Plz plz solve this(maxima/minima)....

Prove that the least perimeter of an isosceles triangle in which circle of radius r can be inscribed is 6 root3 r.

Consider an isosceles triangle ABC, with AB = AC, angle BAC = t and inradius r. Here r is a constant (given) and t is variable. The question is what is the perimeter of ABC in terms of t and r. We can then minimize the perimeter by differentiating over t.

To calculate the perimeter, let the incenter be I and let D,E,F, be perpendiculars from I to BC, CA, AB, respectively. Since AI bisects angle BAC, r/AE = tan(t/2), so AE = r cot(t/2). Similarly, AF = r cot(t/2). Next, the angle ABC is (pi - t)/2, so r/BD = tan ((pi - t)/4). Hence, BD = r cot((pi - t)/4). Also, BD = FB = DC = CE, so all these four segments have length r cot((pi - t)/4).

Hence, the total perimeter of the triangle is:

f(t) = 2r cot(t/2) + 4r cot ((pi - t)/4)

We want to minimize this perimeter. First, we simplify f(t) slightly. Let t/4 = x. Note that 0 <= x <= pi/4. Given this, f(t) can be written as:

g(x) = 2r cot (2x) + 4r cot (pi/4 - x)

We have g'(x) = -4r csc^2 (2x) + 4r csc^2 (pi/4 - x) = 0
=> csc^2 (2x) = csc^2(pi/4 - x)
=> sin^2 (2x) = sin^2 (pi/4 - x)

Since, 0 <= x <= pi/4, sin(2x) and sin(pi/4 - x) are both non-negative.

So, we get sin(2x) = sin (pi/4 - x), and again because 0 <= x <= pi/4,
2x = pi/4 - x.

So x = pi/12 and t = 4x = pi/3. Hence, the triangle is equilateral.
The minimum perimeter is:
2r cot(t/2) + 4r cot ((pi - t)/4) = 2r cot(pi/6) + 4r cot (pi/6) = 6r * sqrt(3), as required.

If a circle is inscribed, the radii are perpendicular to the sides of the triangle.

So start there assuming an equilateral triangle. Then show any non-equilateral triangle will add to the perimeter.