1. tricky optimization question

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

2. Originally Posted by aaasssaaa
Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r.
First see the attached diagram.

1. Use the Intersecting Chord Theorem to get:

$
(b/2)^2=x(2r-x)
$

This is a quadratic in $x$, which we may solve using the quadratic formula:

$
x=r \pm \sqrt{4r^2-b^2)}
$

$
x=r(1 - \sqrt{1-(b/(2r)^2)})
$

2. The area of the triangle is:

$
A = \frac{b\ h}{2}=\frac{b\ (2r-x)}{2}=\frac{b(2r-r \left[ 1-\sqrt{1-(b/2r)^2}\right] )}{2}
$

3. Now to find the isosceles triangle which maximises this area you need for
solve:

$\frac{dA}{db}=0$

and if there are multiple solutions sort through them to identify which corresponds
to a maximum, and then find the other sides of the triangle.

RonL

3. 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 $AB$ and $AC.$

We will maximize the area of $\Delta ABC$
. . which has base $2x$ and height $y + r.$

The area of the triangle is: . $A \:=\:\frac{1}{2}(2x)(y + r) \:=\:x(y + r)$ [1]

Using Pythagorus, we have: . $x^2 + y^2\,=\,r^2\quad\Rightarrow\quad y = \sqrt{r^2 - x^2}$ [2]

Substitute [2] into [1]: . $A \;= \;x\left[(r^2 - x^2)^{\frac{1}{2}} + r\right]$

Differentiate: . $A' \;=\;x\left[\frac{1}{2}(r^2 - x^2)^{-\frac{1}{2}}(-2x)\right] + (r^2-x^2)^{\frac{1}{2}} + r$

Equate to zero: . $\frac{-x^2}{\sqrt{r^2-x^2}} + \sqrt{r^2-x^2} + r \;= \;0$

Multiply by $\sqrt{r^2-x^2}\!:\;\;-x^2 + r^2 - x^2 + r\sqrt{r^2-x^2} \:=\:0$

Then we have: . $r\sqrt{r^2-x^2} \:=\:2x^2 - r^2$

Square both sides: . $r^2(r^2-x^2) \:=\:(2x^2-r^2)^2\quad\Rightarrow\quad r^4 - r^2x^2\:=\:4x^4 - 4r^2x^2 + r^4$

. . $4x^4 - 3r^2x^2\:=\:0\quad\Rightarrow\quad x^2(4x^2 - 3)\:=\:0$

If $x^2=0$, we have $x = 0$ which produces minimum area . . . ha!

If $4x^2-3 = 0$, we have: . $\boxed{x = \frac{\sqrt{3}}{2}r}$

Substitute into [2] and we get: . $\boxed{y = \frac{r}{2}}$

Therefore, the triangle has: . $\begin{array}{cc}\text{base} \\ \text{height}\end{array}\begin{array}{cc}=\\=\end{ array} \begin{array}{cc}2x\\y+r\end{array} \begin{array}{cc}=\\=\end{array}\begin{array}{cc}\ sqrt{3}r \\ \frac{3}{2}r\end{array}$

Note: The triangle happens to be equilateral,
. . . . .which you may have suspected all along.

4. Why did you need to find the negative sign of the quadratic?
Abd how did it turn out to be

x=r(1 - \sqrt{1-(b/(2r)^2)})

5. Originally Posted by liquidanubis3
Why did you need to find the negative sign of the quadratic?
Abd how did it turn out to be