# Help with this extrema problem?

## Jan 9, 2018

A(θ)=sinθ(1+cosθ) , θ=π/3

#### Explanation:

Solution supposed the problem is about isosceles triangle. $r = 1$ so $\left(O B\right) = \left(O C\right) = 1$

From the right triangle $O \hat{B} M$:

• sinθ=(BM)/(OB) $\iff$ (BM)=(OB)*sinθ=1*sinθ

and (BC)=2(BM)=2sinθ ,

• cosθ=(OM)/(OB) $\iff$ (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 ${\cos}^{2} x - {\sin}^{2} x = \cos 2 x$
/ 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 $\iff$

cos2θ+cosθ=0 $\iff$

cos2θ=-cosθ $\iff$

cos2θ=cos(π-θ) $\iff$
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)
$= \frac{1}{2} + \frac{\sqrt{3}}{2} > 0$

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 ${A}_{r e a}$ maximum value when θ=π/3 and the value is $\frac{3 \sqrt{3}}{4}$

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 $\iff$ cosθ+cos^2θ-(1-cos^2θ)=0 $\iff$ (2cosθ-1)(cosθ+1)=0 and since θ$\in$(0,π) we get
cosθ=1/2 so θ=π/3