Solving Maximisation Problem with Sections a, b, c

[tomc612]

Hi,
Ive got a problem I need some help with

View attachment 6076

Ive got sections a, b, c
a) d = \sqrt{{x}^{2}+{R}^{2}}
b) cos\theta = x/\sqrt{{x}^{2}+{R}^{2}}
c) I = 1/{x}^{2}+{R}^{2}
d) x>0

My question is how do i differentiate for I from answer in c) and find 0 and then maximise from there

 

[MarkFL]

Hello and welcome to MHB, tomc612! (Wave)

a) By Pythagoras, we find:

\(\displaystyle d=\sqrt{x^2+r^2}\quad\checkmark\)

b) Using the definition of cosine as adjacent over hypotenuse, we find:

\(\displaystyle \cos(\theta)=\frac{x}{d}=\frac{x}{\sqrt{x^2+r^2}}\quad\checkmark\)

c) Using the given information regarding intensity $I$, we find:

\(\displaystyle I=k\frac{\cos(\theta)}{d^2}=k\frac{\dfrac{x}{\sqrt{x^2+r^2}}}{x^2+r^2}=k\frac{x}{(x^2+r^2)^{\frac{3}{2}}}\)

where $0<k$ is a constant of proportionality.

d) We must have $0\le x$.

e) Can you now use the quotient and chain rules to find \(\displaystyle \d{I}{x}\)?

 

[tomc612]

Hi Mark,
thanks for the help,
I see where I went wrong with the proportions as I only includes I= 1/d rather than I= 1/d. Cos\theta

For the differentiation of DI/Dx do we leave the R^2 in the equation or differentiate that in another equation as part of the chain rule.. as in DI/Dx = DR/DX x DI/DR

Thanks

 

[MarkFL]

We can treat $r$ as a constant, or a parameter of the problem. $x$ can vary as the light is raised or lowered, but the radius of the table will remain the same. :D