Hello, aaasssaaa!
Here's my approach to this problem . . .
Find the dimensions of the isosceles triangle of largest area
that can be inscribed in a circle of radius r. Code:
A
* * *
* : *
* : *
* r: *
:
* : *
* * *
* : \ *
y: \r
* : \ *
B* - - - + - - - *C
* x *
* * *
Draw line segments $\displaystyle AB$ and $\displaystyle AC.$
We will maximize the area of $\displaystyle \Delta ABC$
. . which has base $\displaystyle 2x$ and height $\displaystyle y + r.$
The area of the triangle is: .$\displaystyle A \:=\:\frac{1}{2}(2x)(y + r) \:=\:x(y + r)$ [1]
Using Pythagorus, we have: .$\displaystyle x^2 + y^2\,=\,r^2\quad\Rightarrow\quad y = \sqrt{r^2 - x^2}$ [2]
Substitute [2] into [1]: .$\displaystyle A \;= \;x\left[(r^2 - x^2)^{\frac{1}{2}} + r\right] $
Differentiate: .$\displaystyle A' \;=\;x\left[\frac{1}{2}(r^2 - x^2)^{-\frac{1}{2}}(-2x)\right] + (r^2-x^2)^{\frac{1}{2}} + r$
Equate to zero: .$\displaystyle \frac{-x^2}{\sqrt{r^2-x^2}} + \sqrt{r^2-x^2} + r \;= \;0$
Multiply by $\displaystyle \sqrt{r^2-x^2}\!:\;\;-x^2 + r^2 - x^2 + r\sqrt{r^2-x^2} \:=\:0$
Then we have: .$\displaystyle r\sqrt{r^2-x^2} \:=\:2x^2 - r^2$
Square both sides: .$\displaystyle r^2(r^2-x^2) \:=\:(2x^2-r^2)^2\quad\Rightarrow\quad r^4 - r^2x^2\:=\:4x^4 - 4r^2x^2 + r^4$
. . $\displaystyle 4x^4 - 3r^2x^2\:=\:0\quad\Rightarrow\quad x^2(4x^2 - 3)\:=\:0$
If $\displaystyle x^2=0$, we have $\displaystyle x = 0$ which produces minimum area . . . ha!
If $\displaystyle 4x^2-3 = 0$, we have: .$\displaystyle \boxed{x = \frac{\sqrt{3}}{2}r}$
Substitute into [2] and we get: .$\displaystyle \boxed{y = \frac{r}{2}}$
Therefore, the triangle has: .$\displaystyle \begin{array}{cc}\text{base} \\ \text{height}\end{array}\begin{array}{cc}=\\=\end{ array} \begin{array}{cc}2x\\y+r\end{array} \begin{array}{cc}=\\=\end{array}\begin{array}{cc}\ sqrt{3}r \\ \frac{3}{2}r\end{array}$
Note: The triangle happens to be equilateral,
. . . . .which you may have suspected all along.