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I need to maximize the function $$f(x,\theta) =x\sin\theta(xcos\theta + w - 2x)$$ which defines the area enclosed by a folded plate that forms a canal, where $w$ is the length of the plate, $x$ is the length of each folded piece, $w - 2x$ is the length of the part that isn't folded and $\theta$ is the angle at which the plate is folded.

So I already found the partial derivatives of the function, which are $$ f_{x} = \sin\theta(2x\cos\theta +w- 4x) $$ $$ f_{\theta} = x[x\cos(2\theta) + \cos\theta(w-2x)] $$

And I have to solve the system $$ \sin\theta(2x\cos\theta +w- 4x) = 0 $$ $$ [x\cos(2\theta) + \cos\theta(w-2x)] = 0 $$ but I have no idea how to start. The only solution I could find was $x = 0$ and $\sin\theta = 0$, but this solution is obviously useless because then it wouldn't be a canal but a flat unfolded plate.

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Are there any bounds on $x$ or $\theta$? –  Antonio Vargas May 1 '12 at 22:08
    
I think the bounds here would be $0 < \theta < \pi/2$ and $x < w/2$. But I just realized there were bounds for this function. –  user1002327 May 1 '12 at 22:22
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2 Answers

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Since you reject $\sin\theta=0$, your first equation becomes $$2x\cos\theta+w-4x=0$$ which you can solve for $x$ in terms of $\theta$, $$x={w\over2\cos\theta-4}$$ Now put that expression for $x$ into your second equation, multiply through by the denominator, and replace $\cos2\theta$ with $2\cos^2\theta-1$, and you will have a quadratic equation for $\cos\theta$. Can you take it from there?

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Maybe you can use the Weierstrass substitution method. Then you can transform the variables with $$\cos(\theta)=\frac{1-t^2}{1+t^2}$$ and $$\sin(\theta)=\frac{2 t}{1+t^2}$$ to make your equations (with some simplifications) $$ w (t^2+1)-2 x (3 t^2+1) = 0 $$ $$ x (3 t^4-6 t^2-1)-w (t^2+1) (t^2-1) = 0 $$ which is solved in term of $x$ and $t^2$ as

The second equation yeilds $ x = \frac{w (t^4-1)} { 3 t^4-6 t^2-1 } $ which is inserted into the first one as

$$ w \frac{ (t^2+1)^2 (1-3 t^2)}{3 t^4-6 t^2-1}=0 $$ where from numerator you get the solutions $t^2=-1$ and $t^2=\frac{1}{3} $. You an guess which $t$ to use, and how to get back to $\theta$ from $t$.

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Never heard of that substitution, thanks for that. –  user1002327 May 2 '12 at 1:55
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@user1002327: It is often call the tangent half angle method, and it is often used in robotics where lots of trig functions are used. –  ja72 May 2 '12 at 12:32
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