Optimization!!!!?

Question

Find the angle theta that maximizes the area of isosceles triangle with legs l.

It shows triangle with two legs l and the theta between those two legs.

I know I have Area=1/2B*h
And I know I can split the triangle to make a statement about theta/2 but I'm stuck on how to do all this.....please help!

Answer 1

Area = (1/2)(sin alpha)(b)(c)

Let's say that alpha is the angle that is unique.

To maximimize the area, we must maximize sin alpha (because that's the only thing that changes).

d/d(alpha) sin (alpha) = cos alpha

cos alpha = 0

alpha = 90

Since alpha is 90, this is the special isosceles right traignle.

The isosceles triangle with maximum area is the 45-45-90 right triangle.

Answer 2

the area A= 1/2 B*h
........ Base=B
heigth=h
after trigonometric relations,
and derive ,
sin(theta) =1/2
theta= 30^ degrees.
.

Answer 3

OKay, with t being theta, then the area is l^2*(sin(t/2))*(cos(t/2)).

SO, differinating, and set to 0,
i^2*(.5(cos^2(t/2))- .(5(sin^2(t/2)))=0.
SO, cos(t/2)=sin(t/2).
SO, solving, t= 90 degrees.

Answer 4

b = 2htanΘ/2

A = 0.5bh = h²tanΘ/2

dA/dΘ = 0.5h²sec²(Θ/2)...I think

Therefore, the maximum is where the above = 0