Question

Help with this extrema problem?

Answer 1

`A(θ)=sinθ(1+cosθ)` , `θ=π/3`

Explanation:

Solution supposed the problem is about isosceles triangle.

`r=1` so `(OB)=(OC)=1`

From the right triangle `OhatBM`:

• `sinθ=(BM)/(OB)` `<=>` `(BM)=(OB)*sinθ=1*sinθ`

and `(BC)=2(BM)=2sinθ` ,

• `cosθ=(OM)/(OB)` `<=>` `(OM)=cosθ*(OB)=1*cosθ`

Thus,
`(AM)=(OA)+(OM)=1+cosθ`

so Area will be
`A_(rea)=A(θ)=1/2*(BC)*(AM)=1/2*2sinθ(1+cosθ)` `=`

`sinθ(1+cosθ)`

• For `θ` `in` `(0,π)`

`Α'(θ)=(sinθ(1+cosθ))'` `=`

`(sinθ)'(1+cosθ)+sinθ(1+cosθ)'` `=`

`cosθ(1+cosθ)+sinθ*(-sinθ)` `=`

`cosθ+cos^2θ-sin^2θ` `=`

`cos2θ+cosθ`

(used the trigonometric identity `cos^2x-sin^2x=cos2x`
/ proof can be found here https://socratic.org/questions/how-do-you-prove-cos2x-cos-2x-sin-2-using-other-trigonometric-identities)

`A'(θ)=0` `<=>`

`cos2θ+cosθ=0` `<=>`

`cos2θ=-cosθ` `<=>`

`cos2θ=cos(π-θ)` `<=>`
graph{cosx [-0.883, 3.984, -1.187, 1.246]}

`2θ=π-θ`
`θ` `in` `(0,π)`

so `θ=π/3`

• For `0<` `θ<π/3` for example `θ=π/6` we see that `A'(θ)>0`

as a result `A` is strictly increasing in `[0,π/3]`

• For `π/3<θ<π` for example `θ=π/2` we see that `A'(θ)<0`

as a result `A` is strictly decreasing in `[π/3,π]`

because `A'(π/6)=cos((2π)/6)+cos(π/6)=cos(π/3)+cos(π/6)`
`=1/2+sqrt3/2>0`

`A'(π/2)= cos((2π)/2)+cos(π/2)=cosπ+cos(π/2)=-1<0`

`A` is increasing at `[0,π/3]` and decreasing at `[π/3,π]`
and has a local maximum at `θ_0=π/3` with `A(π/3)=sin(π/3)(1+cos(π/3))=sqrt3/2(1+1/2)=(3sqrt3)/4`

• Therefore, we have `A_(rea)` maximum value when `θ=π/3` and the value is `(3sqrt3)/4`

NOTE: Alternative solution to `A'(θ)=0` which is pretty straightforward and doesn't require further trigonometric examination would be the following
`cosθ+cos^2θ-sin^2θ=0` `<=>` `cosθ+cos^2θ-(1-cos^2θ)=0` `<=>` `(2cosθ-1)(cosθ+1)=0` and since `θ` `in` `(0,π)` we get
`cosθ=1/2` so `θ=π/3`