91Ó°ÊÓ

Calculus is a powerful mathematical tool that helps us investigate and understand changes. It provides the foundation for concepts like limits, derivatives, and integrals.
These are essential for solving optimization problems, where we find the best solution from a set of possible options. In the given exercise, calculus plays a critical role in identifying the maximum product of two numbers that satisfy a certain sum.

• Limits: Calculus examines the behavior of functions as they approach specific points or infinity.
• Derivatives: These measure the rate at which a function's value changes.
• Integrals: Used to find areas under curves or accumulated quantities.

Understanding these concepts allows us to solve questions involving rates of change and areas, crucial for practical applications like physics or economics.

A derivative is a fundamental concept in calculus that represents the rate of change of a function with respect to one of its variables. In simpler terms, it shows how a function changes as its input changes.
The derivative helps in identifying where a function increases or decreases and finding its maximum or minimum points, known as optimization.
In the context of the exercise, the derivative of the objective function, the product of two variables, was calculated to find the point where this product is maximized.

• The derivative function is expressed as: \[ Z'(X) = \frac{(108-X)}{2} - \frac{X}{2} \]
• Setting this derivative to zero allows us to find the critical points where changes like maxima or minima occur.

By solving \(Z'(X) = 0\), we find that \(X = 54\), a critical step towards solving our optimization problem.

In optimization problems, the objective function is the function you want to maximize or minimize. It represents the goal that you are trying to achieve. For this exercise, the goal is to find the maximum product of two numbers with a given sum condition.
The objective function can be a formula designed based on the conditions given in the problem.
Here, it is represented by the product of the two numbers \(X\) and \(Y\), denoted as \(Z = XY\).
To adapt this to a single variable, a substitution is done using the relation derived from the sum condition:

• Substitute \(Y\) using \(Y = (108 - X) / 2\)
• Yields the objective function: \[ Z = X \times \frac{(108-X)}{2} \]

This step ensures the function is ready for calculating its derivative, a crucial strategy in finding its optimal value.

In calculus, a critical point of a function is where the derivative is zero or undefined. Identifying these points is essential to find potential maxima or minima of a function.
After setting the derivative of our objective function to zero, these critical points could lead to optimal values, essential in solving maximum or minimum problems in calculus.
In our exercise, the critical point is determined from the equation:

• Solve \(Z'(X) = 0\)
• Leads to \(X = 54\) as a critical point

To confirm whether this critical point results in a maximum or minimum, further analysis such as the second derivative test or considering the function's shape helps.
Here, the function's downward-opening parabola indicates \(X = 54\) provides a maximum product. Substituting back, we find \(Y = 27\), finalizing the optimal solution.