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maximum of this equation?

need help finding the maximum value for y

y = (5.0) sin 120x + (7.0) sin 240x.
(edited 12 years ago)
Reply 1
Avatar for just george
just george
I'm a bit confused, is the equation y=5sin(120x)+7sin(240x) y=5sin(120x)+7sin(240x) ?
and do you mean you want the maximum x value or maximum y?
Reply 2
Avatar for stubear
stubear
OP
11
that's the equation, and got to find maximum y
Reply 3
Avatar for just george
just george
Well you know that the gradient of the curve is dydx \frac{dy}{dx}. So where dydx=0\frac{dy}{dx}=0 you have a stationary point. You can then substitute the x value of the stationary points into d2ydx2 \frac{d^2y}{dx^2} to determine the nature of the stationary points. If d2ydx2>0 \frac{d^2y}{dx^2}>0 then the stationary point is a minimum. If d2ydx2<0 \frac{d^2y}{dx^2}<0 the stationary point is a maximum. :smile:

Does that help?
Reply 4
Avatar for stubear
stubear
OP
11
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
Reply 5
Avatar for just george
just george
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?
Reply 6
Avatar for stubear
stubear
OP
11
forgot about that

I'm stuck again

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

sorry if I'm being thick
Reply 7
Avatar for just george
just george
hehe no it's not a problem just gotta keep practising and you'l get there :smile:

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 :smile:
Reply 8
Avatar for dugdugdug
dugdugdug
14
Original post by just george
Well you know that the gradient of the curve is dydx \frac{dy}{dx}. So where dydx=0\frac{dy}{dx}=0 you have a stationary point. You can then substitute the x value of the stationary points into d2ydx2 \frac{d^2y}{dx^2} to determine the nature of the stationary points. If d2ydx2>0 \frac{d^2y}{dx^2}>0 then the stationary point is a minimum. If d2ydx2<0 \frac{d^2y}{dx^2}<0 the stationary point is a maximum. :smile:

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.
Reply 9
Avatar for stubear
stubear
OP
11
Original post by just george
hehe no it's not a problem just gotta keep practising and you'l get there :smile:

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 :smile:


thanks mate I thought trig equations din;t work like normal quadratics...
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 :smile: sorry not a great explaination i know :frown:
No probs. If my memory serves me well, it's a point of inflexion. I miss this stuff, :-)
(edited 12 years ago)
Reply 12
Avatar for stubear
stubear
OP
11
solved it apparently the solution is ymax = 7

is that obvious from the question?

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