347HomeworkSolutionsSP05

# 347HomeworkSolutionsSP05 - Math 347 Honors Spring 2009...

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Unformatted text preview: Math 347 Honors, Spring 2009, Problem Solutions HW# 1 1.4 The square has the largest area among all rectangles with a given perimeter. The problem may be rewritten as follows: “ Given p > 0, prove that among all rectangles with perimeter p , the square of side p/ 4 is the one with the largest area.” Proof (without calculus): Let p be a positive real number, and let R be a rectangle with perimeter p . If R has side-lengths x,y , then x + y = p/ 2 and the area of R is xy = x ( p/ 2- x ). So the problem is to determine what x gives the maximum of x ( p/ 2- x ). This can be done in an elementary way since x ( p/ 2- x ) has a parabolic shape, or with calculus, or with the AGM inequality (see 1.28 (i)). Either way, the maximum occurs at x = p/ 4. This corresponds to y = p/ 4, in which case R is a square. Alternatively, if a rectangle has perimeter p , then the sides have lengths p/ 4- y and p/ 4 + y for some y satisfying 0 ≤ y ≤ p/ 4. The area is then ( p/ 4 + y )( p/ 4- y ) = p 2 / 16- y 2 ≤ p 2 / 16, with equality only if y = 0, i.e., the rectangle is a square. 1.7 Correction of “If x and y are nonzero real numbers and x > y , then (- 1 /x ) > (- 1 /y ) .” Analysis: The statement is false when y is negative and x is positive, since- 1 /x is negative and- 1 /y is positive, so (- 1 /x ) < (- 1 /y ). Corrected Statement: “If x and y are nonzero real numbers, x > y and x and y have the same sign, then (- 1 /x ) > (- 1 /y ).” Proof: If x > y and x,y have the same signs, then dividing by (- xy ), which is a negative number, gives (- 1 /y ) < (- 1 /x ). It is also correct to modify the statement to “If x and y are nonzero real numbers and x > y > 0, then (- 1 /x ) > (- 1 /y ).” 1.20 Roots of a quadratic. If r and s are the roots of the quadratic ax 2 + bx + c = 0, then ax 2 + bx + c = a ( x- r )( x- s ) (i.e., the polynomials are the same). Expanding the right side gives ax 2 + bx + c = ax 2- a ( r + s ) x + ars . Therefore r + s =- b/a and rs = c/a . 1.27 Determine the set of real solutions to | x/ ( x + 1) | ≤ 1 . The solutions set is S = { x ∈ R : x ≥ - 1 2 } . Let T = { x ∈ R : | x/ ( x +1) | ≤ 1 } . If x ∈ T , multiplying by | x +1 | then squaring gives x 2 ≤ ( x +1) 2 = x 2 +2 x +1. This is equivalent to x ≥ - 1 2 , and it follows that x ∈ S . This proves T ⊆ S . Conversely, if x ∈ S , then 2 x + 1 ≥ 0. Adding x 2 to both sides gives ( x + 1) 2 ≥ x 2 . Now for any y ∈ R , p y 2 = | y | . So taking square roots in ( x + 1) 2 ≥ x 2 gives | x + 1 | ≥ | x | . Since x 6 =- 1, we can divide by | x + 1 | , giving 1 ≥ | x/ ( x + 1) | , thus x ∈ T . Therefore S ⊆ T and we conclude that S = T . 1.28 Applications of the AGM inequality . (i) Let c and x be real numbers. By the AGM inequality, x ( c- x ) ≤ x + c- x 2 2 = c 2 / 4 ....
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## This note was uploaded on 01/27/2010 for the course MATH 347 taught by Professor ? during the Spring '09 term at University of Illinois at Urbana–Champaign.

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347HomeworkSolutionsSP05 - Math 347 Honors Spring 2009...

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