Question

Find the area of the largest isosceles triangle that can be inscribed in a circle of radius 6 inches.

Answer 1

From the diagram, the height of the triangle is $6+h$

We can as well gfrac{det the base using Pythagoras theorem.

$b=2\sqrt{36-h^2}$

Thus, the area of the triangle is

$A=\frac{1}{2}(2\sqrt{36-h^2})(6+h)=(\sqrt{36-h^2}(6+h)$

Find the derivative of A and set it to 0.

$\frac{dA}{dh}=\sqrt{36-h^2}-\frac{h(6+h)}{\sqrt{36-h^2}}=0\\ 36-h^2-h(6+h)=0\\ 36-2h^2-6h=0\\ h^2+3h-18=0\\ (h-3)(h+6)=0\\ h=3,h=-6$

Thus h=-6,3 are the critical points.

$\frac{d^2A}{dh^2}$ at h=-6 >0, $\frac{d^2A}{dh^2}$ at h=3<0.

Thus, $h=3$ is the maximum point.

The largest area of the triangle is

$A=(\sqrt{36-3^2})(6+3)=27\sqrt{3} in^2$