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Math Help - Maximum area of a triangle

  1. #1
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    Maximum area of a triangle

    Given a right angle triangle with integral lengths for all 3 sides. Let a, b ,c be the sides of the triangle, and a<b<c and a+c=81. Find the maximum area of this triangle under these conditions.

    I can't think of a way to do this without using calculus. I got 47 as the length of b that will give the largest area.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by chengbin View Post
    I can't think of a way to do this without using calculus. I got 47 as the length of b that will give the largest area.
    Since c is the largest side, it is the hypotenuse.

    Therefore

    A=\frac{1}{2}b\cdot{a}

    You know that c=\sqrt{a^2+b^2} so a+\sqrt{a^2+b^2}=81Solving for a

    (\sqrt{a^2+b^2})^2=(81-a)^2
    a^2+b^2=81^2-162a+a^2

    b^2=81^2-162a

    a=\frac{81^2-b^2}{162}

    Now substitute and maximize
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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    Since c is the largest side, it is the hypotenuse.

    Therefore

    A=\frac{1}{2}b\cdot{a}

    You know that c=\sqrt{a^2+b^2} so a+\sqrt{a^2+b^2}=81Solving for a

    (\sqrt{a^2+b^2})^2=(81-a)^2
    a^2+b^2=81^2-162a+a^2

    b^2=81^2-162a

    a=\frac{81^2-b^2}{162}

    Now substitute and maximize
    That's exactly the way I did it, and I got 47. But this requires calculus. I need a way to do this problem without using calculus.

    BTW, can/would manipulating the values of a and c change the area?
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    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by chengbin View Post
    That's exactly the way I did it, and I got 47. But this requires calculus. I need a way to do this problem without using calculus.

    BTW, can/would manipulating the values of a and c change the area?
    I thought you said you wanted to do it WITH calculus. Sorry. I must have read your post wrong.
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    Quote Originally Posted by VonNemo19 View Post
    I thought you said you wanted to do it WITH calculus. Sorry. I must have read your post wrong.
    Any ideas?
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by chengbin View Post
    Any ideas?
    Intuitively I Know that a right triangle will have it's largest area when a=b. Maybe there's something there.....


    Maybe still do the substitution but only evaluate at the endpoints of the feasable domain. I'm sure that'll work.
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  7. #7
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    Quote Originally Posted by VonNemo19 View Post
    Maybe still do the substitution but only evaluate at the endpoints of the feasable domain. I'm sure that'll work.
    We don't know the endpoints of the domain.

    Anyway, this was on my friend's worksheet from this tutor center. I seriously doubt they taught calculus, that's why I have to exclude the use of calculus, even though that this solution is my first instinct when I look at that question.
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  8. #8
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    Hello, chengbin!

    Edit: I corrected my typos . . . Thanks, yeongil!

    This is more of a Discrete Math problem, not Calculus.
    I did some experimenting and came up with a solution, sort of.


    Given a right angle triangle with integral lengths for all 3 sides.
    Let a, b ,c be the sides of the triangle, and a<b<c and a+c=81.
    Find the maximum area of this triangle under these conditions.
    We want three positive integers a,b,c where: . \begin{array}{cc}{\color{blue}(a)} & a<b<c \\ {\color{blue}(b)} & a^2+b^2 \:=\:c^2 \\ {\color{blue}(c)} & a+c \:=\:81 \end{array}

    Here's the experimental part . . .

    From (b) we have: . b^2 \:=\:c^2 - a^2 \:=\:(c+a)(c-a)

    Sustituting (c), we have: . b^2 \:=\:81(c-a)
    . . and we see that (c-a) must be a square.

    Since a+c \:=\:81, it narrows down the choices.

    I found four solutions: . (a,b,c) \:=\:\begin{array}{cc}(1) & (40,9,41) \\ (2) & (36,27,45) \\ (3) & (28,45,53) \\ (4) & (16,63,65) \end{array}

    Solutions (1) and (2) violate condition (a),
    . . which leaves us with solutions (3) and (4).

    . . \begin{array}{|c|c|}\hline (a,b,c) & \text{Area} \\ \hline \\[-5mm]<br />
(28,45,53) & {\color{red}630} \\ (16,63,65) & 504 \\ \hline\end{array}



    Hope I didn't miss any . . .
    .
    Last edited by Soroban; July 16th 2009 at 06:07 AM. Reason: Typo ... pointed out by yeongil
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  9. #9
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    @Soroban

    How did you get those numerical values? Guess?

    I don't understand the part where you say a-c must be a square.
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  10. #10
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    Quote Originally Posted by chengbin View Post
    I don't understand the part where you say a-c must be a square.
    Earlier, Soroban said:
    From (b) we have: . a+c)(a-c)" alt="b^2 \:=\:a^2 - c^2 \:=\a+c)(a-c)" />

    Sustituting (c), we have: . b^2 \:=\:81(a-c)
    . . and we see that (a-c) must be a square.
    First, a minor correction: it should be
    c + a)(c - a)" alt="b^2 \:=\:{\color{red}c}^2 - {\color{red}a}^2 \:=\c + a)(c - a)" />
    and
    b^2 \:=\:81({\color{red}c - a}).

    81 times (c - a) is a perfect square (because it equals b^2). Since 81 is a perfect square, then (c - a) must also be a perfect square.


    01
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