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Thread: Optimization Question

  1. #1
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    Optimization Question

    Hi
    I need help on the following question:
    1) A symmetrical gutter is made of metal sheeting 30cm. wide by bending it twice as shown in the cross section. Find an expression for the area $\displaystyle A(\theta)$ of the cross section and hence find the value of $\displaystyle \theta$ for the gutter to have maximum carrying capacity.



    This is what i have try to do so far.
    I know that Area of a square is L*W and Area of 2 triangles is $\displaystyle 2(0.5*x^2 * sin(\theta))$

    so $\displaystyle L*W + x^2 * sin(\theta) = 30$

    Make $\displaystyle W = \frac{30 - x^2(sin(\theta))}{10}$

    $\displaystyle A = L*W$

    $\displaystyle A = 10(\frac{30 - x^2(sin(\theta))}{10})$

    $\displaystyle A = 30 - x^2(sin(\theta))$

    $\displaystyle A' = -x^2(\frac{dA}{d\theta})cos(\theta) + 2xsin(\theta)$

    $\displaystyle A' = \frac{2xsin(\theta)}{-x^2(cos(\theta))}$

    Don't think this is correct.
    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I need help on the following question:
    1) A symmetrical gutter is made of metal sheeting 30cm. wide by bending it twice as shown in the cross section. Find an expression for the area $\displaystyle A(\theta)$ of the cross section and hence find the value of $\displaystyle \theta$ for the gutter to have maximum carrying capacity.


    area of a trapezoid ...

    $\displaystyle A = \frac{h}{2}(b_1+b_2)$

    $\displaystyle h = 10\cos{\theta}$

    $\displaystyle b_1 = 10$

    $\displaystyle b_2 = 10 + 20\sin{\theta}$

    try again.
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  3. #3
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I need help on the following question:
    1) A symmetrical gutter is made of metal sheeting 30cm. wide by bending it twice as shown in the cross section. Find an expression for the area $\displaystyle A(\theta)$ of the cross section and hence find the value of $\displaystyle \theta$ for the gutter to have maximum carrying capacity.



    This is what i have try to do so far.
    I know that Area of a square is L*W and Area of 2 triangles is $\displaystyle 2(0.5*x^2 * sin(\theta))$
    This makes no sense because you don't say what "x" means. Your area should depend on the single variable $\displaystyle \theta$, not both $\displaystyle \theta$ and some undefined "x".

    What you can do is this: you know that the hypotenuse of each right triangle is 10 so, since "$\displaystyle sin(\theta)= \frac{opposite side}{hypotenuse}$", $\displaystyle opposite side= hypotenuse* sin(\theta)= 10sin(\theta)$ and "$\displaystyle cos(\theta)= \frac{near side}{hypotenuse}$", $\displaystyle near side= hypotenuse* cos(\theta)= 10 cos(\theta)$.

    Of course, the area of a right triangle is "(1/2) near side* opposite side" so $\displaystyle A= \frac{1}{2}(10 cos(\theta))(10 sin(\theta))= 50 sin(\theta)cos(\theta)$.

    The base of the central rectangle is 10 and the height is the "near side" of the two right triangles, $\displaystyle 10 cos(\theta)$ so its area is $\displaystyle 100 cos(\theta)$.

    The entire area is $\displaystyle 100 cos(\theta)+ 100 sin(\theta)cos(\theta)= 100 cos(\theta)(1+ sin(\theta))$.

    so $\displaystyle L*W + x^2 * sin(\theta) = 30$

    Make $\displaystyle W = \frac{30 - x^2(sin(\theta))}{10}$

    $\displaystyle A = L*W$

    $\displaystyle A = 10(\frac{30 - x^2(sin(\theta))}{10})$

    $\displaystyle A = 30 - x^2(sin(\theta))$

    $\displaystyle A' = -x^2(\frac{dA}{d\theta})cos(\theta) + 2xsin(\theta)$

    $\displaystyle A' = \frac{2xsin(\theta)}{-x^2(cos(\theta))}$

    Don't think this is correct.
    P.S
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  4. #4
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    so then to find the value of $\displaystyle \theta$ do i make the equation:
    $\displaystyle 100 cos(\theta)(1+sin(\theta)) = 30$
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  5. #5
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    Quote Originally Posted by Paymemoney View Post
    so then to find the value of $\displaystyle \theta$ do i make the equation:
    $\displaystyle 100 cos(\theta)(1+sin(\theta)) = 30$
    why are you setting the area = 30 ???

    you want to find the value of $\displaystyle \theta$ that maximizes the cross-sectional area of the gutter ... i.e. the trapezoid.

    find $\displaystyle \frac{dA}{d\theta}$ and maximize like you were taught.
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  6. #6
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    ok this is what i have done

    $\displaystyle \frac{d(100cos(\theta)(1+sin(\theta)) = 900)}{dx}$

    $\displaystyle 100cos(\theta)cos(\theta)+(1+sin(\theta))-100sin(\theta) = 0$

    $\displaystyle 100cos^2(\theta) -100sin^2(\theta)-100sin(\theta) = 0$

    $\displaystyle 1-2sin^2(\theta) - 100sin(\theta) =0$ this is where i get stuck, what should i do next?
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  7. #7
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    $\displaystyle A = 100\cos{\theta}(1 + \sin{\theta})$

    using the product rule ...

    $\displaystyle
    \frac{dA}{d\theta} = 100\cos{\theta}(\cos{\theta}) + (1+\sin{\theta})(-100\sin{\theta})
    $

    $\displaystyle
    \frac{dA}{d\theta} = 100\cos^2{\theta} - 100\sin{\theta} - 100\sin^2{\theta}
    $

    $\displaystyle
    100\cos^2{\theta} - 100\sin{\theta} - 100\sin^2{\theta} = 0
    $

    $\displaystyle
    (1-\sin^2{\theta}) - \sin{\theta} - \sin^2{\theta} = 0
    $

    $\displaystyle
    1 - \sin{\theta} - 2\sin^2{\theta} = 0
    $

    $\displaystyle
    (1 - 2\sin{\theta})(1 + \sin{\theta}) = 0
    $

    finish it
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