MHB Light Intensity: A & B, Distance D, Minimal Point?

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The intensity of illumination at any point is determined by the intensities of light sources A and B and inversely proportional to the square of the distance from these sources. To find the point along the line connecting A and B where lighting intensity is minimal, one must consider the combined effects of both sources. The relationship can be expressed using the formula I = ksenf/d², with d² = r² + h². A substitution of B = kA can simplify the calculations, where k is a real number between 0 and 1. This approach generalizes previous light source problems to identify the minimal intensity point effectively.
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the intensity of iturninacion at any point is proportional to Ia intensity of light source and varies inversely as the square of the distance from the source. if two sources, A and B intensities, are at a distance D, at what point on the line that joins them, thee intensity of
Lighting will be minimal?. (Note: Supongase that the intensity at any point is the sum of the intensities due to both bulbs)

The same thinking I = ksenf/d2
and sin (h/d)
and d2 = r2+h2
 
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This problem is just a generalization of your other light source problem. I would work this problem first, and then use the formula you derive to answer the other question. I would probably use the substitution:

$$B=kA$$ where $0<k\in\mathbb{R}$
 
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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