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maximum of this equation?

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    need help finding the maximum value for y

    y = (5.0) sin 120x + (7.0) sin 240x.
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    I'm a bit confused, is the equation  y=5sin(120x)+7sin(240x) ?
    and do you mean you want the maximum x value or maximum y?
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    that's the equation, and got to find maximum y
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    Well you know that the gradient of the curve is  \frac{dy}{dx}. So where \frac{dy}{dx}=0 you have a stationary point. You can then substitute the x value of the stationary points into  \frac{d^2y}{dx^2} to determine the nature of the stationary points. If  \frac{d^2y}{dx^2}>0 then the stationary point is a minimum. If  \frac{d^2y}{dx^2}<0 the stationary point is a maximum.

    Does that help?
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    I did that and got know where you still have the equation

    600cos 120x + 1680 cos 240x = 0

    which I have no idea how to solve
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    (Original post by stubear)
    I did that and got know where you still have the equation

    600cos 120x + 1680 cos 240x = 0

    which I have no idea how to solve
    Assuming up to that point is correct (sorry i havent done the workings myself yet) you could say let t=120x

    then you have 600cos(t) + 1680cos(2t) = 0

    could you then solve that by using the double angle formulae?
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    forgot about that

    I'm stuck again

    600cost + 1680cos^2(t)-1680sin^2(t) = 0

    sorry if I'm being thick
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    hehe no it's not a problem just gotta keep practising and you'l get there

    600cos(t) + 1680cos(2t) = 0

    should use: cos(2t) = 2cos^2(t) - 1

    so 600cos(t) + 3360cos^2(t) - 1680 = 0

    then you have a quadratic, so you can substitute the values of a b and c (ax^2+bx+c) into the quadratic formula to find the value for cos(t). You should then be able to solve to find 't', remembering t=120x, so you can then find what x equals
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    (Original post by just george)
    Well you know that the gradient of the curve is  \frac{dy}{dx}. So where \frac{dy}{dx}=0 you have a stationary point. You can then substitute the x value of the stationary points into  \frac{d^2y}{dx^2} to determine the nature of the stationary points. If  \frac{d^2y}{dx^2}>0 then the stationary point is a minimum. If  \frac{d^2y}{dx^2}<0 the stationary point is a maximum.

    Does that help?
    Just like to make a quick point re d2y/dx2.

    What you say is correct for finding if it's a max or min but what if d2y/dx2 = 0? From memory, sometimes it's quicker to find points either side to determine max / min.

    But do correct me if I've forgotten my calculus.
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    (Original post by just george)
    hehe no it's not a problem just gotta keep practising and you'l get there

    600cos(t) + 1680cos(2t) = 0

    should use: cos(2t) = 2cos^2(t) - 1

    so 600cos(t) + 3360cos^2(t) - 1680 = 0

    then you have a quadratic, so you can substitute the values of a b and c (ax^2+bx+c) into the quadratic formula to find the value for cos(t). You should then be able to solve to find 't', remembering t=120x, so you can then find what x equals
    thanks mate I thought trig equations din;t work like normal quadratics...
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    I cant remember what its called when its 0 :L if you imagine how y=tanx cuts through 0, if a graph does that shape but actually goes to a gradient of 0 in the middle, then d^2y/dx^2=0 sorry not a great explaination i know
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    No probs. If my memory serves me well, it's a point of inflexion. I miss this stuff, :-)
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    solved it apparently the solution is ymax = 7

    is that obvious from the question?

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Updated: March 23, 2012
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