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Math Help - Equilateral Triangle + Square

  1. #1
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    Equilateral Triangle + Square

    The sum of the perimeter of a equilateral triangle and square is 10. Find the dimensions of the triangle of and square that produce a minimum total area.

    My Work:

    Perimeter of square is 4x.
    Perimeter of Equilateral is 3y.
    Area of square is x^2.
    Area of Equilateral triangle is [y^2*sqrt(3)]/4.

    Total area is x^2 + [y^2*sqrt(3)/4].
    Total perimeter is 3y + 3x, where 3y + 4x = 10.

    I then solved 10 = 3y + 4x for y and got y = (10-4x)/3.

    Afterward, I plugged y into the total perimeter.

    Let T = total perimeter

    T = x^2 + (1/4)[(10-4x)/3]^2*sqrt(3).

    I then found T prime.

    T' = 2x + (2x-5)/3*sqrt(3)

    Let T' = 0 to find x.

    I found out that x = 5/[6*sqrt(3) + 2]

    I stopped here because it does not make sense to me.

    Can someone get me started in the right direction?
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  2. #2
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    Re: Equilateral Triangle + Square

    I believe you'll find that this extreme point is a minimum not a maximum. Thus your maximum lies at a boundary of your constraint equation.

    You should try plotting the Area and y as a function of x to get a better idea of what is going on.
    Thanks from nycmath
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  3. #3
    MHF Contributor ebaines's Avatar
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    Re: Equilateral Triangle + Square

    Quote Originally Posted by nycmath View Post
    .

    My Work:

    Perimeter of square is 4x.
    Perimeter of Equilateral is 3y.
    Area of square is x^2.
    Area of Equilateral triangle is [y^2*sqrt(3)]/4.

    Total area is x^2 + [y^2*sqrt(3)/4].
    Total perimeter is 3y + 3x, where 3y + 4x = 10.

    I then solved 10 = 3y + 4x for y and got y = (10-4x)/3.
    So far so good.

    Quote Originally Posted by nycmath View Post
    .
    Afterward, I plugged y into the total perimeter.

    Let T = total perimeter

    T = x^2 + (1/4)[(10-4x)/3]^2*sqrt(3).
    This is actually a formula for area, not perimeter. But that's actually what you want, so as long as we remember that T=area we can continue:

    Quote Originally Posted by nycmath View Post
    .
    I then found T prime.

    T' = 2x + (2x-5)/3*sqrt(3)
    This is not right. I get: T' = 2x + (4/9) sqrt(3)(2x-5)

    If you continue from here you'll find a value for x where T' is zero, But then you must check to see whether this gives a min or a max. Since T is a quadratic in x, you can see that the second derivative of T will be positive throughtout the domain, so the point we just calculated is a min, not a max. Hence as romsek noted - the max value for T will occur at the boundary of the domain, where either x=0 or y= 0.
    Thanks from nycmath
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    Re: Equilateral Triangle + Square

    Can you please find T" and I'll take it from there?
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  5. #5
    MHF Contributor ebaines's Avatar
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    Re: Equilateral Triangle + Square

    This is the second post in a day where you are asking for someone to give you the answer for a fairly simple derivative. I suggest you try it yourself, post your attempt, and we'll help if you need it.
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