Applied Optimization

Optimization problems ask for the largest or smallest value of a real-world quantity — maximum area, minimum cost — subject to a constraint. Calculus turns them into a one-variable extremum hunt: write the quantity as \(f(x)\), then use the critical-point machinery you already have.

By the end you'll be able to set up an objective and a constraint from a word problem, reduce to one variable, and find the optimum with derivative tests — checking the feasible domain's endpoints, not just where \(f'(x) = 0\).

Predict: a farmer has 100 m of fence for a rectangular pen against a river — no fence needed on the river side. What width x gives the biggest area? A skinny-deep pen? A shallow-wide one? Something in between? Commit to a guess, then slide x and watch both panels.

The two perpendicular sides each have length x, which leaves y = 100 − 2x for the side along the river. The left panel is the pen to scale; the right panel plots the area A(x) = x(100 − 2x) with your current pen marked on the curve. Hover or focus the curve and points for exact values.

x = 10.0 m · y = 100 − 2x = 80.0 m · A = x·y = 800 m² · fence used: 2x + y = 100 m

The pen, to scale — the river side needs no fence
Area A(x) = x(100 − 2x) — every pen is a point on this curve

Optimization problems translate a real-world quantity to maximize or minimize into a single-variable function using constraints, then apply critical-point and extreme-value methods to find the optimum.

The method: model, reduce, optimize

Every applied optimization problem follows the same script. What's new here isn't calculus — it's the modeling that happens before the calculus.

  1. Identify the objective — the quantity to maximize or minimize (area, cost, volume) — and name the variables.
  2. Write the objective as a formula, and write the constraint equation that ties the variables together (a fixed fence length, a required volume).
  3. Reduce to one variable: solve the constraint for one variable and substitute, so the objective becomes \(f\) of a single variable.
  4. Determine the feasible domain — the physically allowed range (lengths \(\ge 0\), capacity limits).
  5. Find critical points (\(f' = 0\) or \(f'\) undefined) and apply the closed-interval method or the First/Second Derivative Test.
  6. Verify max vs. min, and answer the question actually asked — including units.
Applied

In the pen problem the objective is area and the constraint is fence: \(2x + y = 100\). The constraint is what makes the problem interesting — without it you'd just build an infinite pen. Solving it for \(y = 100 - 2x\) and substituting collapses two unknowns into \(A(x) = x(100 - 2x)\) — the parabola you dragged along above. Real problems (materials, shipping costs, revenue) differ only in what plays the role of fence.

Formal

Once reduced, this is exactly the absolute-extrema problem you've already solved: \(A(x) = 100x - 2x^2\) is continuous on the closed feasible domain \([0, 50]\), so the EVT guarantees an absolute maximum, attained at a critical point or an endpoint. \(A'(x) = 100 - 4x = 0\) at \(x = 25\); compare \(A(0) = 0\), \(A(25) = 1250\), \(A(50) = 0\). The interior critical point wins here — but the endpoint check is part of the method, because in other problems (as in the quiz below) the optimum sits on the boundary.

Don't skip the endpoints

The three classic pitfalls: ignoring the constraint (never reducing to one variable), forgetting the domain (the optimum may sit at an endpoint, or your critical point may be infeasible), and not confirming max vs. min — or answering with \(x\) when the question asked for the area. A critical point with \(f'(x) = 0\) is a candidate, nothing more, until you've compared it against the feasible domain's endpoints and classified it.

Worked example

A farmer has 100 m of fence for a rectangular pen against a river (no fence on the river side). Maximize the area.
Objective & constraint: maximize \(A = xy\) subject to \(2x + y = 100\). Reduce: \(y = 100 - 2x\), so \(A(x) = x(100 - 2x) = 100x - 2x^2\) on the feasible domain \(0 \le x \le 50\). Optimize: \(A'(x) = 100 - 4x = 0 \Rightarrow x = 25\), and \(A''(x) = -4 < 0\), so it's a maximum. Check endpoints: \(A(0) = A(50) = 0\). Answer: \(x = 25\) m, \(y = 50\) m, area \(A = 1250\ \text{m}^2\).

Your turn

Same script, different fence: 60 m of fence for a fully enclosed rectangular pen (no river this time — all four sides need fence). Constraint: \(2x + 2y = 60\), so \(y = 30 - x\). Objective: \(A(x) = x(30 - x) = 30x - x^2\), feasible domain \(0 \le x \le 30\). Now finish it: \(A'(x) = 30 - 2x = 0 \Rightarrow x = \) ____, so \(y = \) ____ and the maximum area is ____ m².

Reveal the answer

\(A'(x) = 30 - 2x = 0 \Rightarrow x = 15\), and \(A''(x) = -2 < 0\) confirms a maximum. Then \(y = 30 - 15 = 15\) and \(A = 15 \cdot 15 = \) 225 m² — with all four sides fenced, the best rectangle is a square. (Endpoints: \(A(0) = A(30) = 0\).) Compare with the river pen above, where the free side made the optimum twice as wide as it is deep.

More info — why this is the closed-interval method in disguise

Steps 4–6 of the method are exactly the absolute-extrema toolkit from the earlier lesson on maxima and minima: find where \(f' = 0\) or is undefined, evaluate \(f\) there and at the endpoints, and take the largest (or smallest) value. Optimization just adds a modeling front end — objective, constraint, reduce — that produces the function and its interval. If your feasible domain is open or unbounded (say \(x > 0\) with no upper limit), the EVT guarantee is gone and you lean on the First or Second Derivative Test to classify the lone critical point instead. The OpenStax chapter in Dive deeper below walks the full strategy on area and surface-area problems.

Check your understanding

Question 1 of 4

A farmer has 80 m of fence for a rectangular pen against a barn (no fence needed on the barn side). What is the maximum possible area?

Question 2 of 4

You've modeled a quantity as \(f(x)\) on the feasible domain \([a, b]\) and found exactly one interior critical point where \(f'(x) = 0\). Before reporting it as the maximum, what must you still do?

Question 3 of 4

A cost function \(C(x) = x^2 - 8x + 20\) models production cost, where capacity limits \(x\) to \(0 \le x \le 3\). What is the minimum cost?

Question 4 of 4

You're designing an open-top box with a square base of side \(x\) and height \(h\) that must hold exactly 500 cm³, using as little material as possible. Which setup is correct?

Recap

  • Optimization = modeling + extrema: identify the objective and the constraint, then solve the constraint and substitute to reduce the objective to one variable.
  • Always write down the feasible domain — physical limits like lengths \(\ge 0\) — before differentiating.
  • Find critical points (\(f' = 0\) or undefined) and compare against the endpoints; the optimum can sit on the boundary, and a critical point outside the domain is irrelevant.
  • Confirm max vs. min (sign change of \(f'\), or \(f''\)), and answer the question asked — the requested quantity, with units.
  • River pen with 100 m of fence: \(A(x) = x(100 - 2x)\), \(A'(x) = 100 - 4x = 0\) at \(x = 25\), giving \(y = 50\) and \(A = 1250\ \text{m}^2\).

Dive deeper

Sources

  • Applied Optimization