Introduction to Function Optimization with Calculus

In calculus, optimizing a function often involves finding its maximum or minimum values under given constraints. This process is particularly useful in various fields such as economics, engineering, and data analysis. In this article, we will explore the optimization of the function f(x) x^a (1 - x^b), where x, a, and b are positive real numbers, using calculus techniques.

Finding the Maximum Value of f(x)

The function f(x) x^a (1 - x^b) is a composite function that benefits from the application of calculus for finding its maximum value over the interval [0, 1]. We will follow a systematic approach to solve this problem:

Step 1: Find the Derivative

We start by taking the natural logarithm of f(x) to simplify the differentiation process:

ln(f(x)) ln(x^a (1 - x^b)) a ln(x) ln(1 - x^b)

Next, we differentiate ln(f(x)) with respect to x:

df(x)/dx d(ln(f(x)))/dx a/x - b/(1 - x)

Step 2: Solve for Critical Points

We set the derivative equal to zero to find the critical points:

a/x - b/(1 - x) 0

Rearranging the equation gives:

a b/x - b/(1 - x)

Cross-multiplying leads to:

a(1 - x) bx

This simplifies to:

a - ax bx

Rearranging further gives:

a a bx

Step 3: Solve for x

Solving the equation for x:

x a/(a b)

Step 4: Evaluate the Function at the Critical Points

We now evaluate f(x) at x a/(a b):

f(a/(a b)) (a/(a b))^a (1 - a/(a b))^b

This simplifies to:

f(a/(a b)) (a/(a b))^a (b/(a b))^b

Conclusion

The maximum value of f(x) x^a (1 - x^b) on the interval [0, 1] occurs at x a/(a b) and is given by:

max f(x) (a/(a b))^a (b/(a b))^b

This value is positive since a and b are positive real numbers. Therefore, the maximum value of f(x) is boxed{(a/(a b))^a (b/(a b))^b}.

Using Weighted AM-GM for Further Insight

We can also use the Arithmetic Mean-Geometric Mean Inequality (AM-GM) to verify the result. By weighted AM-GM on bx with weight a and a(1 - x) with weight b, we get:

(bx)^a (a(1 - x))^b)^{1/ab} ≤ (a(bx) b(a(1 - x)))/(ab)

Cleaning this up a bit, we find:

((bx)^a (a(1 - x))^b)^{1/ab} ≤ (ab)/(ab)

This implies:

(x^a (1 - x)^b ≤ (1/(a^b b^a)) (a^b b^a)^b)/(a^b b^a)

The maximum value is attained when bx a(1 - x). The intermediate value theorem confirms that this is the maximum value on the interval [0, 1].