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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 4, Problema 4.5.38.a

Rectangles beneath a line


a. A rectangle is constructed with one side on the positive x-axis, one side on the positive y-axis, and the vertex opposite the origin on the line y = 10 - 2x. What dimensions maximize the area of the rectangle? What is the maximum area?

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Identify the variables: Let the coordinates of the vertex opposite the origin be (x, y) on the line y = 10 - 2x. The dimensions of the rectangle are x (width) and y (height).
Express the area A of the rectangle as a function of x: Since y = 10 - 2x, the area A can be expressed as A(x) = x * y = x * (10 - 2x).
Simplify the area function: A(x) = 10x - 2x^2. This is a quadratic function in the form of A(x) = -2x^2 + 10x.
Find the critical points: To maximize the area, take the derivative of A(x) with respect to x, set it to zero, and solve for x. The derivative is A'(x) = 10 - 4x.
Solve for x: Set A'(x) = 0, which gives 10 - 4x = 0. Solve for x to find the critical point. Then, use the second derivative test or analyze the function to confirm that this point gives a maximum area.

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Area of a Rectangle

The area of a rectangle is calculated by multiplying its length by its width. In this problem, the dimensions of the rectangle are determined by the coordinates of the vertex on the line y = 10 - 2x. Understanding how to express the area as a function of x is crucial for maximizing it.
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Estimating the Area Under a Curve Using Left Endpoints

Optimization

Optimization in calculus involves finding the maximum or minimum values of a function. To maximize the area of the rectangle, we need to set up the area function based on the dimensions derived from the line equation and then use techniques such as taking derivatives and finding critical points to determine the maximum area.
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Intro to Applied Optimization: Maximizing Area

Derivatives and Critical Points

Derivatives represent the rate of change of a function and are essential for finding critical points, where the function's slope is zero or undefined. In this context, taking the derivative of the area function allows us to identify the x-value that maximizes the area, leading to the optimal dimensions of the rectangle.
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Critical Points
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