91Ó°ÊÓ

The product of two numbers, when their sum is constant, is an interesting concept in optimization. Suppose we have two positive numbers, \(a\) and \(b\), such that their sum equals 1, i.e., \(a + b = 1\). Our goal is to maximize the product \(ab\).

When you approach this, first express one variable in terms of the other. From the constraint, we have \(b = 1 - a\). The product \(ab\) then becomes \(a(1-a) = a - a^2\). This expression is a quadratic function, and finding its maximum value involves calculus techniques.

• Quadratic functions in the form of \(ax^2 + bx + c\) have maxima or minima, depending on the sign of \(a\).
• In our case, \(-a^2 + a\) is a downward opening parabola, indicating a maximum.

Thus, the maximum product is achieved, through calculus, when \(a = b = \frac{1}{2}\), giving a maximum product of \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\). This concept highlights the beauty of balancing terms to reach optimal conditions.

Proving inequalities often involves clever manipulation and substitution. Given our two numbers, \(a\) and \(b\), such that \(a + b = 1\), we must prove that \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right) \geq 9\).

Begin by substituting \(b = 1-a\). The expression becomes \((1 + \frac{1}{a})(1 + \frac{1}{1-a})\). Simplifying gives \(\left(\frac{a+1}{a}\right)\left(\frac{2-a}{1-a}\right)\). Our goal is to show this expression is always greater than or equal to 9 for all positive \(a\) satisfying the constraint.

• Substitute \(a = \frac{1}{2}\), which balances the constraint and maximizes product \(ab\).
• This substitution gives \(\left(\frac{3}{2}\right)^2 = 9\).

This demonstrates the inequality holds precisely at this critical point. Inequalities pose a fantastic challenge in mathematics, bridging algebraic manipulation and logical reasoning.

Differentiation is a powerful tool that helps us find maxima and minima of functions. In this context, we need to find the maximum product of \(ab = a - a^2\), given the constraint \(a + b = 1\).

Here, set \(f(a) = a - a^2\) as our function. Differentiate this with respect to \(a\), yielding \(f'(a) = 1 - 2a\). Locate the stationary point by setting the derivative equal to zero: \(1 - 2a = 0\), leading to \(a = \frac{1}{2}\).

• Evaluate \(f(a)\) at this stationary point to find the extremum value.
• This critical point defines where the maximum occurs when \(a = \frac{1}{2}\), thus \(b = \frac{1}{2}\) as well.

Differentiation provides a systematic way to assess changes in functions, especially for finding optimal solutions under constraints. Applying derivatives not only finds extremum points but also provides insight into the function's behavior.