Mini-Lecture: Domains & The Candidate Test

One of the most common sources of anxiety in Calculus, especially in Optimization problems, is figuring out the "Domain" and then knowing if you actually found the right answer.

We are going to break this down into two skills:

  1. Finding the Domain: What are you allowed to plug in?
  2. The Candidate Test: How to check your answer and verify the winner.

Part 1: Finding the Domain

When solving real-world optimization problems, finding the domain is a two-step process: The Math Check and The Reality Check.

Math
The Algebra Rules
1. Denominators $\neq 0$
2. Even roots $\ge 0$
Reality
The Physical Rules
1. Length/Radius $> 0$
2. Resources are finite.

Example: The Cost Function

Let's look at the function for the cost of materials for a can:

$$ C(r) = k\pi r \left( r + \frac{355}{\pi r^2} \right) $$

1. The Math Check:

We see a fraction: $\frac{355}{\pi r^2}$. Since we cannot divide by zero, $r \neq 0$.

2. The Reality Check:

Because $r$ is a radius, it must be positive. Is there a maximum size? Mathematically, no. You could have a radius of 1 mile (the can would just be microscopically flat to keep the volume constant). So, the only constraint is that the radius must exist.

Conclusion:

Combining $r \neq 0$ and $r$ is positive, our domain is:

Domain: $(0, \infty)$

Visualizing the Domain: This graph shows the cost function. Notice that it only exists for $x > 0$. As $r$ gets close to 0, the cost shoots up (vertical asymptote).

Part 2: The Candidate Test

Once you have your domain, you take the derivative and find your Critical Points. But then the doubt sets in: "Is this the answer? Is this a max or a min? Did I miss something?"

If your domain is a Closed Interval (like $[a, b]$), you use The Candidate Test.

The Candidate Test (Closed Interval Method)

If you are optimizing on a closed interval $[a, b]$, the Absolute Max/Min MUST occur at either:

  1. A Critical Point, OR
  2. An Endpoint ($a$ or $b$).

Strategy: Make a table. Plug the critical points AND the endpoints into the original function. The highest number wins.

Example: The Cardboard Box

We have a square piece of cardboard that is 12 inches wide. We want to cut square corners of size $x$ and fold up the sides to make a box with the maximum Volume.

The Volume function is: $V(x) = x(12-2x)(12-2x)$.

Calculated Derivative: $V'(x) = 12(6-x)(2-x)$.

We find two Critical Points: $x=6$ and $x=2$.

1. Establish the Domain

  • Lower Bound: $x \ge 0$ (Length must be positive).
  • Upper Bound: If we cut $x$ from both sides, $2x$ cannot exceed 12. So $x \le 6$.

Domain: $[0, 6]$

2. Run the Candidate Test

We test our Critical Points ($2, 6$) and our Endpoints ($0, 6$).

Candidate ($x$) Volume $V(x)$ Result
0 (Endpoint) 0 Min
2 (Critical) 128 MAX (Winner)
6 (Endpoint/Crit) 0 Min

Conclusion: By checking the endpoints, we proved that $x=2$ is the Absolute Maximum.

Check Your Understanding: Running the Full Test

Find the Absolute Maximum and Absolute Minimum of the function $f(x) = 2x^3 - 3x^2 - 12x$ on the interval $[-2, 3]$.

Hint: Find the critical points first, then check the table.