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Math Help - Optimization Problem

  1. #1
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    Optimization Problem

    a pentagon with a perimeter of 30 m is to be constructed by adjoining an equilateral triangle to a rectangle. find the dimension of the rectangle and triangle that will maximize the area of the pentagon.

    the image given is a triangle on top of a rectangle with x beside both sides of the triangle and y beside one of the widths of the rectangle. I'm not sure if i should include the line between the two shapes with the triangle or the rectangle. and whether i should add the bottom of the rectangle to the triangle since it is the same size as the triangle sides.

    would it be A(x,y)=(sqrt(3)/4)x^2 + y^2
    3x+4y=10 then solve for y which is y=(10-3x)/4

    then where do i go from there?
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  2. #2
    Member mfetch22's Avatar
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    Quote Originally Posted by Devonmcinerney View Post
    a pentagon with a perimeter of 30 m is to be constructed by adjoining an equilateral triangle to a rectangle. find the dimension of the rectangle and triangle that will maximize the area of the pentagon.

    the image given is a triangle on top of a rectangle with x beside both sides of the triangle and y beside one of the widths of the rectangle. I'm not sure if i should include the line between the two shapes with the triangle or the rectangle. and whether i should add the bottom of the rectangle to the triangle since it is the same size as the triangle sides.

    would it be A(x,y)=(sqrt(3)/4)x^2 + y^2
    3x+4y=10 then solve for y which is y=(10-3x)/4

    then where do i go from there?
    Let me explain a bit. There is one important stipulation that the problem made, it said that the triangle is an "Equilateral Triangle", which will be helpful in this problem; it will save us from some trigonometry. So, if x is the side of the triangle, then x is also the bottom of the triangle. The bottom of the triangle is consequently also the length of the rectangle, since they share a side. Does that make sense?

    So, we know to find the area of a triangle we need the hiegth of the triangle, which we can easily find with the side length. So, if we draw a vertical line down the triangle, and stop it at the top of the rectangle (call this length h), then we know we have the following relationship:

    (\frac{x}{2})^2 + h^2 = x^2

    \frac{x^2}{4} + h^2 = x^2

    h^2 = x^2 - \frac{x^2}{4}

    h^2 = x^2(1-\frac{1}{4})

    h^2 = x^2(\frac{3}{4})

    h = x\sqrt{\frac{3}{4}}

    h = x \frac{\sqrt{3}}{2}

    h = (x/2) (\sqrt{3})

    So, now we have the height and we need to use it to find the area of the sides, as you had shown in your formula for the area, we get something like:

    x = \mathrm{Equilateral \;\; Triangle's \;\; Side \;\; Length}

    y = \mathrm{Height \;\; of \;\; Rectangle}

    therefore:

    A(x, y) = xy + (1/2)(x)(h) = xy + (\frac{x}{2})(\frac{x}{2})( \sqrt{3} )<br />

    = xy + \frac{\sqrt{3} x^2}{4}

    So, theres the formula for the Area, now for the perimeter, which we can call P, and we know that P=30. We also know that:

    P = (x + 2y) + 2x = 3x+2y

    But since P = 30 we have:

    30 = 3x + 2y

    so:

    30-3x = 2y

    and:

    (30 - 3x)(1/2) = y

    thus:

    y = \frac{(3)(10-x)}{2}

    So, plugging this back into the other equation gives us:

    A(x) = (x)(\frac{(3)(10-x)}{2}) + \frac{\sqrt{3} x^2}{4}

    Simplifying gives:

    A(x) = \frac{(3)(10x-x^2)}{2} + (x^2)(\frac{\sqrt{3}}{4})

    and:

    A(x) = (10x-x^2)(\frac{3}{2}) + (x^2)(\frac{\sqrt{3}}{4})

    Now, we need to find the maximum area. So lets start by finding the derivitive:

    \frac{dy}{dx}[A(x)] = A'(x) = (3)(5-x) + (x)(\frac{\sqrt{3}}{2})

    The zero of this equation will tell us where the maximums and minimum of the functions are:

    0 = 15 - 3x + (x)(\frac{\sqrt{3}}{2})

    The solution of the above equation is:

    x=7.029137098...

    To prove that this is a maximum, we need to show that at this point, the second derivitive is negative, and the second derivitive is:

    \frac{dy}{dx}[\frac{dy}{dx}[A(x)]] = A''(x) = -3 + \frac{\sqrt{3}}{2}

    Thus:

    A''(7.029137098...)= - 2.1339745962156...

    The second derivitive is a constant function, so whatever the input may be for x, it will still be negitive. So we know that is a maxiumum. So, now that we know the proper value of x, lets see what the value of the maximum area is:

    Let:

    n = 7.029137098...

    And thus:

    A(n) = 74.113152510569...

    So:

    A(n) \approx 74.11

    And thats the answer. Any questions?
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  3. #3
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    that makes SO much sense. it really helped with my other questions too. thank you so much.
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