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Cylinder and cones (Putnam Exam 1938) Right circular cones

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 40E Chapter 4.4

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 40E

Cylinder and cones (Putnam Exam 1938) Right circular cones of he ? ight? and r? dius? are attached to each end of a right circular cylinder of h? eight ?h and?radius? , forming a double-pointed object. For a given surfa?ce area ?A,what are the dim?ension? s r? and h? that maximize the volume of the object?

Step-by-Step Solution:

Solution 40E Step 1: Consider that the right circular cones of height hand radius rare attached to each end of a right circular cylinder of height h and radius r, forming a double pointed object. Consider the following figure, Step 2 Consider that the volume of the cylinder V crwith radius r and height h is given by the following formula 2 V cr= r h And the volume of the cone V c with radius and height is given by the following formula Therefore, the volume of the double-pointed object becomes V = V cr + V c V c (from above figure ) We have to find the dimensions r and h of given surface area that minimizes the volume of the object consider that that the surface area of the cylinder A=2rh And surface area of the cone A=rl The area of the double-pointed object is Therefore, Here,l is the slant height of the cone and is given by the pythagoras theorem So, the value of the surface area becomes Square on both the sides using the algebraic identity

Step 3 of 4

Chapter 4.4, Problem 40E is Solved
Step 4 of 4

Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

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Cylinder and cones (Putnam Exam 1938) Right circular cones