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    I recently came across this question from the 2011 MAT:

    "A rectangle has perimeter P and area A. The values of P and A satisfy

    (a) P^3 > A
    (b) A^2 > 2P+1
    (c) P^2 > 16A (weak inequality)
    (d) PA > A+P (again, weak inequality) "

    I remember doing this question a few months ago and have rediscovered it, however it is frustrating me because this time I can't do it. Any help would be much appreciated. Thank you in advance.
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    (Original post by 10001)
    I recently came across this question from the 2011 MAT:

    "A rectangle has perimeter P and area A. The values of P and A satisfy

    (a) P^3 > A
    (b) A^2 > 2P+1
    (c) P^2 > 16A (weak inequality)
    (d) PA > A+P (again, weak inequality) "

    I remember doing this question a few months ago and have rediscovered it, however it is frustrating me because this time I can't do it. Any help would be much appreciated. Thank you in advance.
    Could easily be missing something, but to me it seems that when you work through it, the first one is correct (without checking the rest...)

    If you say the sides are x and y, then P = ... and A = ... hence ...
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    (Original post by 10001)
    I recently came across this question from the 2011 MAT:

    "A rectangle has perimeter P and area A. The values of P and A satisfy

    (a) P^3 > A
    (b) A^2 > 2P+1
    (c) P^2 > 16A (weak inequality)
    (d) PA > A+P (again, weak inequality) "

    I remember doing this question a few months ago and have rediscovered it, however it is frustrating me because this time I can't do it. Any help would be much appreciated. Thank you in advance.
    since P can be arbitrarily large for small A we can discount b and d

    consider c with x=y
    equality occurs

    consider c with y=x+1 (we can consider +- any number and it holds for all numbers since x and y have any dimensions we like, and we label y as the larger)

    lhs
    (4x+2)^2

    rhs
    16x(x+1)

    lhs 16x^2+16x+4

    rhs
    16x^2+16x

    therefore lhs is larger and x holds

    Therefore c holds - implying the others do not
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    (Original post by MathMoFarah)
    since P can be arbitrarily large for small A we can discount b and d

    consider c with x=y
    equality occurs

    consider c with y=x+1 (we can consider +- any number and it holds for all numbers since x and y have any dimensions we like, and we label y as the larger)

    lhs
    (4x+2)^2

    rhs
    16x(x+1)

    lhs 16x^2+16x+4

    rhs
    16x^2+16x

    therefore lhs is larger and x holds

    Therefore c holds - implying the others do not
    Thanks a lot!

    I also have another way of solving it now:

    Let x and y represent the length of the sides of the triangle.

    P=2x+2y and A=xy, therefore y=A/x

    Hence P=2x+2(A/x), which rearranges to 2x^2 -Px +2A=0

    Since x and y are positive real numbers, the discriminant of the quadratic equation is greater than or equal to 0 (b^2 - 4ac> 0).

    Therefore P^2 - 16A>0, which again leads to (c) being the answer.
 
 
 
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