Maximizing Angle Theta on a Line AB: Calculus Assignment Help

[menal]

Homework Statement

http://i1183.photobucket.com/albums/x476/Laura_Petrea/4-7-69.gif

"Where should the point P be chosenon the line AB so as to maximize the angle theta"
how would i do this question?

 

[Undoubtedly0]

Greetings! Is your answer supposed to be in terms of the length of AB?

 

[menal]

yes! sorry i forgot to mention that up there =)

 

[Undoubtedly0]

In that case, try extending a horizontal line from point P in order to split up θ into two angles, θ₁ and θ₂, such that θ = θ₁ + θ₂. Call x an unknown length starting at point P and going upwards along AB. Check out the bottom-most angle on the right. Call it φ, and note that θ₁ (the lower part of θ) is

\theta_1 = \varphi = \arctan(\frac{3+x}{5})

because φ and θ₁ are alternate interior angles. Try deriving a similar expression for θ₂, keeping in mind that this part will depend on AB.

 

[menal]

alright so will this work?: θ = θ1 + θ2
tan θ1=BP/BC (where C= the angle between B and P)
θ1=arctan(BP/BC)
tan θ2=AP/AD (where D= the angle between A and P)
θ2=arctan(AP/AD)

sum= θ1 + θ2 --> arctan(BP/BC)+arctan(AP/AD)

then add in the known values:

θ1 + θ2= arctan(BP/2)+arctan(AP/5)

θ= arctan((1/2)BP)+arctan((1/5)AP)

since we know AP+BP=3...solve for BP.
BP=-AP+3 and then substitute into the equation above:

θ= arctan((1/2)(-AP+3))+arctan((1/5)AP)
θ= -arctan((1/2AP-3/2))+arctan((1/5)AP)...then found the derivative

y= -1/(2(1+(1/2)AP-(3/2))^2) + 1/(5(1+(1/25)AP)^2)

i then solved for AP

and got 5+2sqrt(5)=9.472
and 5-2sqrt(5)=.5279
and my answer was .5279

is that correct?

 

[Undoubtedly0]

since we know AP+BP=3...solve for BP.

Was it given in the problem that AP+BP = 3? Just looking at the image, it seems that we only know that AP is currently length 3.

 

[menal]

yes, but in my image on my sheet, AP AND BP total is 3.

 

[menal]

so based on that, would what i did be correct?

 

[Undoubtedly0]

yes, but in my image on my sheet, AP AND BP total is 3.

If this is true, then your result seems fine to me. Good work!