Calculus Honors · AP-aligned study guide

3.4 · Optimization: constraints, domains, and justification

Turn a geometric or contextual goal into a one-variable function, then prove the candidate gives the requested optimum.

AP 5.10–5.11 · Reading: Larson & Edwards §3.7

Learning objectives

Build an objective function using a constraint.
Restrict the domain to physically meaningful values.
Justify a global optimum rather than reporting a critical number.

Separate the objective from the constraint

The objective is the quantity to maximize or minimize, such as area or cost. The constraint connects the variables. For a rectangle with perimeter $\displaystyle 20$ , $\displaystyle 2x+2y=20$ lets you write $\displaystyle y=10-x$ .

The area becomes $\displaystyle A(x)=x(10-x)$ , with meaningful dimensions $\displaystyle 0<x<10$ . A sketch and a domain prevent a mathematically valid but physically impossible solution.

Differentiate the objective and test the candidate

The derivative $\displaystyle A'(x)=10-2x$ vanishes at $\displaystyle x=5$ . Its sign changes from positive to negative there, proving a maximum. The dimensions are $\displaystyle 5$ by $\displaystyle 5$ , with area $\displaystyle 25$ .

On a closed domain, compare endpoints too. On an open domain, justify with monotonicity or boundary behavior. Include units and translate the optimizing input back into the quantity requested.

A minimum can come from competing terms

For $\displaystyle C(x)=x+9/x$ with $\displaystyle x>0$ , increasing $\displaystyle x$ raises the first term but lowers the second. The derivative $\displaystyle 1-9/x^2$ balances these effects at $\displaystyle x=3$ .

The derivative is negative for $\displaystyle 0<x<3$ and positive for $\displaystyle x>3$ , proving an absolute minimum there. Merely solving $\displaystyle C'=0$ would not establish which kind of point it is.

Worked examples

Fixed perimeter

Maximize rectangle area with perimeter $\displaystyle 24$ .

Write $\displaystyle y=12-x$ , so $\displaystyle A(x)=12x-x^2$ for $\displaystyle 0<x<12$ .
Solve $\displaystyle A'(x)=12-2x=0$ to get $\displaystyle x=6$ .
The derivative changes from positive to negative; the optimal dimensions are $\displaystyle 6$ by $\displaystyle 6$ , with area $\displaystyle 36$ .

Minimum cost

Minimize $\displaystyle C(x)=x+16/x$ for $\displaystyle x>0$ .

The derivative is $\displaystyle 1-16/x^2$ .
The only positive critical input is $\displaystyle 4$ .
The derivative changes from negative to positive, so the minimum is $\displaystyle C(4)=8$ .

Common mistakes

Differentiating before using the constraint to eliminate a variable.
Ignoring the physical domain.
Reporting a critical input without proving it gives the requested optimum.

Try it yourself

No calculator required.

1. A rectangle has perimeter $\displaystyle 20$ m. Find the maximum area and justify it.

The downward parabola A(x) = 10x minus x squared is defined for widths between 0 and 10. — Use the derivative to locate the maximum area.

Show worked solution

The area is $\displaystyle A(x)=10x-x^2$ for $\displaystyle 0<x<10$ . Its derivative changes from positive to negative at $\displaystyle x=5$ , so the maximum is $\displaystyle 25$ square meters for dimensions $\displaystyle 5$ by $\displaystyle 5$ .

The downward parabola A(x) = 10x minus x squared is defined for widths between 0 and 10. The area is maximal at width 5, giving area 25. — The area is maximal at width 5, giving area 25.

2. Minimize $\displaystyle C(x)=x+9/x$ on $\displaystyle x>0$ .

Show worked solution

The derivative is $\displaystyle 1-9/x^2$ , negative below $\displaystyle 3$ and positive above $\displaystyle 3$ . Thus the absolute minimum occurs at $\displaystyle x=3$ and is $\displaystyle C(3)=6$ .

3. Find the point on $\displaystyle y=x^2$ closest to $\displaystyle (0,1/2)$ .

Show worked solution

Minimize squared distance $\displaystyle D(x)=x^2+(x^2-1/2)^2=x^4+1/4$ . This is minimized at $\displaystyle x=0$ , so the point is $\displaystyle (0,0)$ and the distance is $\displaystyle 1/2$ . Squared distance has the same minimizer as distance.

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