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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 sidelengths 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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 Spring '09
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 Math, Angles, Natural number, Prime number, Problem Solutions HW

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