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We have the following formula for area $$A = r^2(\sinθ\cosθ-\sqrt{3}\sin(θ)^2)$$

We then need to find what value θ will give maximum area, so we differentiate to get; $$ \frac{\mathrm{d}A}{\mathrm{d}θ} = r^2((-\sinθ)^2+(\cosθ^2-2\sqrt{3}\sin(θ)^2\cosθ) $$

but how do I simplify this to find the turning points and hence the maximum value of θ?

So far I have simplified to;

$$ 1(1+2\sqrt{\cos\theta}) = (\tan\theta)^2 $$

but I'm not sure if this is the right way to go about it as I have no idea where to go from here.

Thanks in advance!

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Is the original formula $A = r^2(\sin(\theta)\cos(\theta)-\sqrt{3}(\sin(\theta))^2)$? I can't tell where the square is... –  msteve Jul 20 at 2:53
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You have I think the wrong derivative. With the right one it will be not hard, you will need to solve a quadratic in $\tan\theta$. –  André Nicolas Jul 20 at 2:56
    
That's very like me, would you be able to point out where I might have gone wrong? I differentiated as a product for the first bit (the sinθ.cosθ), was that incorrect? –  user157020 Jul 20 at 2:59
    
It is the differentiation of the last term, which is basically $\sin^2\theta$. The derivative of that is $2\sin\theta\cos\theta$, either by product rule or chain rule. –  André Nicolas Jul 20 at 3:12
    
Because it is not typeset in LaTeX, I can't tell whether you differentiated the first part right. The derivative of $\sin\theta\cos\theta$ is $-\sin^2\theta+\cos^2\theta$. Product rule. Now I see you got it wrong, sign error. –  André Nicolas Jul 20 at 3:14

3 Answers 3

I do not know whether you set up the problem correctly, so cannot guarantee that $r$ is constant, and that therefore we must maximize $\sin\theta\cos\theta-\sqrt{3}\sin^2\theta$. An exact description of the actual problem would be useful.

But let us maximize. The derivative of $\sin\theta\cos\theta$, by the Product Rule, is $(\sin\theta)(-\sin\theta)+(\cos\theta)(\cos\theta)$, which is $\cos^2\theta-\sin^2\theta$.

The derivative of $\sin^2\theta$, by the Product Rule or Chain Rule, is $2\sin\theta\cos\theta$.

So our derivative is $\cos^2\theta-\sin^2\theta-2\sqrt{3}\sin\theta\cos\theta$.

Set this equal to $0$. We cannot have $\cos\theta=0$, so we can divide by $\cos^2\theta$, and change signs, obtaining $\tan^2\theta+2\sqrt{3}\tan\theta-1=0$. This is a quadratic equation in $\tan\theta$. The Quadratic Formula works out very nicely here, as does completing the square.

Remark: Because of uncertainty about the placement of brackets, I do not know whether you mean $\sqrt{3}\sin^2\theta$ or $\sqrt{3}\sin(\theta^2)$. If it is the latter, then the derivative of the last part would be $2\sqrt{3}\theta\cos(\theta^2)$, and we would obtain an equation that is hopeless to solve exactly.

There is in the post a problem with the treatment of parentheses. That probably is a large part of the reason for the difficulties in computation.

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Hint

Continuing in the same spirit as André Nicolas, assuming that $$A=r^2 \Big(\sin (t) \cos (t)-\sqrt{3} \sin ^2(t) \Big)$$ deriving what is inside the brackets with respect to $t$ gives $$A'=r^2 \Big(-\sin ^2(t)+\cos ^2(t)-2 \sqrt{3} \sin (t) \cos (t) \Big)$$ which can be rearranged as $$A'=r^2 \Big(\cos (2 t)-\sqrt{3} \sin (2 t) \Big)$$ So, the maximum of A corresponds to the solution of $$\tan(2t)=\frac{1}{\sqrt{3}}$$ that is to say $$2t=\frac{\pi}{6}+2k\pi$$ At this point, in order to simplify further calculations, it looks better to use $\sin (t) \cos (t)=\frac{1}{2}\sin(2t)$ and $\sin ^2(t)=\frac{1}{2} (1-\cos (2 t))$ which allow to rewrite $$A=r^2 \Big(\frac{1}{2} \sin (2 t)+\frac{1}{2} \sqrt{3} \cos (2 t)-\frac{\sqrt{3}}{2}\Big)$$ what could have been done from the very beginning.

Further simplifications could be done but I let them to you.

I am sure that you can take from here.

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A little Trigonometric manipulation at the start will ease of the calculation

$$\frac{2A}{r^2}=2\sin\theta\cos\theta-\sqrt3(2\sin^2\theta)=\sin2\theta-\sqrt3(1-\cos2\theta)$$

$$\implies\frac{2A}{r^2}=\sin2\theta+\sqrt3\cos2\theta-\sqrt3$$

$$\frac2{r^2}\frac{d(A)}{d\theta}=2\cos2\theta-2\sqrt3\sin\theta$$

I think the rest should not be too difficult to deal with

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@user157020, How about this? –  lab bhattacharjee Jul 20 at 5:02

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