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Calculus is a branch of mathematics that helps us understand change and motion. When dealing with optimization problems, calculus becomes essential, especially when we aim to find maximum or minimum values given certain conditions. In this particular problem, we are trying to find the dimensions of a box that will maximize its volume while being constrained by cost. This involves setting up equations derived from the given problem statement and then using calculus techniques such as derivatives to solve these equations. Optimization in calculus usually entails identifying critical points and using the first and second derivative tests to determine whether these points are maxima or minima. Therefore, understanding how calculus applies to optimization ensures we can solve these types of problems efficiently and accurately.

The volume of a box is calculated by multiplying the area of its base by its height. For a rectangular box with a square base, the formula is quite straightforward. If the side of the square base is given by length \(x\) and the height is \(h\), the volume \(V\) is calculated as \(V = x^2h\). This basic geometric principle is central to solving the problem, as we aim to maximize this volume. However, the challenge lies in balancing the box's dimensions considering the cost constraints given in the problem. The more material used, particularly expensive material, the higher the cost, which is a limiting factor. Thus, a critical aspect of this type of problem is expressing volume and cost constraints in a manageable mathematical form for optimization.

Derivatives are a tool in calculus that allow for finding the rate of change of a function. They are imperative in optimization problems, as they help locate critical points where functions reach their maximum or minimum values. In our box volume problem, the function we wish to optimize is the volume \(V = 9x - \frac{x^3}{2}\). We start by finding the first derivative \(\frac{dV}{dx} = 9 - \frac{3}{2}x^2\) to determine the rate at which the volume changes with respect to \(x\). To find the critical points, we set this derivative to zero and solve for \(x\), identifying potential points for maximum volume. Additionally, the second derivative \(\frac{d^2V}{dx^2} = -3x\) helps confirm these points as maxima because it gives us information about the concavity of the function. Negative concavity indicates a local maximum, thereby confirming the optimization.

Constraints in optimization problems limit the potential solutions to those that satisfy additional conditions, such as cost or boundary restrictions. In our scenario, the main constraint is the total construction cost of \$36. This forces us to balance material use on the box’s top and sides. The setup leads to the constraint equation \(2x^2 + 4xh = 36\), which establishes a relationship between the base size and height of the box. Constraints are not only limitations but also guides to structuring solutions that are both mathematically possible and financially feasible. Solving such a constraint equation helps express one variable in terms of another, simplifying the problem. Once the relationship between \(x\) and \(h\) is established through rearranging the constraint equation, it can be substituted into the volume equation for further processing. Thus, handling constraints effectively is key to arriving at the optimal solution in real-world applications.