Mathfolis

Optimization

Unit 5 · Analytical Applications of Differentiation

General Approach

Optimization problems ask: what value of x makes quantity Q as large (or small) as possible? The key steps are always the same — identify, constrain, differentiate, verify.

Optimization Strategy

1. Objective function
Q=f(x) — the quantity to maximize or minimizeQ = f(x) \text{ — the quantity to maximize or minimize}
2. Constraint
g(x,y)=k  ⟹  y=h(x) — reduces to one variableg(x, y) = k \implies y = h(x) \text{ — reduces to one variable}
3. Critical points
Q′(x)=0 (and check domain endpoints if closed interval)Q'(x) = 0 \text{ (and check domain endpoints if closed interval)}
4. Verify max/min
Q′′(x)<0⇒maxQ′′(x)>0⇒minQ''(x) < 0 \Rightarrow \text{max} \qquad Q''(x) > 0 \Rightarrow \text{min}
Caution: On a closed interval, the maximum or minimum may occur at an endpoint, not at a critical point. Always check all candidates.
Type 1

Geometric Optimization

Maximize or minimize a geometric quantity — area, perimeter, volume, or surface area. Write the objective function, apply the constraint to reduce to one variable, then optimize.

Example 1

A farmer has 120 ft of fencing to enclose a rectangular field along a river. No fence is needed along the river. Find the dimensions that maximize the enclosed area.

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Type 2

Applied Optimization

Real-world problems involving cost, revenue, distance, or time. Identify the quantity to optimize, set up the constraint, reduce to one variable, and verify the extremum.

Example 2

A square piece of cardboard measuring 12 in × 12 in has equal squares of side x cut from each corner. The sides are folded up to form an open box. Find the value of x that maximizes the volume.

V(x)=x(12−2x)2,x∈(0, 6)V(x) = x(12-2x)^2, \quad x \in (0,\, 6)

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