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Optimization problems

  1. Nov 18, 2009 #1
    1. The problem statement, all variables and given/known data
    a) Show that of all the rectangles with a given area, the one with the smallest perimeter is a square.
    b) Show that of all the rectangles with a given perimeter, the one with the greatest are is a square.


    2. Relevant equations
    As=x2
    AR=xy
    Ps = 4x
    PR= 2x+2y

    3. The attempt at a solution

    I don't know how to do this any help?
     
  2. jcsd
  3. Nov 19, 2009 #2

    CompuChip

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    Suppose that you have a rectangle with sides x and y.
    So for a), the area A is fixed, which allows you to express y in terms of A and x.
    Now the equation for the perimeter can be written in terms of x only, this will give you an equation which you can optimise for x.

    The same is true for b, if you assume a fixed perimeter p. Perhaps b is even easier to start with, because you get the quadratic equation immediately.

    Hint: what does the graph of f(x) = ax2 + bx + c look like and what is the x value of the maximum or minimum?

    (That you will get the smallest perimeter in a but the largest area in b can also be seen from the formulas, by the way: when is the function value at the x that I asked you about in the hint a maximum? And when is it a minimum?).
     
  4. Nov 21, 2009 #3
    for the rectangle

    P=2x+2y
    A=xy

    y=A/x

    p=2x+2A/x

    take the derivative

    p'=2+ (A'-xA)/x^2

    solve for zero

    2x^2 - Ax + A' = 0

    x = [A +- sqrt(A^2 -8A')/4]

    y= A / [A +- sqrt(A^2 -8A')/4]

    P = 2[A +- sqrt(A^2 -8A')/4]+ A/[A +- sqrt(A^2 -8A')/4]


    for square

    P=4x
    A=x^2

    x= sqrt(A)

    P=4sqrt(A)
     
  5. Nov 22, 2009 #4

    CompuChip

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    Note that the area of the rectangle is fixed as it is given. So A' = ... ?
    This will simplify your equation greatly.

    By the way, note that for a function
    f(x) = a x2 + b x + c
    the extremum (minimum or maximum, depending on the sign of a) is always at
    x = - b / 2a
    (following from f'(x) = 2ax + b = 0, or the fact that the extremum lies exactly between the two roots x = - b / 2a + sqrt(D) / 2a and x = - b / 2a - sqrt(D) / 2a of f(x)).

    For the second one, note that the rectangular perimeter 2x + 2y is fixed to the value p. You don't know a priori that x = y, but what can you say about the area A = xy ?
     
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