Module 4 · Lesson 4.6

Optimization — Derivatives Solving Real-World Problems

Turn a design constraint into a single-variable function, then use the derivative to locate the dimensions that minimize material cost.

Intuition

Many practical problems ask for the “best” choice under a fixed constraint: the cheapest can that holds a given volume, the strongest beam for a given amount of wood, the fastest route that stays inside a budget. The constraint lets you eliminate one variable, leaving a function of a single free variable whose minimum or maximum is found by setting the derivative to zero.

The classic packaging problem is a perfect illustration. An open-top box must hold a fixed volume; the amount of material (surface area) depends on the base side length. Calculus finds the side length that uses the least cardboard.

Core concept

Write the quantity to optimize as a function of one variable by substituting the constraint. Differentiate, set the derivative equal to zero, and test the critical points (and endpoints if the domain is closed). The winning value is the optimal trade-off.

Volume fixed$x^{2}h=32$
Surface$S=x^{2}+4xh$
Eliminate $h$$S=x^{2}+128/x$
$S'=0$$2x-128/x^{2}=0$
Optimum$x=4$, $h=2$

Interactive Lab

Adjust the base side length $x$. The height is forced by the volume constraint $x^{2}h=32$; the surface area $S$ is plotted live so you can see the minimum.

Height $h$: 2.00 Surface $S$: 48.00
x S
Volume constraint $x^{2}h=32$. Surface $S=x^{2}+4xh=x^{2}+128/x$.

Green marker is the calculus minimum ($x=4$, $S=48$). Drag the amber probe to compare other designs; any other $x$ uses more material for the same volume.

Key Idea

$$ S'(x)=0 \quad\Rightarrow\quad x\text{ that minimizes (or maximizes) }S $$

After the constraint has reduced the problem to a single-variable function $S(x)$, the optimal design occurs at a critical point of $S$. Solve $S'(x)=0$, discard any solutions outside the physical domain, and compare the remaining candidates (including endpoints when the interval is closed). The smallest value of $S$ is the minimal material cost.

$x$The free design variable (base side length)
$h$The dependent variable eliminated by the volume constraint
$S(x)$Quantity to minimize (surface area / material cost)
$S'(x)=0$The equation that locates the optimal trade-off

Think Further

  1. Why must you check the second-derivative test or the first-derivative sign chart after solving $S'(x)=0$?
  2. If the box were required to have a lid, how would the surface-area formula change and what would the new optimal $x$ be?
  3. In an AI training budget, the “constraint” might be a fixed number of GPU hours. What quantity would you differentiate to decide the best model size versus training length trade-off?