Optimisation is where calculus earns its keep: the cheapest box, the best angle, the least metal, the fastest route. And linearization is the bridge back — the fact that near any point a curve is nearly its tangent line powers everything from error estimates to Newton's method. This chapter pairs the two halves of applied calculus.
1 · The optimization master method
Every optimization problem, from trivial to fiendish, runs the same loop:
- Variable: name the quantity you may change (x, r, h, θ).
- Constraint: translate the fixed condition ('perimeter 40', 'volume 32') into an equation linking your variables.
- Objective: write the quantity to extremize (area, surface cost, time) as a function of one variable, using the constraint to eliminate the other.
- Differentiate and solve: find critical points of the one-variable objective.
- Verify the endpoints: on a closed interval, compare critical values against boundary values; on open domains, check the limit at the edges.
- Answer the actual question: dimensions, not just x.
2 · An easy warm-up: fixed perimeter
A rectangle has perimeter 40. What dimensions maximize its area? Let the width be x; the height is (40 − 2x)/2 = 20 − x. Objective: A(x) = x(20 − x) = 20x − x². A′(x) = 20 − 2x = 0 gives x = 10, and A″(x) = −2 < 0 confirms a maximum. The rectangle is a square, 10 × 10, area 100 — perimeter fixed, square wins.
3 · A harder problem: the least-material box
Open-top box, square base, volume 32 cm³ — minimize surface area
- Variable: side of the square base, a > 0.
- Constraint: a²h = 32, so h = 32/a².
- Objective: metal used S(a) = a² + 4ah = a² + 4a·(32/a²) = a² + 128/a.
- Differentiate: S′(a) = 2a − 128/a². Set to zero: 2a³ = 128 → a³ = 64 → a = 4.
- Verify: S″(a) = 2 + 256/a³, and S″(4) = 6 > 0 → minimum. Or note S → ∞ as a → 0⁺ and as a → ∞.
- Dimensions: a = 4 cm, h = 32/16 = 2 cm; minimum surface area S(4) = 16 + 32 = 48 cm².
4 · Optimization with implicit differentiation: cone in a sphere
A right circular cone is inscribed in a sphere of radius R. Find the cone of maximum volume. Let r be the base radius and h the height measured from the bottom of the sphere. The geometry gives r² + (h − R)² = R², i.e. r² = 2Rh − h². Differentiate implicitly: 2r·dr + 2(h − R)·dh = 0, relating the rates of change directly.
Volume: V = (1/3)πr²h = (1/3)π(2Rh − h²)h = (1/3)π(2Rh² − h³). Then V′(h) = (1/3)π(4Rh − 3h²) = (1/3)πh(4R − 3h). Critical points: h = 0 (rejected) and h = 4R/3. Check second derivative or endpoints: V(4R/3) is maximal since V(0) = V(2R) = 0 with V ≥ 0 between. Then r² = 2R(4R/3) − 16R²/9 = 8R²/9, so r = (2√2/3)R, and V_max = (1/3)π·(8R²/9)·(4R/3) = 32πR³/81. The implicit-differentiation route shines when variables are entangled and solving one outright is messy.
5 · Linearization and the differential
Near a point a, the tangent line is an excellent stand-in for the curve:
L(x) = f(a) + f′(a)(x − a)
Example: √x near a = 4. L(x) = 2 + (x − 4)/4. So √4.1 ≈ 2 + 0.1/4 = 2.025 (true value ≈ 2.02485). Also eˣ ≈ 1 + x near 0, and sin x ≈ x near 0 — the small-angle rule, resurrected. Linearization is trustworthy only close to a; the further you wander, the worse the lie.
The differential dy = f′(x)dx is the tangent-line change, packaged as algebra. It answers: if x shifts by dx, how much does f move along the tangent? Real functions curve, so dy approximates the true change Δf. The error is controlled by the second derivative, or by the largest slope on the interval: if |f′(t)| ≤ M for t between x and x + dx, then
|Δf| ≤ M·|dx|
and the sharper Taylor form |Δf − dy| ≤ ½·M₂·dx² holds with M₂ bounding |f″| on the interval.
Example: estimate √9.1 to five decimals — with a = 9, dx = 0.1, f′(x) = 1/(2√x) ≤ 1/6 on [9, 10], so the error is at most ½·(1/108)·(0.1)² ≈ 5×10⁻⁵. Small enough to trust every displayed digit.
6 · Newton's method
Newton's method for √2
- Setup: solve f(x) = x² − 2 = 0. f′(x) = 2x. Iteration: x_{n+1} = x_n − (x_n² − 2)/(2x_n).
- Start at x₀ = 1: x₁ = 1 − (1 − 2)/2 = 1 + ½ = 1.5.
- Iterate: x₂ = 1.5 − (2.25 − 2)/3 = 1.5 − 1/12 ≈ 1.41667. Error already below 0.003.
- One more: x₃ ≈ 1.41422 — correct to five decimals, and the number of correct digits roughly doubles each step (quadratic convergence).
Newton's method is just linearization used backwards: walk to where the tangent line would cross the axis, then re-linearize. It can fail — a zero derivative, a poor starting guess, or a cycle can all strand you — which is why a grapher check before iterating is cheap insurance.
Do not confuse the two linearization errors. dy = f′(x)dx is the first-order (tangent) estimate; |Δf − dy| ≤ ½Mdx² is the correction that tells you how far the true change can drift from dy. And in Newton's method, x_{n+1} = x_n − f(x_n)/f′(x_n) — divide by f′, not f, and never by f evaluated at the wrong index.
L(x) = f(a) + f′(a)(x − a)
dy = f′(x)dx
|Δf − dy| ≤ ½·M·dx², with |f′| ≤ M on the interval
Newton: x_{n+1} = x_n − f(x_n) / f′(x_n)
eˣ ≈ 1 + x near 0; sin x ≈ x near 0
Common mistakes
- Minimizing or maximizing before expressing the objective as a function of a single variable.
- Forgetting to verify interior critical points against endpoints or the open-domain limits.
- Accepting h = 4R/3-type algebra without checking r² comes out positive.
- Applying L(x) = f(a) + f′(a)(x − a) with a typo in the expansion — keep the (x − a), not (x + a).
- Using dy as if it were the exact Δf for large dx — linearization is local.
- Newton starting at a point where f′ is zero or nearly zero, and watching the iterates explode.