Optimization Problems
Set up and solve optimization problems: define variables, write the objective function, express any constraints, reduce to one variable, find and classify critical points, and verify with endpoints or the second derivative test.
A lesson of Calculus, in Applications of Derivatives, about 10 minutes.
By the end of this lesson, you will be able to set up and solve applied optimization problems using calculus, confirm global extrema using closed interval or infinity boundary analysis, and interpret results in context.
The Big Idea
Optimization is finding the maximum or minimum value of a quantity. In calculus, the process is: model the quantity as a function, find critical points via $f' = 0$, then confirm the global extreme using the closed interval method or behavior at the boundaries.
The Optimization Procedure
- Identify what to maximize or minimize (the objective function)
- Define variables for all quantities
- Write the objective as a function of one variable (use constraint equations to eliminate extras)
- Find the domain (physical restrictions)
- Differentiate and set $f' = 0$
- Verify using the second derivative test or endpoint comparison
- Answer the question with units and context
Example 1 — Fencing (Maximize Area)
200 feet of fence encloses three sides of a rectangle (the fourth side is a river). Find the maximum area.
Let $x$ = width, $y$ = length. Constraint: $2x + y = 200 \implies y = 200 - 2x$.
$A(x) = xy = x(200 - 2x) = 200x - 2x^2$; domain: $0 < x < 100$.
$A'(x) = 200 - 4x = 0 \implies x = 50$
$A''(50) = -4 < 0$ → maximum.
$A(50) = 50 \times 100 = 5000$ sq ft. Dimensions: $50 \times 100$ ft.
Example 2 — Box Without Lid
A box with a square base and open top must have volume $32000$ cm³. Minimize the material used (surface area).
Let $x$ = side of square base, $h$ = height.
Constraint: $V = x^2 h = 32000 \implies h = 32000/x^2$
Surface area: $S = x^2 + 4xh = x^2 + 4x \cdot \frac{32000}{x^2} = x^2 + \frac{128000}{x}$
$S'(x) = 2x - \frac{128000}{x^2} = 0 \implies 2x^3 = 128000 \implies x^3 = 64000 \implies x = 40$ cm
$S''(x) = 2 + \frac{256000}{x^3} > 0$ → minimum.
$h = 32000/1600 = 20$ cm. Dimensions: $40 \times 40 \times 20$ cm.
Example 3 — Distance to a Curve
Find the point on the parabola $y = x^2$ nearest to the point $(3, 0)$.
Distance: $d = \sqrt{(x-3)^2 + y^2} = \sqrt{(x-3)^2 + x^4}$
Minimize $f(x) = (x-3)^2 + x^4$ (same critical points as distance, easier algebra):
$f'(x) = 2(x-3) + 4x^3 = 0$
$4x^3 + 2x - 6 = 0 \implies 2x^3 + x - 3 = 0 \implies (x-1)(2x^2+2x+3) = 0$