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Of all the rectangles with a given area, the one with smallest perimeter is a square. The following is a proposed proof for the given statement. 1. Let the rectangle have sides x and y and area A, so A = xy or y = . 2. The problem is to maximize the perimeter; P(x) = 2x + 2y = 2x+ 24. 2A 3. Now P'(x) = 2 -- 2(x? - A) %3D 4. So the critical number is x = A. 5. Since P'(x) < 0 for 0 < x< VĀ and P'(x) > 0 for x > VĀ, there is an absolute minimum at x = VA. A 6. The sides of the rectangle are v A and = vA, so the rectangle is a square. Identify the error(s) in the proposed proof. (Select all that apply.) O The first sentence assumes A = xy when this is not the case. O The second sentence should say minimize instead of maximize. O In the third sentence, the derivative P'(x) is incorrect: it should be P'(x) = 2 - 24 = 2(x - A) O In the fourth sentence, the critical number is incorrect; it should be x = VA instead of x = A. O The fifth sentence should say absolute maximum instead of absolute minimum. A A O The sixth sentence has an incorrect calculation; it should say E= A instead of %3D 5) Consider the following statement. Of all the rectangles with a given perimeter, the one with greatest area is a square. The following is a proposed proof for the given statement. 1. Let p be the perimeter and x and y the lengths of the sides, so p = 2x + 2y = 2y = p - 2x = y=ip - x. 2. The area is A(x) = x 3. Now setting A'(x) = 0 2x 4. Since A (x) = -2 < 0, there is an absolute minimum for A when = =p by the second derivative test. 5. The sides of the rectangle are p and Ip -Ip =p. so the

Of all the rectangles with a given area, the one with smallest perimeter is a square. The following is a proposed proof for the given statement. 1. Let the rectangle have sides x and y and area A, so A = xy or y = . 2. The problem is to maximize the perimeter; P(x) = 2x + 2y = 2x+ 24. 2A 3. Now P'(x) = 2 -- 2(x? - A) %3D 4. So the critical number is x = A. 5. Since P'(x) < 0 for 0 < x< VĀ and P'(x) > 0 for x > VĀ, there is an absolute minimum at x = VA. A 6. The sides of the rectangle are v A and = vA, so the rectangle is a square. Identify the error(s) in the proposed proof. (Select all that apply.) O The first sentence assumes A = xy when this is not the case. O The second sentence should say minimize instead of maximize. O In the third sentence, the derivative P'(x) is incorrect: it should be P'(x) = 2 - 24 = 2(x - A) O In the fourth sentence, the critical number is incorrect; it should be x = VA instead of x = A. O The fifth sentence should say absolute maximum instead of absolute minimum. A A O The sixth sentence has an incorrect calculation; it should say E= A instead of %3D 5) Consider the following statement. Of all the rectangles with a given perimeter, the one with greatest area is a square. The following is a proposed proof for the given statement. 1. Let p be the perimeter and x and y the lengths of the sides, so p = 2x + 2y = 2y = p - 2x = y=ip - x. 2. The area is A(x) = x 3. Now setting A'(x) = 0 2x 4. Since A (x) = -2 < 0, there is an absolute minimum for A when = =p by the second derivative test. 5. The sides of the rectangle are p and Ip -Ip =p. so the

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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The image contains two mathematical problems and their proposed proofs regarding rectangles and squares. (a) Problem Statement: Of all the rectangles with a given area, the one with the smallest perimeter is a square. Proposed Proof: 1. Let the rectangle have sides \( x \) and \( y \) and area \( A \), so \( A = xy \) or \( y = \frac{A}{x} \). 2. The problem is to maximize the perimeter; \( P(x) = 2x + 2y = 2x + \frac{2A}{x} \). 3. Now \( P'(x) = 2 - \frac{2A}{x^2} \). 4. So the critical number is \( x = \sqrt{A} \). 5. Since \( P'(x) < 0 \) for \( 0 < x < \sqrt{A} \) and \( P'(x) > 0 \) for \( x > \sqrt{A} \), there is an absolute minimum at \( x = \sqrt{A} \). 6. The sides of the rectangle are \( \sqrt{A} \) and \( \frac{A}{\sqrt{A}} = \sqrt{A} \), so the rectangle is a square. Errors in the Proposed Proof: - The first sentence assumes \( A = xy \) when this is not the case. - The second sentence should say "minimize" instead of "maximize". - The third sentence's derivative \( P'(x) \) is incorrect. - The critical number in the fourth sentence should be \( x = \sqrt{A} \) instead of \( x = A \). - The fifth sentence should say "absolute minimum" instead of "absolute minimum". - The sixth sentence has an incorrect calculation; it should say \( \frac{A}{\sqrt{A}} = A \) instead of \( \frac{A}{\sqrt{A}} = \sqrt{A} \). (b) Problem Statement: Of all the rectangles with a given perimeter, the one with the greatest area is a square. Proposed Proof: 1. Let \( p \) be the perimeter and \( x \) and \( y \) the lengths of the sides, so \( p = 2x + 2y \) leading to \( 2y = p - 2x \) resulting
Transcribed Image Text:The image contains two mathematical problems and their proposed proofs regarding rectangles and squares. (a) Problem Statement: Of all the rectangles with a given area, the one with the smallest perimeter is a square. Proposed Proof: 1. Let the rectangle have sides \( x \) and \( y \) and area \( A \), so \( A = xy \) or \( y = \frac{A}{x} \). 2. The problem is to maximize the perimeter; \( P(x) = 2x + 2y = 2x + \frac{2A}{x} \). 3. Now \( P'(x) = 2 - \frac{2A}{x^2} \). 4. So the critical number is \( x = \sqrt{A} \). 5. Since \( P'(x) < 0 \) for \( 0 < x < \sqrt{A} \) and \( P'(x) > 0 \) for \( x > \sqrt{A} \), there is an absolute minimum at \( x = \sqrt{A} \). 6. The sides of the rectangle are \( \sqrt{A} \) and \( \frac{A}{\sqrt{A}} = \sqrt{A} \), so the rectangle is a square. Errors in the Proposed Proof: - The first sentence assumes \( A = xy \) when this is not the case. - The second sentence should say "minimize" instead of "maximize". - The third sentence's derivative \( P'(x) \) is incorrect. - The critical number in the fourth sentence should be \( x = \sqrt{A} \) instead of \( x = A \). - The fifth sentence should say "absolute minimum" instead of "absolute minimum". - The sixth sentence has an incorrect calculation; it should say \( \frac{A}{\sqrt{A}} = A \) instead of \( \frac{A}{\sqrt{A}} = \sqrt{A} \). (b) Problem Statement: Of all the rectangles with a given perimeter, the one with the greatest area is a square. Proposed Proof: 1. Let \( p \) be the perimeter and \( x \) and \( y \) the lengths of the sides, so \( p = 2x + 2y \) leading to \( 2y = p - 2x \) resulting
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