Explanation Given: Number of woods for the frame of the kites is 6. Formula used: Pythagorean Theorem: If c denotes the length of the hypotenuse and a and b denote the length of the other two sides of a right angle triangle, then the Pythagorean theorem can be expressed as the Pythagorean equation: a 2 + b 2 = c 2 Calculation Consider the figure In Figure 1, In triangle A B C , A C D By Pythagoras theorem, B C = a 2 − x 2 = C D In triangle B C E , C D E By Pythagoras theorem, C E = a 2 − x 2 x 2 + b 2 − a 2 Area of triangle A B C = 1 2 (base)(height) The area is A = Area of triangle A B C + area of triangle A C D + area of triangle B C E + area of triangle C D E A = 1 2 ( 2 a 2 − x 2 ) x + 1 2 ( 2 a 2 − x 2 ) x 2 + b 2 − a 2 = a 2 − x 2 ( x + x 2 + b 2 − a 2 ) Differentiate with respect to x , d A d x = a 2 − x 2 ( 1 + x x 2 + b 2 − a 2 ) + ( x + x 2 + b 2 − a 2 ) ( − x a 2 − x 2 ) For critical points, d A d x = 0 a 2 − x 2 ( 1 + x x 2 + b 2 − a 2 ) + ( x + x 2 + b 2 − a 2 ) ( − x a 2 − x 2 ) = 0 ( x + x 2 + b 2 − a 2 ) ( x a 2 − x 2 ) = a 2 − x 2 ( x 2 + b 2 − a 2 + x x 2 + b 2 − a 2 ) x a 2 − x 2 = a 2 − x 2 x 2 + b 2 − a 2 Simplify that, x ( x 2 + b 2 − a 2 ) = a 2 − x 2 x 2 ( x 2 + b 2 − a 2 ) = a 4 + x 4 − 2 a 2 x 2 x 4 + b 2 x 2 − a 2 x 2 = a 4 + x 4 − 2 a 2 x 2 b 2 x 2 + a 2 x 2 − a 4 = 0 Then¸ x 2 ( a 2 + b 2 ) = a 4 x = a 2 a 2 + b 2 Differentiate d A d x with respect to x d 2 A d x 2 = { − x a 2 − x 2 + x 2 + b 2 − a 2 ( a 2 − x 2 + − x 2 a 2 − x 2 ) − x a 2 − x 2 x x 2 + b 2 − a 2 x 2 + b 2 − a 2 + ( 1 + x x 2 + b 2 − a 2 ) ( − x a 2 − x 2 ) + ( x + x 2 + b 2 − a 2 ) ( − a 2 − x 2 + x ( − x a

Essential Calculus: Early Transcendentals; MAC 2311 Sequence| MAC 2281Sequence USF (Essential Calculus)

2nd Edition

ISBN: 9781285101552

Author: James Stewart

Publisher: Cenage

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Chapter 4.5, Problem 50E

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To find: The length of the diagonal pieces.

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Chapter 4.5, Problem 49E

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Find all values x = a where the function is discontinuous. For each value of x, give the limit of the function as x approaches a. Be sure to note when the limit doesn't exist. 8+x f(x) = x(x-1) (Use a comma to separate answers as needed.) OA. The function f is discontinuous at the single value x = OB. The function f is discontinuous at the single value x = OC. The function f is discontinuous at the two values x = OD. The function f is discontinuous at the two values x = not oo or -0. OE. The function f is discontinuous at the two values x = The limit is The limit does not exist and is not oo or - co. The limits for both values do not exist and are not co or - co. The limit for the smaller value is The limit for the larger value does not exist and is The limit for the smaller value does not exist and is not co or - co. The limit for the larger

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Chapter 4 Solutions

Essential Calculus: Early Transcendentals; MAC 2311 Sequence| MAC 2281Sequence USF (Essential Calculus)

Ch. 4.1 - Explain the difference between an absolute minimum...Ch. 4.1 - Suppose f is a continuous function defined on a...Ch. 4.1 - For each of the numbers a, b, c, d, r, and s,...Ch. 4.1 - For each of the numbers a, b, c, d, r, and s,...Ch. 4.1 - Use the graph to state the absolute and local...Ch. 4.1 - Use the graph to state the absolute and local...Ch. 4.1 - Sketch the graph of a function f that is...Ch. 4.1 - 710 Sketch the graph of a function f that is...Ch. 4.1 - 710 Sketch the graph of a function f that is...Ch. 4.1 - 710 Sketch the graph of a function f that is...

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The length of the diagonal pieces.

Solution Summary: The author explains how the Pythagorean theorem can be expressed by the following formula: c denotes the length of the hypotenuse and a and

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To find: The length of the diagonal pieces.

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Chapter 4 Solutions