Optimization of Mathematical Functions: Finding Maximum Values and Derivatives

In this article, we delve into the process of finding the maximum values of mathematical functions, focusing on specific functions and their optimizations. We explore the principles of function analysis and how to use derivatives to find critical points and determine the maximum value of a function. We also discuss the relevance of these techniques in the context of real-world applications.

Introduction to Function Optimization

Mathematical optimization plays a crucial role in many fields, from engineering to economics. By understanding how to find the maximum value of a function, one can effectively solve problems and make informed decisions. This article will guide you through the process of optimizing a specific function, f(x), by finding its maximum value.

Function Analysis and Optimization

Consider the function defined in the following equations:

f(x) a - b

where

a 8x - x^2

b x^2 - 14x 48 (x - 6)(x - 8)

To find the maximum value, we need to determine the intervals where the function is defined and analyze the behavior of the function within those intervals.

Function f(x) Defined Over [6,8]

The function is defined over the interval [6,8] because the first quadratic expression a 8x - x^2 is non-negative for x ∈ [0,8] and the second expression b x^2 - 14x 48 is non-negative for x ∈ [6,8]. We can rewrite f(x) as:

f(x) sqrt{8 - x}(sqrt{x} - sqrt{x - 6})

Here, both f_1(x) 6sqrt{8 - x} and f_2(x) frac{1}{sqrt{x}sqrt{x - 6}} are decreasing positive functions. Therefore, f(x) is a decreasing function over the interval [6,8].

The maximum value of f(x) is at x 6, where:

f(6) sqrt{12}

The minimum value of f(x) is at x 8, where:

f(8) 0

Alternative Interpretation: Difference of Semi-Circles

The function can be interpreted as the difference between two semi-circular functions, specifically:

y sqrt{8x - x^2} is a concave down semi-circle with center at (4,0) and radius 4.

y sqrt{14x - x^2 - 48} is a concave down semi-circle with center at (7,0) and radius 1.

The difference between these two semi-circles is strictly decreasing, and hence the maximum difference occurs at x 6, which is:

sqrt{12}

Using Derivatives to Find Maximum Value

To confirm the maximum value, we can also use derivatives. The function can be rewritten as:

f(x) sqrt{4 - x - 4^2} - sqrt{1 - x - 7^2}

The domain of the function is:

Df [6,8]

The derivative of the function is:

f'(x) -frac{x - 4}{sqrt{16 - x - 4^2}} cdot frac{x - 7}{sqrt{1 - x - 7^2}}

The derivative is valid only for x ∈ [6,8]. Solving for the critical points:

frac{16 - x - 4^2}{x - 4^2} frac{1 - x - 7^2}{x - 7^2}

(4x - 28 pm (x - 4)Rightarrow x in {frac{32}{5}, 8}

Since frac{32}{5} otin [7,8], the derivative has no zero in the interval. Therefore, the derivative retains its sign on the entire interval [6,8], being negative over the entire interval. Hence, the function strictly decreases and the maximum value is at x 6, where:

f(6) 2sqrt{3}

Conclusion

Understanding the principles of function optimization is essential in various fields. By applying these techniques, one can find the maximum values of functions, making informed decisions and solving complex problems. The process we have outlined in this article provides a clear and concise methodology for optimizing functions, which can be applied to a wide range of scenarios.