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Math Help - intermediate and extreme value theorms?

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    intermediate and extreme value theorms?

    I need help with starting the question below. I believe I have to use either intermediate-value or extreme-value theorems, however, I am not sure. The question is as follows:

    Fix positive number P. Let R denote the set of all rectangles with perimeter P. Prove that there is a member of R that has maximum area. What are the dimensions of the rectangle of maximum area? HINT: Express the area of an arbitrary element of R as a function of length of one of the sides.



    No idea where to begin. I understand it wants the max area and the sides of it. However, I don't know what to do.
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    Re: intermediate and extreme value theorms?

    Quote Originally Posted by Barthayn View Post
    I need help with starting the question below. I believe I have to use either intermediate-value or extreme-value theorems, however, I am not sure. The question is as follows:

    Fix positive number P. Let R denote the set of all rectangles with perimeter P. Prove that there is a member of R that has maximum area. What are the dimensions of the rectangle of maximum area? HINT: Express the area of an arbitrary element of R as a function of length of one of the sides.



    No idea where to begin. I understand it wants the max area and the sides of it. However, I don't know what to do.
    P is a constant ...

    P = 2L + 2W

    W = \frac{P-2L}{2}

    A = LW

    A = L\left(\frac{P-2L}{2}\right) =

    note that A is quadratic in L ... you should find the max area occurs when the rectangle is a square.
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    Re: intermediate and extreme value theorms?

    I still don't understand where to go from here. I know to understand where to get the lengths of the sides of the rectangle I should take A(L) = A'(L). However, I do not see how this is proof and where this will lead to.
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    Re: intermediate and extreme value theorms?

    it's a optimization problem ... find the value of L in terms of P that maximizes the area of the rectangle.
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    Re: intermediate and extreme value theorms?

    I understand that. This is what I do:

    A = LW
    P = 2L+2W
    A(L) = 1/2LP - L^2
    A'(L) = 1/2P - 2L
    A'(L) = 1/2P - 2L = 0
    A'(L) => 1/4P

    From here, now knowing that L = 1/4 P I do not know where to go. I get 1/4P if I sub it into P = 2L+2W and zero if I put it into A(L) = 1/2LP - L^2
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    Re: intermediate and extreme value theorms?

    Quote Originally Posted by Barthayn View Post
    I understand that. This is what I do:

    A = LW
    P = 2L+2W
    A(L) = 1/2LP - L^2
    A'(L) = 1/2P - 2L
    A'(L) = 1/2P - 2L = 0
    A'(L) => 1/4P

    From here, now knowing that L = 1/4 P I do not know where to go. I get 1/4P if I sub it into P = 2L+2W and zero if I put it into A(L) = 1/2LP - L^2
    if the length is P/4, what does that say about the lengths of the other 3 sides?
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    Re: intermediate and extreme value theorms?

    it says that there are all 1/4 P. !!!! It is a perfect square!!! Therefore it has maximum area. However, how do I prove that this claim is true? (my university always wants people to prove it mathematically, otherwise you mention a proven theorem that was taught in class).
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    Re: intermediate and extreme value theorms?

    would the first or second derivative test for extrema suffice?
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    Re: intermediate and extreme value theorms?

    not sure. I would say the first derivative because it tells you when something is at a maximum (in this case the area). How would I should it to be that case though?
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    Re: intermediate and extreme value theorms?

    Quote Originally Posted by Barthayn View Post
    not sure. I would say the first derivative because it tells you when something is at a maximum (in this case the area). How would I should it to be that case though?
    show A'(L) > 0 for L < P/4 and A'(L) < 0 for L > P/4

    second derivative test may be easier ... all you would have to show is that A''(P/4) < 0
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