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2 Maximizing rectangle perimeters All rectangles with an

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

Solution for problem 46E Chapter 4.1

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 46E

2 Maximizing rectangle perimeters All rectangles with an area of 64 m have a perimeter given ? by P(? ) = 2 x + 128/x. ?wher? is the length of one. side of the rectangle. Find the absolute minimum value of the perimeter function. What are the Dimensions of the rectangle with minimum perimeter?

Step-by-Step Solution:
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STEP_BY_STEP SOLUTION Step-1 Let f be a continuous function defined on an open interval containing a 1 number ‘c’.The number ‘c’ is critical value ( or critical number ). If f (c) 1 = 0 or f (c) is undefined. A critical point on that graph of f has the form (c,f(c)). Step-2 When an output value of a function is a maximum or a minimum over the entire domain of the function, the value is called the absolute maximum or the absolute minimum. Let f be a func tion with domain D and let c be a fixed constant in D . Then the output value f( ) is the 1. Absolute maximum value of f on D if and only if f(x) f(c) , for all x in D. 2. Absolute minimum value of f on D if and only if f(c) f(x) , for all x in D. Step_3 Given that ; x is the length of one side of the rectangle. Area of the rectangle is 64 m 2 128 Perimeter of the given rectangle is P(x) = 2x+ x .Clearly P(x) is a rational function and it is continuous for all x ,except zero. But here x is length of a rectangle , so it takes only positive values .Hence the function is continuous on this particular domain. Now , we have to find out the absolute minimum value of the perimeter function .For that we have to evaluate the function at critical points.so, the critical points satisfies the equation . Given function is; P(x) = 2x+ 128 . For the critical points we have to differentiate the x function with respect to n .[ Since from the step-1] d d 128 dx (P(x)) = dx ( 2x+ x ). 1 d d 128 d d P (x) = dx( 2x) + dx( x ) ,since dx(u+v) = dx(u)+ = 2 d ( ) + 128 d ( ),s ince d (Cx) = C dx dx x dx = 2(1) +128( 2 ), since d (x ) = n x n1 , x dx = 2 - 1x8 . 1 128 Since , from the definition f (c) = 0 = 2 - c 2 - 128 = 0 c2 That is , 128 = 2 64 = c 2 c Therefore , c = -8 , 8. Hence ,x = 8 lies in the domain , since x is the length of a rectangle it should be positive number. Therefore , we can say that P(x) has minimum perimeter value at x = 8m. 128 So, the minimum value of the perimeter is ; P(8) = (2(8) + 8 )m = (16 + 16)m = (32)m. Therefore , the minimum value of the perimeter is 32m.

Step 2 of 3

Chapter 4.1, Problem 46E is Solved
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Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. The full step-by-step solution to problem: 46E from chapter: 4.1 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. This full solution covers the following key subjects: perimeter, rectangle, minimum, maximizing, function. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. Since the solution to 46E from 4.1 chapter was answered, more than 364 students have viewed the full step-by-step answer. The answer to “2 Maximizing rectangle perimeters All rectangles with an area of 64 m have a perimeter given ? by P(? ) = 2 x + 128/x. ?wher? is the length of one. side of the rectangle. Find the absolute minimum value of the perimeter function. What are the Dimensions of the rectangle with minimum perimeter?” is broken down into a number of easy to follow steps, and 54 words.

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