91Ó°ÊÓ

Area constraints refer to the limitation that the total floor space available for the booth must not exceed 300 square feet. For any optimization problem in calculus, it is crucial to first identify such constraints as they guide the rest of the solution process.
For this exercise, let us imagine a rectangular area where the width is denoted by \( x \), and the length, which is the sum of the two sides, is \( y \). When invoking area constraints, we can set up an equation: \( x \cdot y = 300 \).
This constraint helps in eliminating one variable by expressing \( y \) in terms of \( x \) (or vice versa), simplifying our calculations for further steps. Area constraints provide the mathematical boundary that shapes how we build our cost function and approach the overall problem.

The cost function is central to optimization problems, as it quantifies the resources consumed, in this case, the dollars spent. Here, we need a function, \( C(x, y) \), that represents the total cost of constructing the booth.
Given the specific materials required, the cost of the back display board is \( 30 \times x \) dollars, and the sides, which require gathered fabric, cost \( 4y \) dollars in total. Thus, the cost function becomes \( C = 30x + 4y \).
Defining this cost function helps us see how different dimensions impact the cost. It integrates the variables \( x \) and \( y \) tied to both the structure and expense. It provides a basis for further analysis using calculus.

Differentiation is a powerful calculus tool used to find the minimum or maximum values of a function. By differentiating the cost function with respect to one of its variables, we can determine at what point the cost is minimized.
In this problem, we have expressed \( y \) in terms of \( x \) due to the area constraints, leading to a simplified cost function: \( C = 30x + \frac{1200}{x} \). Now, differentiate: \( \frac{dC}{dx} = 30 - \frac{1200}{x^2} \).
The next step involves finding the critical points; set the derivative equal to zero: \( 30 - \frac{1200}{x^2} = 0 \). Solving this equation helps locate where the cost reaches a minimum, crucial for deriving an optimal solution.

Calculating the minimum cost involves evaluating the critical points found from differentiation. Once \( \frac{dC}{dx} = 0 \) gives us a critical point \( x = \sqrt{40} \approx 6.32 \) feet for the booth's width, substitute back to find \( y \).
Substituting \( x \) into \( y = \frac{300}{x} \) results in \( y \approx 47.43 \) feet. With these dimensions, the calculated minimum cost is derived from \( C = 30 \times 6.32 + 4 \times 47.43 \approx 382.68 \).
This calculated cost confirms that the combination of \( x \) and \( y \) truly provides the minimal expenditure while respecting all constraints, ensuring a cost-effective solution to the problem.