Calculus I · Chapter 6
Chapter 6: Applications of the Derivative
Absolute extrema · Optimization · Elasticity of demand · Implicit differentiation · Related rates · Differentials & approximation
This is where calculus earns its keep. The same derivative that describes the shape of a graph now answers practical questions: What's the most profit possible? The least cost? The largest box? How fast is the shadow moving? This chapter is the toolkit of applied optimization and rates.
Section 6.1
Absolute Extrema
Definition: Absolute (Global) Extrema
The absolute maximum is the single largest value $f$ attains on its domain; the absolute minimum is the smallest. Unlike relative extrema (which are local), these compare against every point.
Theorem: Extreme Value Theorem & the Closed-Interval Method
A continuous function on a closed interval $[a,b]$ always attains an absolute max and min. To find them:
- Find all critical numbers in $(a, b)$.
- Evaluate $f$ at each critical number and at both endpoints $a$ and $b$.
- The largest value is the absolute max; the smallest is the absolute min.
📘 Example 6.1: Closed-Interval Method
Find the absolute extrema of $f(x) = x^3 - 3x^2$ on $[-1, 4]$.
- $f'(x) = 3x^2 - 6x = 3x(x - 2)$ ⇒ critical numbers $x = 0, 2$ (both in the interval).
- Evaluate: $f(-1) = -4$, $f(0) = 0$, $f(2) = -4$, $f(4) = 16$.
- Largest is $16$; smallest is $-4$.
Section 6.2
Applications of Extrema (Optimization)
The Optimization Strategy
- Identify the quantity to maximize or minimize; write it as a function.
- Use a constraint equation to reduce to one variable.
- Determine the realistic domain.
- Differentiate, find critical numbers, and confirm it's the max/min you want.
- Answer the question asked (with units).
A classic optimization: an open-top box folded from a sheet by cutting squares of side $x$ from the corners. Drag $x$, the volume rises, peaks, then falls. The graph on the right tracks volume vs. $x$; the green dot marks the maximizing cut. Calculus finds that peak exactly.
V = ,
📘 Example 6.2: Maximizing the Volume of a Box
Squares of side $x$ are cut from the corners of a $12 \times 12$ sheet, and the sides folded up. What $x$ maximizes the volume?
- Volume: $V(x) = x(12 - 2x)^2$, domain $0 < x < 6$.
- Expand: $V = 144x - 48x^2 + 4x^3$. Then $V'(x) = 144 - 96x + 12x^2 = 12(x-2)(x-6)$.
- Critical numbers $x = 2, 6$; only $x = 2$ is interior.
- $V(2) = 2(8)^2 = 128$. Endpoints give $V \to 0$, so $x = 2$ is the max.
📘 Example 6.3: Maximizing Profit
A firm's revenue is $R(x) = 60x - x^2$ and cost is $C(x) = 10x + 100$ (in dollars, $x$ items). Maximize profit.
- Profit: $P(x) = R - C = 60x - x^2 - 10x - 100 = -x^2 + 50x - 100$.
- $P'(x) = -2x + 50 = 0 \Rightarrow x = 25$.
- $P''(x) = -2 < 0$ ⇒ concave down ⇒ maximum.
- $P(25) = -625 + 1250 - 100 = 525$.
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Section 6.3
Elasticity of Demand
Definition: Elasticity of Demand
For a demand function $q = f(p)$, the elasticity measures the percent change in quantity per percent change in price:
$$E = -\frac{p}{q}\cdot\frac{dq}{dp}.$$
The minus sign makes $E$ positive (since demand slopes down). It tells a business whether raising prices will raise or lower revenue.
Reading Elasticity
| Case | Name | Effect of a price increase |
|---|---|---|
| $E > 1$ | Elastic | revenue falls (demand very responsive) |
| $E = 1$ | Unit elastic | revenue is maximized |
| $E < 1$ | Inelastic | revenue rises (demand barely moves) |
A linear demand curve. The elastic region (high price, $E>1$) is orange; inelastic (low price, $E<1$) is blue; the midpoint is unit-elastic ($E=1$), where revenue peaks. Drag the price marker and watch $E$ and total revenue $R = pq$ respond.
