximize its area, and state the maximum area.

EXERCISE 5.3

Profit is maximized where marginal revenue equals marginal cost — set $\Pi'(x)=0$ and solve.

$R(x)=80x-0.05x^2$ , $C(x)=20x+1500$ . $\Pi(x)=60x-0.05x^2-1500$ . $\Pi'(x)=60-0.1x=0 \implies x=600$ . $\Pi''(x)=-0.1<0$ , confirming a max. $\Pi(600)=60(600)-0.05(600)^2-1500=36{,}000-18{,}000-1500=16{,}500$ .

Given $R(x)=80x-0.05x^2$ and $C(x)=20x+1500$ , find the quantity that maximizes profit and the maximum profit value.

EXERCISE 5.4

Apply Newton's update formula twice in sequence, using the previous iterate as the input to the next.

$f(x)=x^2-10$ , $f'(x)=2x$ , $x_0=3$ . $x_1 = 3-\dfrac{9-10}{6}=3-\dfrac{-1}{6}=3+0.1\overline{6}\approx3.1667$ . $x_2 = 3.1667 - \dfrac{3.1667^2-10}{2(3.1667)} = 3.1667-\dfrac{0.0281}{6.3334}\approx3.1667-0.00444\approx3.1623$ . Compare to $\sqrt{10}\approx3.16228$ — accurate to three decimals after two iterations.

Use Newton's method with $x_0=3$ to approximate $\sqrt{10}$ (i.e. find a root of $f(x)=x^2-10$ ). Compute two iterations.

EXERCISE 5.5

Find where $f'(x)=0$ for the given function, then check whether Newton's method starting exactly there is even defined.

$f(x)=x^3-3x$ , $f'(x)=3x^2-3$ . At $x_0=1$ : $f'(1)=3(1)-3=0$ . The update formula $x_1=x_0-\dfrac{f(x_0)}{f'(x_0)}$ requires dividing by $f'(1)=0$ , which is undefined. Newton's method cannot proceed from this starting point — a different initial guess is required.

Show that Newton's method fails to produce a next iterate for $f(x)=x^3-3x$ starting from $x_0=1$ . Explain why using the update formula directly.

EXERCISE 5.6

The absolute minimum of total cost per unit occurs where the derivative of the average-cost function is zero — form the average cost first, then differentiate.

Total cost $C(x)=0.01x^2+20x+4500$ . Average cost $\bar{C}(x)=\dfrac{C(x)}{x}=0.01x+20+\dfrac{4500}{x}$ . Differentiate: $\bar{C}'(x)=0.01-\dfrac{4500}{x^2}$ . Set to $0$ : $0.01=\dfrac{4500}{x^2} \implies x^2=450{,}000 \implies x=\sqrt{450{,}000}\approx670.8$ . Since $\bar{C}''(x)=\dfrac{9000}{x^3}>0$ for $x>0$ , this is a minimum. Producing roughly $671$ units minimizes average cost per unit.

Total production cost is $C(x)=0.01x^2+20x+4500$ . Find the production quantity $x>0$ that minimizes the average cost per unit $\bar{C}(x)=C(x)/x$ .

06 · Chapter Summary

Concept	Formula / Rule
Extreme Value Theorem	Continuous $f$ on $[a,b]$ attains an absolute max and min
Closed interval method	Compare $f$ at interior critical points and both endpoints
Optimization word problem	Constraint $\to$ solve for one variable $\to$ single-variable function $\to$ differentiate
Marginal cost / revenue	$MC=C'(x)$ , $MR=R'(x)$
Profit maximization	$\Pi'(x)=0 \iff MR(x)=MC(x)$
Newton's method	$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}$ ; fails if $f'(x_n)=0$

Next: Chapter 06 — Integrals: Definition and FTC, which introduces the reverse process of differentiation and the theorem connecting the two.