# Maximum Radius of Circle Optimization Problem with Solution

Use the derivative to find the size of an angle of a right triangle so that the radius of the circle inscribed is maximum; for a constant hypotenuse.

## Problem

ABC is a right triangle and $$r$$ is the radius of the inscribed circle.
a) Express $$r$$ in terms of angle $$x$$ and the length of the hypotenuse $$h$$.
b) Assume that $$h$$ is constant and $$x$$ varies; find $$x$$ for which $$r$$ is maximum.

Solution to Problem:

a) Let M, N, and P be the points of tangency of the circle and the sides of the triangle. $$OM$$, $$ON$$, and $$OP$$ are perpendicular to $$CB$$, $$CA$$, and $$AB$$ respectively.

Triangles COM and CON are right triangles and have two congruent sides: $$CO$$ and $$OM$$ and $$ON$$; the two triangles are therefore congruent. We denote the size of angle $$MCN$$ by $$x$$ and write $\tan\left(\frac{x}{2}\right) = \frac{r}{CM}$ Triangles BOM and BOP are right triangles and have two congruent sides: $$BO$$ and $$OM$$ and $$OP$$; the two triangles are therefore congruent. We denote the size of angle $$MBP$$ by $$y$$ and write $\tan\left(\frac{y}{2}\right) = \frac{r}{BM}$ Note that
$$y + x = 90$$ which gives $$y / 2 = 45 - x / 2$$
Substitute $$y / 2$$ by $$45 - x / 2$$ in the equation $$\tan\left(\dfrac{y}{2}\right) = \dfrac{r}{BM}$$ to obtain $\tan\left(45 - \frac{x}{2}\right) = \frac{r}{BM}$ We now solve the equation $$\tan\left(\dfrac{x}{2}\right) = \dfrac{r}{CM}$$ for $$CM$$ and solve the equation $$\tan\left(45 - \dfrac{x}{2}\right) = \dfrac{r}{BM}$$ for $$BM$$ to obtain $$CM = \dfrac{r}{\tan\left(\dfrac{x}{2}\right)}$$ and $$BM = \dfrac{r}{\tan\left(45 - \dfrac{x}{2}\right)}$$
We now use the fact that $$h = CM + BM$$ to write the equation
$$h = \dfrac{r}{\tan\left(\dfrac{x}{2}\right)} + \dfrac{r}{\tan\left(45 - \dfrac{x}{2}\right)}$$
$= r \left[ \frac{1}{\tan\left(\frac{x}{2}\right)} + \frac{1}{\tan\left(45 - \frac{x}{2}\right)} \right]$
We now use trigonometric identities to simplify the above equation. The first identity we will use is $\tan\left(45 - \frac{x}{2}\right) = \frac{\tan(45) - \tan\left(\frac{x}{2}\right)}{1 + \tan(45)\tan\left(\frac{x}{2}\right)}$ $= \frac{1 - \tan\left(\frac{x}{2}\right)}{1 + \tan\left(\frac{x}{2}\right)}$ The formula for $$h$$ is now given by
$h = r \left[ \frac{1}{\tan\left(\frac{x}{2}\right)} + \frac{1 - \tan\left(\frac{x}{2}\right)}{1 + \tan\left(\frac{x}{2}\right)} \right]$ We now use the identity $$\tan\left(\dfrac{x}{2}\right) = \dfrac{\sin\left(\dfrac{x}{2}\right)}{\cos\left(\dfrac{x}{2}\right)}$$ and other identities to rewrite $$h$$ as follows $h = r \left[ \frac{\cos\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)} + \frac{\cos\left(\frac{x}{2}\right) + \sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right) - \sin\left(\frac{x}{2}\right)} \right]$ $= r \left[ \frac{1}{\sin\left(\frac{x}{2}\right) (\cos\left(\frac{x}{2}\right) - \sin\left(\frac{x}{2}\right))} \right]$ We now solve the above equation for $$r$$ to obtain $r = h \sin\left(\frac{x}{2}\right) (\cos\left(\frac{x}{2}\right) - \sin\left(\frac{x}{2}\right)) = h (\sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) - \sin^2\left(\frac{x}{2}\right))$ We now use the identities $$\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right) = \dfrac{1}{2} \sin (x)$$ and $$\sin^2\left(\dfrac{x}{2}\right) = \dfrac{1}{2} - \dfrac{1}{2}\cos(x)$$ to write $$r$$ as follows $r = \frac{h}{2} (\sin(x) + \cos(x) - 1)$ b) Now that we have calculated $$r$$ as a function of $$x$$, and $$h$$ assumed constant, let us find the first derivative of $$r$$ with respect to $$x$$ $\frac{dr}{dx} = \frac{h}{2} (\cos(x) - \sin(x))$ Let us find a critical point for $$r$$ by solving the equation $$\dfrac{dr}{dx} = 0$$. $\frac{h}{2} (\cos(x) - \sin(x)) = 0$ gives $\cos(x) - \sin(x) = 0$ since $$h$$ is constant and not equal to 0. $\cos(x) = \sin(x)$ Square both sides $\cos^2(x) = \sin^2(x)$ $\cos^2(x) = 1 - \cos^2(x)$ Solve for $$\cos(x)$$ to obtain $\cos(x) = \pm \frac{1}{\sqrt{2}} \text{ or } - \frac{1}{\sqrt{2}}$ $$x$$ is an acute angle and therefore the solution to our equation is given by $x = \frac{\pi}{4} = 45^\circ$ The graph of $$\dfrac{dr}{dx}$$ is shown below and $$\dfrac{dr}{dx}$$ is positive for $$x \lt \dfrac{\pi}{4}$$ and negative for $$x > \dfrac{\pi}{4}$$, therefore $$r$$ has a maximum at $$x = \dfrac{\pi}{4} = 45^\circ$$.

$$x = 45^\circ$$ is the value of angle $$ACB$$ for which the radius $$r$$ is maximum for a given value $$h$$ of the length of the hypotenuse. Note that in this case triangle ABC is isosceles.