MHB Not sure how to even start this question. Maximizing theta

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To find the maximum angle θ formed by the car C and the recording lights A and B, the distance from the intersection to the car, labeled as x, is crucial. The relationship between the angles and distances can be expressed using trigonometric identities, specifically involving tangent functions. By setting up the equations for the angles ACO and BCO, and applying differentiation, the optimal distance can be calculated. The solution reveals that the car is located at a distance of 100√6 meters from the intersection when θ is maximized. This approach emphasizes the importance of breaking down the problem into manageable parts and applying calculus effectively.
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Two highways intersect at a right angle. Two recording lights are located at points A and B, 200 meters and 300 meters away from the intersection on one of the highway. A car, C, approaching the intersection along the other highway, is being tracked by the two lights.Let θ be the angle ACB(see figure below). Find how far from the intersection the car is when the angle θ is a maximum?
 

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Corey9078 said:
Two highways intersect at a right angle. Two recording lights are located at points A and B, 200 meters and 300 meters away from the intersection on one of the highway. A car, C, approaching the intersection along the other highway, is being tracked by the two lights.Let θ be the angle ACB(see figure below). Find how far from the intersection the car is when the angle θ is a maximum?
You may wish to start with five things:

Label the distance C-Intersection 'x'
Label the angle A-C-Intersection $\phi$

The lower triangle.

$\tan(\phi) = \dfrac{200}{x}$

The larger triangle.

$\tan(\phi+\theta) = \dfrac{300}{x}$

Trigonometry

$\tan(\phi+\theta) = \dfrac{\tan(\phi) + \tan(\theta)}{1-\tan(\phi)\tan(\theta)}$

You should have done this immediately - even if it doesn't lead anywhere. Don't let the easy parts get away from you because you have some fear about the entire solution. WRITE DOWN what you do know. Worry about what you don't know, later.
 
Alternatively,

Let the point of intersection of the two highways be $O$. Then

$$\angle{BCO}-\angle{ACO}=\theta$$

$$\arctan(BO/CO)-\arctan(AO/CO)=\theta$$

Using a well-known trigonometric identity (found here), we have

$$\theta=\arctan\frac{BO/CO-AO/CO}{1+AO\cdot BO/CO^2}$$

Now use values for $AO$ and $BO$ per the given information and you've got a function in terms of $CO$. Differentiate, set equal to 0 and solve for $CO$ for the final answer. (It's not as bad as it looks - you only need to set the numerator equal to $0$ and solve for CO - I get $100\sqrt{6}$).
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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