HW6 Solutions

# HW6 Solutions - ASSIGNMENT 6 SOLUTIONS Problem 1 Here is a...

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Unformatted text preview: ASSIGNMENT 6 - SOLUTIONS Problem 1. Here is a guesstimate. In both parts, we are supposed to find a distinguished rectangle among all rectangles inscribed in a given circle. For reasons of symmetry, the only distinguished rectangle that comes to mind is a square. Heuristically, we expect a square to be the optimal shape for both a) and b). Now let’s check using calculus. Call x and y the dimensions of the rectangle. Since both diagonals are diameters, we have the Pythagorean relation x 2 + y 2 = (2 r ) 2 = 4 r 2 , hence y = √ 4 r 2- x 2 . a) We have to maximize the perimeter P = 2 x + 2 y . After substituting y , we get to the function P ( x ) = 2( x + ð 4 r 2- x 2 ) with feasible domain (0 , 2 r ). The derivative is P Í ( x ) = 2 1- x √ 4 r 2- x 2 ; we find the critical numbers from P Í ( x ) = 0 or P Í ( x ) DNE. The first gives x = r √ 2; the second has no solution within the feasible domain. To check that the critical number x = r √ 2 is indeed a point where P ( x ) achieves its absolute maximum, we look at the sign of P Í ( x ). Between 0 and r √ 2, P Í ( x ) is positive hence P ( x ) is increasing. Between r √ 2 and r , P Í ( x ) is negative hence P ( x ) is decreasing. We conclude that P ( r √ 2) = 4 r √ 2 is the largest possible perimeter, and it is realized by the square with side-length r √ 2....
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## This note was uploaded on 07/31/2011 for the course MATH 102 taught by Professor Maryamnamazi during the Spring '10 term at University of Victoria.

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HW6 Solutions - ASSIGNMENT 6 SOLUTIONS Problem 1 Here is a...

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