[stubear]

need help finding the maximum value for y

y = (5.0) sin 120x + (7.0) sin 240x.

[just george]

I'm a bit confused, is the equation ?
and do you mean you want the maximum x value or maximum y?

[stubear]

that's the equation, and got to find maximum y

[just george]

Well you know that the gradient of the curve is . So where you have a stationary point. You can then substitute the x value of the stationary points into to determine the nature of the stationary points. If then the stationary point is a minimum. If the stationary point is a maximum.

Does that help?

[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

[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?

[stubear]

forgot about that

I'm stuck again

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

sorry if I'm being thick

[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

[dugdugdug]

(Original post by just george)
Well you know that the gradient of the curve is . So where you have a stationary point. You can then substitute the x value of the stationary points into to determine the nature of the stationary points. If then the stationary point is a minimum. If 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.

[stubear]

(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...

[just george]

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

[dugdugdug]

No probs. If my memory serves me well, it's a point of inflexion. I miss this stuff, :-)

[stubear]

solved it apparently the solution is ymax = 7

is that obvious from the question?