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Critical points play a vital role when you're dealing with optimization problems in calculus. They help identify where functions might reach their maximum or minimum values. In this architectural problem, we need to minimize the cost of building a structure given certain constraints.
To solve it, we first find the derivative of the cost function, and then set this derivative to zero. This helps identify potential critical points, which are candidates for minimum or maximum values.
In our example, we derived the cost function and found its derivative as follows:

• Cost function: \( C(y) = 2y + \frac{4050}{y} \)
• Derivative: \( C'(y) = 2 - \frac{4050}{y^2} \)
• Setting \( C'(y) = 0 \), leads us to critical points by solving \( 2 - \frac{4050}{y^2} = 0 \).
• This simplifies to \( y^2 = 2025 \), and thus \( y = 45 \).

Finding this critical point is essential as it indicates where the function might experience a change in its trend.
The critical point needs further verification to check if it indeed represents a minimum or maximum.

The derivative represents the rate of change of a function concerning its variable. Here, the derivative of the cost function helps us figure out how cost changes with respect to the dimensions of the building's length \(y\).
By finding the derivative of the cost function \( C(y) = 2y + \frac{4050}{y} \), we get:

• \( C'(y) = 2 - \frac{4050}{y^2} \)

Calculating this derivative allows us to determine where the slope of our cost function changes. When the derivative \( C'(y) \) is zero, it indicates that the function changes direction at that point. This particular scenario, where the rate of increase is zero, highlights potential points of optimization, possibly a cost minimum.
Understanding derivatives in optimization problems is key, as it allows one to identify trends and extreme points of a function.

The area constraint is a practical condition that must be met in design problems like this architectural one. Constraints like these connect different elements and create relationships between variables. In this case, the area of the building is fixed at 1350 square meters.
The equation representing this constraint is given by:

• \( xy = 1350 \)

This equation shows the relationship between the length \(y\) and the width \(x\) of the building. It provides one of the variables in terms of the other to solve the optimization problem effectively.
By rearranging, we can express \( x \) as \( x = \frac{1350}{y} \), allowing us to substitute back into the cost function and facilitate calculations with a single variable.
Handling these constraints properly is crucial as they form the basis for ensuring that solutions not only optimize but also remain practical and feasible in real-world conditions.

The cost function in this exercise is a mathematical representation of the total wall cost which depends on the dimensions of the building. Constructing an accurate cost function is the first step in any optimization problem.
We start by examining how different walls contribute to the total cost. In this problem, the cost function takes into account that the front wall is twice as expensive as the others:

• \( C = 2y + 3x \)
• Substitute \( x = \frac{1350}{y} \) to get: \( C(y) = 2y + \frac{4050}{y} \)

This function reflects the cost per meter and provides a formula to calculate the minimal cost for specific dimensions.
Understanding the components of a cost function allows us to know what factors influence costs and makes it clear where efficiencies can be achieved. Optimizing it means detecting those conditions under which minimal costs can be realized, making this function an essential tool for any effective cost management practice.