- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
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- Introduction to Trigonometric Functions38m
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- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals3h 25m
5. Graphical Applications of Derivatives
Applied Optimization
Problem 4.5.38.a
Textbook Question
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?

1
Identify the variables: Let the coordinates of the vertex opposite the origin be (x, y). The rectangle will have dimensions x (width) and y (height).
Express y in terms of x using the line equation: Since the vertex lies on the line y = 10 - 2x, we have y = 10 - 2x.
Write the area function: The area A of the rectangle is given by A = x * y. Substitute y = 10 - 2x into this equation to get A = x(10 - 2x) = 10x - 2x^2.
Find the critical points: To maximize the area, take the derivative of A with respect to x, set it to zero, and solve for x. The derivative is A'(x) = 10 - 4x. Set A'(x) = 0 to find the critical points: 10 - 4x = 0.
Solve for x and determine the maximum area: Solve 10 - 4x = 0 to find x = 2.5. Substitute x = 2.5 back into the area function A = 10x - 2x^2 to find the maximum area. Also, verify that this critical point is a maximum by using the second derivative test or analyzing the behavior of A'(x).
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