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 4.5.40.a
Textbook Question
Folded boxes
a. Squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring 5 ft by 8 ft. The resulting piece of cardboard is then folded into a box without a lid. Find the volume of the largest box that can be formed in this way.
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1
Define the dimensions of the box after cutting out the squares: the length will be (8 - 2x) ft, the width will be (5 - 2x) ft, and the height will be x ft.
Write the volume V of the box as a function of x: V(x) = length * width * height = (8 - 2x)(5 - 2x)(x).
Expand the volume function V(x) to express it in a standard polynomial form.
Determine the domain of x, which is constrained by the dimensions of the cardboard: 0 < x < 2.5 ft.
Find the critical points of V(x) by taking the derivative V'(x), setting it to zero, and solving for x to identify the maximum volume.
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