Question

Find the area of largest isosceles triangle that can be inscribed in a circle of radius 4.

Answer 1

I have to use calculus to solve this problem. Where should I start?

Answer 2

I had a problem similar 2 this on my old exam! ahh! Give me a minute haha. I know this!! Lol

Answer 3

Let Q be the apex angle.
Draw the altitude of the triangle (bisecting the apex angle) through the center of the circle.
This altitude has length 4 + h, where h = 4 cos(Q) and r = 4
Base of triangle has length: 8 sin(Q)
So the area is:
\[A = \frac{ 1 }{ 2 }(4+4\cos(Q)(8\sin(Q) = 24\sin(Q) + 12\sin(2Q)\]
\[\frac{ dA }{ dQ } = 24cos(Q) + 24 cos(2Q)\]
Set this expression to zero to find the optimal triangle.
\[\cos(Q)+2\cos^2(Q)-1=0\]
(2 cos(Q) - 1)(cos (Q) + 1 ) = 0

Answer 4

I don't get it, why is altitude 4 + h?

Answer 5

the height=r+h

Answer 6

I'm sorry, but can you sketch it for me?

Answer 7

"height=r+h"

What does h represent?

Answer 8

I used So, b=cos(theta) and h=rsin(theta)

Answer 9

Ah, I see. Thanks.

Answer 10

Sorry, this type of problem is hard to explain. I am using an my old exam to figure out how to solve this. I will definitely post more if I can find a better way to explain.

Answer 11

But I used area as a function of half the angle

Answer 12

Ok, thank you, I appreciate it.

Answer 13

so the area is
\[A = \frac{ 1 }{ 2 }b(r+h)\]
Substituted for b and h (r=4) and get A as a function of theta, so then I could solve the problem by differentiating. Also, I used b=cos(theta) and h=rsin(theta)

Answer 14

@geerky42 My answer is A = 12√3 unit²
My way is to take advantage of the symmetry characteristic between circle and isosceles triangle.
Let me know if you want further details :)