E = ,
📘 Example 6.4: Computing Elasticity
Demand is $q = 200 - 5p$. Find the elasticity at $p = 10$ and interpret.
- $\dfrac{dq}{dp} = -5$; at $p = 10$, $q = 200 - 50 = 150$.
- $E = -\dfrac{p}{q}\cdot\dfrac{dq}{dp} = -\dfrac{10}{150}(-5) = \dfrac{50}{150} = \dfrac{1}{3}$.
Section 6.4
Implicit Differentiation
Some equations, like $x^2 + y^2 = 25$, don't isolate $y$ neatly. Implicit differentiation finds $\frac{dy}{dx}$ anyway, by differentiating both sides and treating $y$ as a function of $x$.
The Method
- Differentiate both sides with respect to $x$.
- Every time you differentiate a $y$-term, multiply by $\dfrac{dy}{dx}$ (the chain rule).
- Collect all $\dfrac{dy}{dx}$ terms on one side and solve.
📘 Example 6.5: Implicit Differentiation
Find $\dfrac{dy}{dx}$ for $x^2 + y^2 = 25$.
Differentiate both sides
$$2x + 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}.$$
At $(3, 4)$: $\dfrac{dy}{dx} = -\dfrac{3}{4}$, the slope of the tangent to the circle there.
$\dfrac{dy}{dx} = -\dfrac{x}{y}$; at $(3,4)$ the slope is $-\tfrac{3}{4}$
Section 6.5
Related Rates
When two or more quantities are linked by an equation and both change with time, their rates are linked too. Related rates problems find one rate from another, differentiating the relationship with respect to time $t$.
The Strategy
- Draw a picture; name the changing quantities.
- Write an equation relating them.
- Differentiate both sides with respect to $t$ (chain rule everywhere).
- Substitute known values and the given rate; solve for the unknown rate.
The classic sliding-ladder problem. The top slides down the wall at a fixed rate; the bottom slides outward. Because $x^2 + y^2 = L^2$ is fixed, the two rates are related by $x\frac{dx}{dt} + y\frac{dy}{dt} = 0$. Notice the bottom races away faster and faster as the ladder flattens. Click to play.
📘 Example 6.6: The Sliding Ladder
A $10$-ft ladder leans on a wall. The bottom is pulled away at $2$ ft/s. How fast is the top sliding down when the bottom is $6$ ft from the wall?
- Relation: $x^2 + y^2 = 100$. At $x = 6$: $y = \sqrt{100 - 36} = 8$.
- Differentiate w.r.t. $t$: $2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$.
- Substitute $x=6$, $y=8$, $\dfrac{dx}{dt}=2$: $\;12(2) + 16\dfrac{dy}{dt} = 0$.
- Solve: $\dfrac{dy}{dt} = -\dfrac{24}{16} = -1.5$ ft/s.
Section 6.6
Differentials: Linear Approximation
Definition: The Differential
For $y = f(x)$, the differential is
$$dy = f'(x)\,dx.$$
It estimates the actual change $\Delta y$ for a small change $dx$ in $x$. Geometrically, $dy$ is the rise along the tangent line, a straight-line stand-in for the curve near a point.
Linear Approximation Formula
$$f(x + \Delta x) \approx f(x) + f'(x)\,\Delta x.$$
Use a known, easy point to estimate a nearby hard one. The tangent line is the best linear approximation to the curve.
📘 Example 6.7: Approximating a Square Root
Estimate $\sqrt{26}$ using differentials.
- Let $f(x) = \sqrt{x}$, with the easy point $x = 25$ and $\Delta x = 1$.
- $f'(x) = \dfrac{1}{2\sqrt{x}}$, so $f'(25) = \dfrac{1}{10}$.
- $\sqrt{26} \approx \sqrt{25} + \dfrac{1}{10}(1) = 5 + 0.1 = 5.1$.
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Looking Ahead
We've now squeezed the derivative for everything it can tell us. The final chapter, Chapter 7, runs the machine in reverse: given a rate, recover the total. That reversal, integration, turns out to compute area, and the Fundamental Theorem of Calculus reveals these two ideas to be one.