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Math Help - 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 A(\theta) of the cross section and hence find the value of \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 2(0.5*x^2  * sin(\theta))

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

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

    A = L*W

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

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

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

    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 A(\theta) of the cross section and hence find the value of \theta for the gutter to have maximum carrying capacity.


    area of a trapezoid ...

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

    h = 10\cos{\theta}

    b_1 = 10

    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 A(\theta) of the cross section and hence find the value of \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 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 \theta, not both \theta and some undefined "x".

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

    Of course, the area of a right triangle is "(1/2) near side* opposite side" so 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, 10 cos(\theta) so its area is 100 cos(\theta).

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

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

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

    A = L*W

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

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

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

    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 \theta do i make the equation:
    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 \theta do i make the equation:
    100 cos(\theta)(1+sin(\theta)) = 30
    why are you setting the area = 30 ???

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

    find \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

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

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

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

    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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    A = 100\cos{\theta}(1 + \sin{\theta})

    using the product rule ...

     <br />
\frac{dA}{d\theta} = 100\cos{\theta}(\cos{\theta}) + (1+\sin{\theta})(-100\sin{\theta})<br />

     <br />
\frac{dA}{d\theta} = 100\cos^2{\theta} - 100\sin{\theta} - 100\sin^2{\theta}<br />

     <br />
100\cos^2{\theta} - 100\sin{\theta} - 100\sin^2{\theta} = 0<br />

     <br />
(1-\sin^2{\theta}) - \sin{\theta} - \sin^2{\theta} = 0<br />

     <br />
1 - \sin{\theta} - 2\sin^2{\theta} = 0<br />

     <br />
(1 - 2\sin{\theta})(1 + \sin{\theta}) = 0<br />

    finish it
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