Rectangular Area Optimization
This page covers optimization problems related to maximizing rectangular areas and minimizing surface area for rectangular prisms. Two main examples are explored using calculus techniques.
Example 1: Maximizing Rectangular Area
A rectangular area needs to be fenced with 280m of fencing. The goal is to find the dimensions that maximize the enclosed area.
Definition: Optimization involves finding the maximum or minimum value of a function, often using calculus techniques like derivatives.
The area function is derived as Ax = 140x - x², where x is the length of the rectangle.
Example: To find the critical points, set A'x = 0:
140 - 2x = 0
x = 70
The maximum area occurs when x = 70m, resulting in a square enclosure.
Highlight: The greatest area enclosed is 4900m² with dimensions 70m by 70m.
Example 2: Minimizing Surface Area of Rectangular Prism
This problem involves minimizing the surface area of an open rectangular bin with a square base and a fixed volume of 108m³.
Vocabulary: Surface area S is the total area of all surfaces of a three-dimensional object.
The surface area function is derived as Sx = x² + 432/x, where x is the length of the base side.
Example: To find the critical points, set S'x = 0:
2x - 432/x² = 0
x³ = 216
x = 6
The minimum surface area occurs when x = 6m.
Highlight: The minimum surface area is 108m² when the length and width are 6m and the height is 3m.
These examples demonstrate the practical application of maximum area of a rectangle formula and maximum area calculator concepts in real-world scenarios, showcasing how calculus can be used to solve fencing optimization for rectangular area problems efficiently.