Extreme Value Problems: Volume and Function Analysis

This page covers two main topics in extreme value problems: volume calculations and analyzing points on a function.

Volume Calculation

The problem involves calculating the maximales Volumen einer Schachtel (maximum volume of a box) given certain constraints.

Vocabulary: Extremwertaufgaben - Extreme value problems

The volume formula is given as V = ab, where a and b are dimensions of the box. The surface area constraint is expressed as O₂ = a² + 4ab = 100.

Example: The calculation process involves:

1. Expressing b in terms of a: b = (100 - a²) / 4a
2. Substituting this into the volume formula: V(a) = a(100 - a²) / 4
3. Simplifying and differentiating to find the maximum

The solution process involves several steps of algebraic manipulation and calculus:

1. V'(a) = (100 - 3a²) / 4 is set to zero to find the critical point
2. Solving 100 - 3a² = 0 yields a² = 33.33
3. Taking the square root gives a ≈ 5.77

Highlight: The final result for the optimal dimension a is approximately 2.89

Points on a Function

The second part of the page deals with analyzing points on a quadratic function.

Definition: The function is defined as f(a) = -a² + 25

Three points are considered: A(0,y), B(a,0), and C(a,y).

The problem involves finding the coordinates of point P, which appears to be the highest point of the parabola.

Vocabulary: HP - Highest Point (Höchstpunkt)

The coordinates of P are calculated:

• x-coordinate: 2.89
• y-coordinate: -2.89² + 25 ≈ 16.65

Highlight: The highest point P has coordinates approximately (2.89, 16.65)

This page demonstrates the application of Extremwertaufgaben techniques to both practical (volume optimization) and theoretical (function analysis) problems, showcasing the versatility of these mathematical methods.