# Thread: Help with a constrained extrema problem.

1. ## Help with a constrained extrema problem.

Hi there. (This is going to be a long post)

I want to show that for any triangle that is inscribed in a circle with a radius of R, the largest perimeter of the triangle comes from an equilateral triangle in the circle.

If i drew a triangle in the circle i noticed that the length of the origin to any vertex of the triangle is R.

So applying that this is what I did, I can basically split the triangle into 3 parts to find each side of the triangle from the law of cosines:

c^2 = a^2 + b^2 - 2ab cos C

For example is c was a side of the triangle I can get its length through R. Here is my bad mspaint interpretation.

c^2 = R^2 + R^2 - 2R^2 cos C

and R = x^2+y^2

so i can do that for the other sides as well, and form the perimeter equation

P(a,b,c) = a + b + c

I can now sub in R as x^2 + y^2 in the perimeter equation to get it in terms of x and y, but after doing that and simplifying, then solving it with the larangian with the constraint as x^y + y^2 = R, I dont get an expression for lamda in terms of x and y. All i get is a constant in terms of the angles cos a, cos b, and cos c.

I guess it would make sense that the critical point would be in terms of the angles rather than x and y, because if cos a = cos b = cos c then it is an equilateral triangle. But what do i do here?

2. ## Re: Help with a constrained extrema problem.

Originally Posted by Kuma
Hi there. (This is going to be a long post)

I want to show that for any triangle that is inscribed in a circle with a radius of R, the largest perimeter of the triangle comes from an equilateral triangle in the circle.

If i drew a triangle in the circle i noticed that the length of the origin to any vertex of the triangle is R.

So applying that this is what I did, I can basically split the triangle into 3 parts to find each side of the triangle from the law of cosines:

c^2 = a^2 + b^2 - 2ab cos C

For example is c was a side of the triangle I can get its length through R. Here is my bad mspaint interpretation.

c^2 = R^2 + R^2 - 2R^2 cos C

and R = x^2+y^2

so i can do that for the other sides as well, and form the perimeter equation

P(a,b,c) = a + b + c

I can now sub in R as x^2 + y^2 in the perimeter equation to get it in terms of x and y, but after doing that and simplifying, then solving it with the larangian with the constraint as x^y + y^2 = R, I dont get an expression for lamda in terms of x and y. All i get is a constant in terms of the angles cos a, cos b, and cos c.

I guess it would make sense that the critical point would be in terms of the angles rather than x and y, because if cos a = cos b = cos c then it is an equilateral triangle. But what do i do here?
First: You need only consider the unit circle with $R=1$.

Second: You can consider one vertex fixed leaving only two free vertices to worry about.

Third: Each of the two free vertices is characterised by the angle (measured anti-clockwise) between the fixed vertex and the free vertex.

CB