Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- 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 4.5.61
Textbook Question
Metal rain gutters A rain gutter is made from sheets of metal 9 in wide. The gutters have a 3-in base and two 3-in sides, folded up at an angle Θ (see figure). What angle Θ maximizes the cross-sectional area of the gutter? <IMAGE>
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1
Define the cross-sectional area A of the gutter as a function of the angle Θ. The area can be expressed as A = base * height, where the base is 3 inches and the height can be determined using trigonometric functions based on the angle Θ.
Express the height of the gutter in terms of the angle Θ. Since the sides are folded up at an angle Θ, the height can be represented as height = 3 * sin(Θ).
Substitute the expression for height into the area function to get A(Θ) = 3 * (3 * sin(Θ)) = 9 * sin(Θ).
To find the angle Θ that maximizes the area, take the derivative of A with respect to Θ, A'(Θ), and set it equal to zero to find critical points.
Analyze the critical points and the second derivative test to determine which angle Θ provides the maximum area.
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