Maxima and Minima Problem No 13 - Application of Derivatives - Diploma Maths - II | Summary and Q&A

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April 11, 2022
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Maxima and Minima Problem No 13 - Application of Derivatives - Diploma Maths - II

TL;DR

Find the dimensions of an open box with a square base to maximize its volume.

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Questions & Answers

Q: What is the objective of the problem?

The objective is to find the dimensions of an open box's square base to maximize its volume.

Q: How is the area of the box calculated?

The area of the box consists of the five faces: the base (X^2) and four sides (X*Y), where Y is the height. The total area should be equal to 192 square centimeters.

Q: How is the volume of the box calculated?

The volume is calculated by multiplying the length, breadth, and height of the box together.

Q: How is the equation for volume manipulated to find a maximum?

The equation for volume is differentiated with respect to X, and then the derivative is set equal to zero to find critical points.

Q: How is the maximum volume determined?

The second-order derivative is calculated by differentiating the derivative with respect to X again. If the second derivative is greater than zero, it indicates a maximum volume.

Q: What are the dimensions of the box that maximize the volume?

The dimensions of the box that maximize the volume are 8 centimeters for length and breadth, and 4 centimeters for height.

Summary & Key Takeaways

  • A box with a square base needs to be made with an open top, and the area of the material for making the box is given as 192 square centimeters.

  • The goal is to determine the dimensions of the box that will maximize its volume.

  • By considering the side length of the square base as X, the dimensions of the box are found to be 8 centimeters for length and breadth, and 4 centimeters for height.

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