What if the domain is an open interval and there are no endpoints?

Use the First Derivative Test or the Second Derivative Test to show that the single critical point provides an absolute extremum on the entire open interval. If $f''(x)$ maintains the same sign for all $x$ in the domain, the local extremum is the absolute extremum.

What if there are multiple critical points?

Evaluate the objective function at each critical point within the domain, as well as the endpoints. Compare all resulting values and select the greatest or least value accordingly to find the absolute maximum or minimum.

Can the constraint equation have multiple valid substitutions?

Yes. You can isolate either variable in the constraint equation. It is standard practice to substitute the variable that results in the simplest algebraic expression for the objective function to minimize differentiation errors.

How to solve an optimization problem

An optimization problem requires finding the absolute maximum or minimum of an objective function subject to given constraints. This method applies whenever you need to optimize a measurable physical quantity, such as minimizing cost or maximizing volume, over a specific domain.

The setup

Identify the objective function to be maximized or minimized and the constraint equation that limits the variables. Express the objective function in terms of a single variable by isolating one variable in the constraint equation and substituting it into the objective function.

The steps

Write the objective function $f(x)$ in terms of one variable. 2. Determine the valid domain of $f(x)$ based on the physical constraints of the problem. 3. Compute the first derivative $f'(x)$ . 4. Find all critical points by setting $f'(x) = 0$ and solving for $x$ , and identify any points where $f'(x)$ is undefined. 5. Evaluate $f(x)$ at all critical points and at the endpoints of the domain.

Checking the result

Use the Second Derivative Test by computing $f''(x)$ . If $f''(c) > 0$ , the critical point $c$ is a local minimum; if $f''(c) < 0$ , it is a local maximum. Ensure that the absolute extremum occurs within the physically valid domain of the problem by comparing these local extrema to the endpoint values.

Common errors

Failing to check the endpoints of a closed interval, which often contain the absolute extrema. Forgetting to substitute the constraint equation, leading to improper differentiation of a multivariable function with respect to a single variable.

Worked example

A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area?

Let $x$ be the width of the field perpendicular to the river and $y$ be the length parallel to the river. The constraint is $2x + y = 2400$ . The objective function for area is $A = xy$ . Substitute $y = 2400 - 2x$ into $A$ to get the single-variable function: $A(x) = x(2400 - 2x) = 2400x - 2x^2$ The physical domain is $0 \leq x \leq 1200$ . Differentiate to find $A'(x) = 2400 - 4x$ . Set $A'(x) = 0$ to get the critical point $x = 600$ . Evaluate the area at the endpoints and the critical point: $A(0) = 0$ , $A(1200) = 0$ , and $A(600) = 600(2400 - 1200) = 720000$ . The second derivative is $A''(x) = -4$ , which is strictly negative, confirming $x = 600$ yields a maximum. The optimal dimensions are width $x = 600$ ft and length $y = 1200$ ft.

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References: Calculus Volume 1 by OpenStax · Calculus: Early Transcendentals 8th Edition by James Stewart